Practice · Chain the given inequality twice · Sensim Math

Variant 1 easy

For nonnegative numbers $m$ and $n$ , it is known that
$2^m+2^n\le1+2^{m+n}$ , and equality holds when $m=0$ or $n=0$ .
If $a$ , $b$ , and $c$ are nonnegative numbers with $a+b+c=3$ , what is the largest possible value of $2^a+2^b+2^c$ ?

Your answer

Show solution

1 · Understandwhat's really being asked

We are given a fact: for nonnegative $m$ , $n$ , $2^m+2^n\le1+2^{m+n}$ , with equality when $m=0$ or $n=0$ . For nonnegative $a$ , $b$ , $c$ with $a+b+c=3$ , find the largest value of $2^a+2^b+2^c$ .

Givens

$2^m+2^n\le1+2^{m+n}$ for nonnegative $m$ , $n$ ; equality when $m=0$ or $n=0$ .
$a$ , $b$ , $c$ are nonnegative and $a+b+c=3$ .

Unknowns

The largest possible value of $2^a+2^b+2^c$ .

Constraints

An upper bound is the answer only if some allowed $a$ , $b$ , $c$ actually reach it.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #13 Convert to Algebra#6 Guess and Check

The given fact handles two powers at a time. Use it on $a$ and $b$ , then treat $a+b$ as a single exponent and use it again with $c$ . That leaves only $2^{a+b+c}=2^3$ .

3 · Execute5 carry out the plan

1Use the fact on $a$ and $b$

#7 Identify Subproblems 7.EE.B.4

a

and

b

are nonnegative, so the fact applies with

m=a

,

n=b

.

2^a+2^b\le1+2^{a+b}

This merges two powers into one, at the cost of a

1

.

2Add $2^c$ to both sides

#13 Convert to Algebra 7.EE.B.4

Adding the same number to both sides keeps the inequality.

2^a+2^b+2^c\le1+2^{a+b}+2^c

Now the right side has only two powers of

2

.

3Use the fact on $a+b$ and $c$

#7 Identify Subproblems 7.EE.B.4

a+b

is also nonnegative, so apply the fact with

m=a+b

,

n=c

.

2^{a+b}+2^c\le1+2^{a+b+c}

Treat

a+b

as one block.

4Chain the inequalities

#13 Convert to Algebra 8.EE.A.1

Put the two steps together and use

a+b+c=3

.

2^a+2^b+2^c\le1+1+2^{a+b+c}=2+2^3=10

2^{a+b+c}=2^3=8

.

5Check the bound is reached

#6 Guess and Check 8.EE.A.1

Equality needs

a=0

or

b=0

, and

a+b=0

or

c=0

. Try

a=0

,

b=0

,

c=3

.

2^0+2^0+2^3=1+1+8=10

2^0=1

, so the bound

10

is actually attained.

Answer:

10

4 · Reviewdoes it hold up?

Other splits give less: $(2,1,0)$ gives $4+2+1=7$ , $(1,1,1)$ gives $2+2+2=6$ . Putting all of the $3$ into one exponent makes one term large, which makes sense for repeated multiplying.

Another way: The fact itself says $(2^m-1)(2^n-1)\ge0$ . Thinking of it this way, the sum is largest when the exponents are as unequal as possible: try whole-number splits $(3,0,0)$ , $(2,1,0)$ , $(1,1,1)$ to get $10$ , $7$ , $6$ ; the chained inequality shows no split, whole or not, beats $10$ .

Standardsmin grade 8

7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Adding to both sides of an inequality and chaining two inequalities.
8.EE.A.1 Know and apply the properties of integer exponents — Combining $2^{a+b}$ and $2^c$ and using $2^0=1$.

💡Takeaway. A fact about two numbers can handle three -- glue two of them into one block and use it twice.

Variant 2 easy

For nonnegative numbers $m$ and $n$ , it is known that
$3^m+3^n\le1+3^{m+n}$ , and equality holds when $m=0$ or $n=0$ .
If $a$ , $b$ , and $c$ are nonnegative numbers with $a+b+c=3$ , what is the largest possible value of $3^a+3^b+3^c$ ?

Your answer

Show solution

1 · Understandwhat's really being asked

We are given a fact: for nonnegative $m$ , $n$ , $3^m+3^n\le1+3^{m+n}$ , with equality when $m=0$ or $n=0$ . For nonnegative $a$ , $b$ , $c$ with $a+b+c=3$ , find the largest value of $3^a+3^b+3^c$ .

Givens

$3^m+3^n\le1+3^{m+n}$ for nonnegative $m$ , $n$ ; equality when $m=0$ or $n=0$ .
$a$ , $b$ , $c$ are nonnegative and $a+b+c=3$ .

Unknowns

The largest possible value of $3^a+3^b+3^c$ .

Constraints

An upper bound is the answer only if some allowed $a$ , $b$ , $c$ actually reach it.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #13 Convert to Algebra#6 Guess and Check

The given fact handles two powers at a time. Use it on $a$ and $b$ , then treat $a+b$ as a single exponent and use it again with $c$ . That leaves only $3^{a+b+c}=3^3$ .

3 · Execute5 carry out the plan

1Use the fact on $a$ and $b$

#7 Identify Subproblems 7.EE.B.4

a

and

b

are nonnegative, so the fact applies with

m=a

,

n=b

.

3^a+3^b\le1+3^{a+b}

This merges two powers into one, at the cost of a

1

.

2Add $3^c$ to both sides

#13 Convert to Algebra 7.EE.B.4

Adding the same number to both sides keeps the inequality.

3^a+3^b+3^c\le1+3^{a+b}+3^c

Now the right side has only two powers of

3

.

3Use the fact on $a+b$ and $c$

#7 Identify Subproblems 7.EE.B.4

a+b

is also nonnegative, so apply the fact with

m=a+b

,

n=c

.

3^{a+b}+3^c\le1+3^{a+b+c}

Treat

a+b

as one block.

4Chain the inequalities

#13 Convert to Algebra 8.EE.A.1

Put the two steps together and use

a+b+c=3

.

3^a+3^b+3^c\le1+1+3^{a+b+c}=2+3^3=29

3^{a+b+c}=3^3=27

.

5Check the bound is reached

#6 Guess and Check 8.EE.A.1

Equality needs

a=0

or

b=0

, and

a+b=0

or

c=0

. Try

a=0

,

b=0

,

c=3

.

3^0+3^0+3^3=1+1+27=29

3^0=1

, so the bound

29

is actually attained.

Answer:

29

4 · Reviewdoes it hold up?

Other splits give less: $(2,1,0)$ gives $9+3+1=13$ , $(1,1,1)$ gives $3+3+3=9$ . Putting all of the $3$ into one exponent makes one term large, which makes sense for repeated multiplying.

Another way: The fact itself says $(3^m-1)(3^n-1)\ge0$ . Thinking of it this way, the sum is largest when the exponents are as unequal as possible: try whole-number splits $(3,0,0)$ , $(2,1,0)$ , $(1,1,1)$ to get $29$ , $13$ , $9$ ; the chained inequality shows no split, whole or not, beats $29$ .

Standardsmin grade 8

7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Adding to both sides of an inequality and chaining two inequalities.
8.EE.A.1 Know and apply the properties of integer exponents — Combining $3^{a+b}$ and $3^c$ and using $3^0=1$.

💡Takeaway. A fact about two numbers can handle three -- glue two of them into one block and use it twice.

Variant 3 easy

For nonnegative numbers $m$ and $n$ , it is known that
$3^m+3^n\le1+3^{m+n}$ , and equality holds when $m=0$ or $n=0$ .
If $a$ , $b$ , and $c$ are nonnegative numbers with $a+b+c=4$ , what is the largest possible value of $3^a+3^b+3^c$ ?

Your answer

Show solution

1 · Understandwhat's really being asked

We are given a fact: for nonnegative $m$ , $n$ , $3^m+3^n\le1+3^{m+n}$ , with equality when $m=0$ or $n=0$ . For nonnegative $a$ , $b$ , $c$ with $a+b+c=4$ , find the largest value of $3^a+3^b+3^c$ .

Givens

$3^m+3^n\le1+3^{m+n}$ for nonnegative $m$ , $n$ ; equality when $m=0$ or $n=0$ .
$a$ , $b$ , $c$ are nonnegative and $a+b+c=4$ .

Unknowns

The largest possible value of $3^a+3^b+3^c$ .

Constraints

An upper bound is the answer only if some allowed $a$ , $b$ , $c$ actually reach it.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #13 Convert to Algebra#6 Guess and Check

The given fact handles two powers at a time. Use it on $a$ and $b$ , then treat $a+b$ as a single exponent and use it again with $c$ . That leaves only $3^{a+b+c}=3^4$ .

3 · Execute5 carry out the plan

1Use the fact on $a$ and $b$

#7 Identify Subproblems 7.EE.B.4

a

and

b

are nonnegative, so the fact applies with

m=a

,

n=b

.

3^a+3^b\le1+3^{a+b}

This merges two powers into one, at the cost of a

1

.

2Add $3^c$ to both sides

#13 Convert to Algebra 7.EE.B.4

Adding the same number to both sides keeps the inequality.

3^a+3^b+3^c\le1+3^{a+b}+3^c

Now the right side has only two powers of

3

.

3Use the fact on $a+b$ and $c$

#7 Identify Subproblems 7.EE.B.4

a+b

is also nonnegative, so apply the fact with

m=a+b

,

n=c

.

3^{a+b}+3^c\le1+3^{a+b+c}

Treat

a+b

as one block.

4Chain the inequalities

#13 Convert to Algebra 8.EE.A.1

Put the two steps together and use

a+b+c=4

.

3^a+3^b+3^c\le1+1+3^{a+b+c}=2+3^4=83

3^{a+b+c}=3^4=81

.

5Check the bound is reached

#6 Guess and Check 8.EE.A.1

Equality needs

a=0

or

b=0

, and

a+b=0

or

c=0

. Try

a=0

,

b=0

,

c=4

.

3^0+3^0+3^4=1+1+81=83

3^0=1

, so the bound

83

is actually attained.

Answer:

83

4 · Reviewdoes it hold up?

Other splits give less: $(3,1,0)$ gives $27+3+1=31$ , $(2,2,0)$ gives $9+9+1=19$ , $(2,1,1)$ gives $9+3+3=15$ , and $\left(\frac{4}{3},\frac{4}{3},\frac{4}{3}\right)$ gives about $13.0$ . Putting all of the $4$ into one exponent makes one term large, which makes sense for repeated multiplying.

Another way: The fact itself says $(3^m-1)(3^n-1)\ge0$ . Thinking of it this way, the sum is largest when the exponents are as unequal as possible: try whole-number splits $(4,0,0)$ , $(3,1,0)$ , $(2,2,0)$ , $(2,1,1)$ to get $83$ , $31$ , $19$ , $15$ ; the chained inequality shows no split, whole or not, beats $83$ .

Standardsmin grade 8

7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Adding to both sides of an inequality and chaining two inequalities.
8.EE.A.1 Know and apply the properties of integer exponents — Combining $3^{a+b}$ and $3^c$ and using $3^0=1$.

💡Takeaway. A fact about two numbers can handle three -- glue two of them into one block and use it twice.

Variant 4 easy

For nonnegative numbers $m$ and $n$ , it is known that
$2^m+2^n\le1+2^{m+n}$ , and equality holds when $m=0$ or $n=0$ .
If $a$ , $b$ , and $c$ are nonnegative numbers with $a+b+c=4$ , what is the largest possible value of $2^a+2^b+2^c$ ?

Your answer

Show solution

1 · Understandwhat's really being asked

We are given a fact: for nonnegative $m$ , $n$ , $2^m+2^n\le1+2^{m+n}$ , with equality when $m=0$ or $n=0$ . For nonnegative $a$ , $b$ , $c$ with $a+b+c=4$ , find the largest value of $2^a+2^b+2^c$ .

Givens

$2^m+2^n\le1+2^{m+n}$ for nonnegative $m$ , $n$ ; equality when $m=0$ or $n=0$ .
$a$ , $b$ , $c$ are nonnegative and $a+b+c=4$ .

Unknowns

The largest possible value of $2^a+2^b+2^c$ .

Constraints

An upper bound is the answer only if some allowed $a$ , $b$ , $c$ actually reach it.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #13 Convert to Algebra#6 Guess and Check

The given fact handles two powers at a time. Use it on $a$ and $b$ , then treat $a+b$ as a single exponent and use it again with $c$ . That leaves only $2^{a+b+c}=2^4$ .

3 · Execute5 carry out the plan

1Use the fact on $a$ and $b$

#7 Identify Subproblems 7.EE.B.4

a

and

b

are nonnegative, so the fact applies with

m=a

,

n=b

.

2^a+2^b\le1+2^{a+b}

This merges two powers into one, at the cost of a

1

.

2Add $2^c$ to both sides

#13 Convert to Algebra 7.EE.B.4

Adding the same number to both sides keeps the inequality.

2^a+2^b+2^c\le1+2^{a+b}+2^c

Now the right side has only two powers of

2

.

3Use the fact on $a+b$ and $c$

#7 Identify Subproblems 7.EE.B.4

a+b

is also nonnegative, so apply the fact with

m=a+b

,

n=c

.

2^{a+b}+2^c\le1+2^{a+b+c}

Treat

a+b

as one block.

4Chain the inequalities

#13 Convert to Algebra 8.EE.A.1

Put the two steps together and use

a+b+c=4

.

2^a+2^b+2^c\le1+1+2^{a+b+c}=2+2^4=18

2^{a+b+c}=2^4=16

.

5Check the bound is reached

#6 Guess and Check 8.EE.A.1

Equality needs

a=0

or

b=0

, and

a+b=0

or

c=0

. Try

a=0

,

b=0

,

c=4

.

2^0+2^0+2^4=1+1+16=18

2^0=1

, so the bound

18

is actually attained.

Answer:

18

4 · Reviewdoes it hold up?

Other splits give less: $(3,1,0)$ gives $8+2+1=11$ , $(2,2,0)$ gives $4+4+1=9$ , $(2,1,1)$ gives $4+2+2=8$ , and $\left(\frac{4}{3},\frac{4}{3},\frac{4}{3}\right)$ gives about $7.6$ . Putting all of the $4$ into one exponent makes one term large, which makes sense for repeated multiplying.

Another way: The fact itself says $(2^m-1)(2^n-1)\ge0$ . Thinking of it this way, the sum is largest when the exponents are as unequal as possible: try whole-number splits $(4,0,0)$ , $(3,1,0)$ , $(2,2,0)$ , $(2,1,1)$ to get $18$ , $11$ , $9$ , $8$ ; the chained inequality shows no split, whole or not, beats $18$ .

Standardsmin grade 8

7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Adding to both sides of an inequality and chaining two inequalities.
8.EE.A.1 Know and apply the properties of integer exponents — Combining $2^{a+b}$ and $2^c$ and using $2^0=1$.

💡Takeaway. A fact about two numbers can handle three -- glue two of them into one block and use it twice.

Variant 5 medium

For nonnegative numbers $m$ and $n$ , it is known that
$5^m+5^n\le1+5^{m+n}$ , and equality holds when $m=0$ or $n=0$ .
If $a$ , $b$ , and $c$ are nonnegative numbers with $a+b+c=3$ , what is the largest possible value of $5^a+5^b+5^c$ ?

Your answer

Show solution

1 · Understandwhat's really being asked

We are given a fact: for nonnegative $m$ , $n$ , $5^m+5^n\le1+5^{m+n}$ , with equality when $m=0$ or $n=0$ . For nonnegative $a$ , $b$ , $c$ with $a+b+c=3$ , find the largest value of $5^a+5^b+5^c$ .

Givens

$5^m+5^n\le1+5^{m+n}$ for nonnegative $m$ , $n$ ; equality when $m=0$ or $n=0$ .
$a$ , $b$ , $c$ are nonnegative and $a+b+c=3$ .

Unknowns

The largest possible value of $5^a+5^b+5^c$ .

Constraints

An upper bound is the answer only if some allowed $a$ , $b$ , $c$ actually reach it.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #13 Convert to Algebra#6 Guess and Check

The given fact handles two powers at a time. Use it on $a$ and $b$ , then treat $a+b$ as a single exponent and use it again with $c$ . That leaves only $5^{a+b+c}=5^3$ .

3 · Execute5 carry out the plan

1Use the fact on $a$ and $b$

#7 Identify Subproblems 7.EE.B.4

a

and

b

are nonnegative, so the fact applies with

m=a

,

n=b

.

5^a+5^b\le1+5^{a+b}

This merges two powers into one, at the cost of a

1

.

2Add $5^c$ to both sides

#13 Convert to Algebra 7.EE.B.4

Adding the same number to both sides keeps the inequality.

5^a+5^b+5^c\le1+5^{a+b}+5^c

Now the right side has only two powers of

5

.

3Use the fact on $a+b$ and $c$

#7 Identify Subproblems 7.EE.B.4

a+b

is also nonnegative, so apply the fact with

m=a+b

,

n=c

.

5^{a+b}+5^c\le1+5^{a+b+c}

Treat

a+b

as one block.

4Chain the inequalities

#13 Convert to Algebra 8.EE.A.1

Put the two steps together and use

a+b+c=3

.

5^a+5^b+5^c\le1+1+5^{a+b+c}=2+5^3=127

5^{a+b+c}=5^3=125

.

5Check the bound is reached

#6 Guess and Check 8.EE.A.1

Equality needs

a=0

or

b=0

, and

a+b=0

or

c=0

. Try

a=0

,

b=0

,

c=3

.

5^0+5^0+5^3=1+1+125=127

5^0=1

, so the bound

127

is actually attained.

Answer:

127

4 · Reviewdoes it hold up?

Other splits give less: $(2,1,0)$ gives $25+5+1=31$ , $(1,1,1)$ gives $5+5+5=15$ . Putting all of the $3$ into one exponent makes one term large, which makes sense for repeated multiplying.

Another way: The fact itself says $(5^m-1)(5^n-1)\ge0$ . Thinking of it this way, the sum is largest when the exponents are as unequal as possible: try whole-number splits $(3,0,0)$ , $(2,1,0)$ , $(1,1,1)$ to get $127$ , $31$ , $15$ ; the chained inequality shows no split, whole or not, beats $127$ .

Standardsmin grade 8

7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Adding to both sides of an inequality and chaining two inequalities.
8.EE.A.1 Know and apply the properties of integer exponents — Combining $5^{a+b}$ and $5^c$ and using $5^0=1$.

💡Takeaway. A fact about two numbers can handle three -- glue two of them into one block and use it twice.

Variant 6 medium

For nonnegative numbers $m$ and $n$ , it is known that
$3^m+3^n\le1+3^{m+n}$ , and equality holds when $m=0$ or $n=0$ .
If $a$ , $b$ , and $c$ are nonnegative numbers with $a+b+c=5$ , what is the largest possible value of $3^a+3^b+3^c$ ?

Your answer

Show solution

1 · Understandwhat's really being asked

We are given a fact: for nonnegative $m$ , $n$ , $3^m+3^n\le1+3^{m+n}$ , with equality when $m=0$ or $n=0$ . For nonnegative $a$ , $b$ , $c$ with $a+b+c=5$ , find the largest value of $3^a+3^b+3^c$ .

Givens

$3^m+3^n\le1+3^{m+n}$ for nonnegative $m$ , $n$ ; equality when $m=0$ or $n=0$ .
$a$ , $b$ , $c$ are nonnegative and $a+b+c=5$ .

Unknowns

The largest possible value of $3^a+3^b+3^c$ .

Constraints

An upper bound is the answer only if some allowed $a$ , $b$ , $c$ actually reach it.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #13 Convert to Algebra#6 Guess and Check

The given fact handles two powers at a time. Use it on $a$ and $b$ , then treat $a+b$ as a single exponent and use it again with $c$ . That leaves only $3^{a+b+c}=3^5$ .

3 · Execute5 carry out the plan

1Use the fact on $a$ and $b$

#7 Identify Subproblems 7.EE.B.4

a

and

b

are nonnegative, so the fact applies with

m=a

,

n=b

.

3^a+3^b\le1+3^{a+b}

This merges two powers into one, at the cost of a

1

.

2Add $3^c$ to both sides

#13 Convert to Algebra 7.EE.B.4

Adding the same number to both sides keeps the inequality.

3^a+3^b+3^c\le1+3^{a+b}+3^c

Now the right side has only two powers of

3

.

3Use the fact on $a+b$ and $c$

#7 Identify Subproblems 7.EE.B.4

a+b

is also nonnegative, so apply the fact with

m=a+b

,

n=c

.

3^{a+b}+3^c\le1+3^{a+b+c}

Treat

a+b

as one block.

4Chain the inequalities

#13 Convert to Algebra 8.EE.A.1

Put the two steps together and use

a+b+c=5

.

3^a+3^b+3^c\le1+1+3^{a+b+c}=2+3^5=245

3^{a+b+c}=3^5=243

.

5Check the bound is reached

#6 Guess and Check 8.EE.A.1

Equality needs

a=0

or

b=0

, and

a+b=0

or

c=0

. Try

a=0

,

b=0

,

c=5

.

3^0+3^0+3^5=1+1+243=245

3^0=1

, so the bound

245

is actually attained.

Answer:

245

4 · Reviewdoes it hold up?

Other splits give less: $(4,1,0)$ gives $81+3+1=85$ , $(3,2,0)$ gives $27+9+1=37$ , $(3,1,1)$ gives $27+3+3=33$ , and $\left(\frac{5}{3},\frac{5}{3},\frac{5}{3}\right)$ gives about $18.7$ . Putting all of the $5$ into one exponent makes one term large, which makes sense for repeated multiplying.

Another way: The fact itself says $(3^m-1)(3^n-1)\ge0$ . Thinking of it this way, the sum is largest when the exponents are as unequal as possible: try whole-number splits $(5,0,0)$ , $(4,1,0)$ , $(3,2,0)$ , $(3,1,1)$ to get $245$ , $85$ , $37$ , $33$ ; the chained inequality shows no split, whole or not, beats $245$ .

Standardsmin grade 8

7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Adding to both sides of an inequality and chaining two inequalities.
8.EE.A.1 Know and apply the properties of integer exponents — Combining $3^{a+b}$ and $3^c$ and using $3^0=1$.

💡Takeaway. A fact about two numbers can handle three -- glue two of them into one block and use it twice.

Variant 7 medium

For nonnegative numbers $m$ and $n$ , it is known that
$2^m+2^n\le1+2^{m+n}$ , and equality holds when $m=0$ or $n=0$ .
If $a$ , $b$ , and $c$ are nonnegative numbers with $a+b+c=5$ , what is the largest possible value of $2^a+2^b+2^c$ ?

Your answer

Show solution

1 · Understandwhat's really being asked

We are given a fact: for nonnegative $m$ , $n$ , $2^m+2^n\le1+2^{m+n}$ , with equality when $m=0$ or $n=0$ . For nonnegative $a$ , $b$ , $c$ with $a+b+c=5$ , find the largest value of $2^a+2^b+2^c$ .

Givens

$2^m+2^n\le1+2^{m+n}$ for nonnegative $m$ , $n$ ; equality when $m=0$ or $n=0$ .
$a$ , $b$ , $c$ are nonnegative and $a+b+c=5$ .

Unknowns

The largest possible value of $2^a+2^b+2^c$ .

Constraints

An upper bound is the answer only if some allowed $a$ , $b$ , $c$ actually reach it.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #13 Convert to Algebra#6 Guess and Check

The given fact handles two powers at a time. Use it on $a$ and $b$ , then treat $a+b$ as a single exponent and use it again with $c$ . That leaves only $2^{a+b+c}=2^5$ .

3 · Execute5 carry out the plan

1Use the fact on $a$ and $b$

#7 Identify Subproblems 7.EE.B.4

a

and

b

are nonnegative, so the fact applies with

m=a

,

n=b

.

2^a+2^b\le1+2^{a+b}

This merges two powers into one, at the cost of a

1

.

2Add $2^c$ to both sides

#13 Convert to Algebra 7.EE.B.4

Adding the same number to both sides keeps the inequality.

2^a+2^b+2^c\le1+2^{a+b}+2^c

Now the right side has only two powers of

2

.

3Use the fact on $a+b$ and $c$

#7 Identify Subproblems 7.EE.B.4

a+b

is also nonnegative, so apply the fact with

m=a+b

,

n=c

.

2^{a+b}+2^c\le1+2^{a+b+c}

Treat

a+b

as one block.

4Chain the inequalities

#13 Convert to Algebra 8.EE.A.1

Put the two steps together and use

a+b+c=5

.

2^a+2^b+2^c\le1+1+2^{a+b+c}=2+2^5=34

2^{a+b+c}=2^5=32

.

5Check the bound is reached

#6 Guess and Check 8.EE.A.1

Equality needs

a=0

or

b=0

, and

a+b=0

or

c=0

. Try

a=0

,

b=0

,

c=5

.

2^0+2^0+2^5=1+1+32=34

2^0=1

, so the bound

34

is actually attained.

Answer:

34

4 · Reviewdoes it hold up?

Other splits give less: $(4,1,0)$ gives $16+2+1=19$ , $(3,2,0)$ gives $8+4+1=13$ , $(3,1,1)$ gives $8+2+2=12$ , and $\left(\frac{5}{3},\frac{5}{3},\frac{5}{3}\right)$ gives about $9.5$ . Putting all of the $5$ into one exponent makes one term large, which makes sense for repeated multiplying.

Another way: The fact itself says $(2^m-1)(2^n-1)\ge0$ . Thinking of it this way, the sum is largest when the exponents are as unequal as possible: try whole-number splits $(5,0,0)$ , $(4,1,0)$ , $(3,2,0)$ , $(3,1,1)$ to get $34$ , $19$ , $13$ , $12$ ; the chained inequality shows no split, whole or not, beats $34$ .

Standardsmin grade 8

7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Adding to both sides of an inequality and chaining two inequalities.
8.EE.A.1 Know and apply the properties of integer exponents — Combining $2^{a+b}$ and $2^c$ and using $2^0=1$.

💡Takeaway. A fact about two numbers can handle three -- glue two of them into one block and use it twice.

Variant 8 medium

For nonnegative numbers $m$ and $n$ , it is known that
$2^m+2^n\le1+2^{m+n}$ , and equality holds when $m=0$ or $n=0$ .
If $a$ , $b$ , and $c$ are nonnegative numbers with $a+b+c=6$ , what is the largest possible value of $2^a+2^b+2^c$ ?

Your answer

Show solution

1 · Understandwhat's really being asked

We are given a fact: for nonnegative $m$ , $n$ , $2^m+2^n\le1+2^{m+n}$ , with equality when $m=0$ or $n=0$ . For nonnegative $a$ , $b$ , $c$ with $a+b+c=6$ , find the largest value of $2^a+2^b+2^c$ .

Givens

$2^m+2^n\le1+2^{m+n}$ for nonnegative $m$ , $n$ ; equality when $m=0$ or $n=0$ .
$a$ , $b$ , $c$ are nonnegative and $a+b+c=6$ .

Unknowns

The largest possible value of $2^a+2^b+2^c$ .

Constraints

An upper bound is the answer only if some allowed $a$ , $b$ , $c$ actually reach it.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #13 Convert to Algebra#6 Guess and Check

The given fact handles two powers at a time. Use it on $a$ and $b$ , then treat $a+b$ as a single exponent and use it again with $c$ . That leaves only $2^{a+b+c}=2^6$ .

3 · Execute5 carry out the plan

1Use the fact on $a$ and $b$

#7 Identify Subproblems 7.EE.B.4

a

and

b

are nonnegative, so the fact applies with

m=a

,

n=b

.

2^a+2^b\le1+2^{a+b}

This merges two powers into one, at the cost of a

1

.

2Add $2^c$ to both sides

#13 Convert to Algebra 7.EE.B.4

Adding the same number to both sides keeps the inequality.

2^a+2^b+2^c\le1+2^{a+b}+2^c

Now the right side has only two powers of

2

.

3Use the fact on $a+b$ and $c$

#7 Identify Subproblems 7.EE.B.4

a+b

is also nonnegative, so apply the fact with

m=a+b

,

n=c

.

2^{a+b}+2^c\le1+2^{a+b+c}

Treat

a+b

as one block.

4Chain the inequalities

#13 Convert to Algebra 8.EE.A.1

Put the two steps together and use

a+b+c=6

.

2^a+2^b+2^c\le1+1+2^{a+b+c}=2+2^6=66

2^{a+b+c}=2^6=64

.

5Check the bound is reached

#6 Guess and Check 8.EE.A.1

Equality needs

a=0

or

b=0

, and

a+b=0

or

c=0

. Try

a=0

,

b=0

,

c=6

.

2^0+2^0+2^6=1+1+64=66

2^0=1

, so the bound

66

is actually attained.

Answer:

66

4 · Reviewdoes it hold up?

Other splits give less: $(5,1,0)$ gives $32+2+1=35$ , $(4,2,0)$ gives $16+4+1=21$ , $(4,1,1)$ gives $16+2+2=20$ . Putting all of the $6$ into one exponent makes one term large, which makes sense for repeated multiplying.

Another way: The fact itself says $(2^m-1)(2^n-1)\ge0$ . Thinking of it this way, the sum is largest when the exponents are as unequal as possible: try whole-number splits $(6,0,0)$ , $(5,1,0)$ , $(4,2,0)$ , $(4,1,1)$ to get $66$ , $35$ , $21$ , $20$ ; the chained inequality shows no split, whole or not, beats $66$ .

Standardsmin grade 8

7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Adding to both sides of an inequality and chaining two inequalities.
8.EE.A.1 Know and apply the properties of integer exponents — Combining $2^{a+b}$ and $2^c$ and using $2^0=1$.

💡Takeaway. A fact about two numbers can handle three -- glue two of them into one block and use it twice.

Variant 9 hard

For nonnegative numbers $m$ and $n$ , it is known that
$2^m+2^n\le1+2^{m+n}$ , and equality holds when $m=0$ or $n=0$ .
If $a$ , $b$ , and $c$ are nonnegative numbers with $a+b+c=7$ , what is the largest possible value of $2^a+2^b+2^c$ ?

Your answer

Show solution

1 · Understandwhat's really being asked

We are given a fact: for nonnegative $m$ , $n$ , $2^m+2^n\le1+2^{m+n}$ , with equality when $m=0$ or $n=0$ . For nonnegative $a$ , $b$ , $c$ with $a+b+c=7$ , find the largest value of $2^a+2^b+2^c$ .

Givens

$2^m+2^n\le1+2^{m+n}$ for nonnegative $m$ , $n$ ; equality when $m=0$ or $n=0$ .
$a$ , $b$ , $c$ are nonnegative and $a+b+c=7$ .

Unknowns

The largest possible value of $2^a+2^b+2^c$ .

Constraints

An upper bound is the answer only if some allowed $a$ , $b$ , $c$ actually reach it.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #13 Convert to Algebra#6 Guess and Check

The given fact handles two powers at a time. Use it on $a$ and $b$ , then treat $a+b$ as a single exponent and use it again with $c$ . That leaves only $2^{a+b+c}=2^7$ .

3 · Execute5 carry out the plan

1Use the fact on $a$ and $b$

#7 Identify Subproblems 7.EE.B.4

a

and

b

are nonnegative, so the fact applies with

m=a

,

n=b

.

2^a+2^b\le1+2^{a+b}

This merges two powers into one, at the cost of a

1

.

2Add $2^c$ to both sides

#13 Convert to Algebra 7.EE.B.4

Adding the same number to both sides keeps the inequality.

2^a+2^b+2^c\le1+2^{a+b}+2^c

Now the right side has only two powers of

2

.

3Use the fact on $a+b$ and $c$

#7 Identify Subproblems 7.EE.B.4

a+b

is also nonnegative, so apply the fact with

m=a+b

,

n=c

.

2^{a+b}+2^c\le1+2^{a+b+c}

Treat

a+b

as one block.

4Chain the inequalities

#13 Convert to Algebra 8.EE.A.1

Put the two steps together and use

a+b+c=7

.

2^a+2^b+2^c\le1+1+2^{a+b+c}=2+2^7=130

2^{a+b+c}=2^7=128

.

5Check the bound is reached

#6 Guess and Check 8.EE.A.1

Equality needs

a=0

or

b=0

, and

a+b=0

or

c=0

. Try

a=0

,

b=0

,

c=7

.

2^0+2^0+2^7=1+1+128=130

2^0=1

, so the bound

130

is actually attained.

Answer:

130

4 · Reviewdoes it hold up?

Other splits give less: $(6,1,0)$ gives $64+2+1=67$ , $(5,2,0)$ gives $32+4+1=37$ , $(5,1,1)$ gives $32+2+2=36$ , and $\left(\frac{7}{3},\frac{7}{3},\frac{7}{3}\right)$ gives about $15.1$ . Putting all of the $7$ into one exponent makes one term large, which makes sense for repeated multiplying.

Another way: The fact itself says $(2^m-1)(2^n-1)\ge0$ . Thinking of it this way, the sum is largest when the exponents are as unequal as possible: try whole-number splits $(7,0,0)$ , $(6,1,0)$ , $(5,2,0)$ , $(5,1,1)$ to get $130$ , $67$ , $37$ , $36$ ; the chained inequality shows no split, whole or not, beats $130$ .

Standardsmin grade 8

7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Adding to both sides of an inequality and chaining two inequalities.
8.EE.A.1 Know and apply the properties of integer exponents — Combining $2^{a+b}$ and $2^c$ and using $2^0=1$.

💡Takeaway. A fact about two numbers can handle three -- glue two of them into one block and use it twice.

Variant 10 hard

For nonnegative numbers $m$ and $n$ , it is known that
$2^m+2^n\le1+2^{m+n}$ , and equality holds when $m=0$ or $n=0$ .
If $a$ , $b$ , and $c$ are nonnegative numbers with $a+b+c=8$ , what is the largest possible value of $2^a+2^b+2^c$ ?

Your answer

Show solution

1 · Understandwhat's really being asked

We are given a fact: for nonnegative $m$ , $n$ , $2^m+2^n\le1+2^{m+n}$ , with equality when $m=0$ or $n=0$ . For nonnegative $a$ , $b$ , $c$ with $a+b+c=8$ , find the largest value of $2^a+2^b+2^c$ .

Givens

$2^m+2^n\le1+2^{m+n}$ for nonnegative $m$ , $n$ ; equality when $m=0$ or $n=0$ .
$a$ , $b$ , $c$ are nonnegative and $a+b+c=8$ .

Unknowns

The largest possible value of $2^a+2^b+2^c$ .

Constraints

An upper bound is the answer only if some allowed $a$ , $b$ , $c$ actually reach it.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #13 Convert to Algebra#6 Guess and Check

The given fact handles two powers at a time. Use it on $a$ and $b$ , then treat $a+b$ as a single exponent and use it again with $c$ . That leaves only $2^{a+b+c}=2^8$ .

3 · Execute5 carry out the plan

1Use the fact on $a$ and $b$

#7 Identify Subproblems 7.EE.B.4

a

and

b

are nonnegative, so the fact applies with

m=a

,

n=b

.

2^a+2^b\le1+2^{a+b}

This merges two powers into one, at the cost of a

1

.

2Add $2^c$ to both sides

#13 Convert to Algebra 7.EE.B.4

Adding the same number to both sides keeps the inequality.

2^a+2^b+2^c\le1+2^{a+b}+2^c

Now the right side has only two powers of

2

.

3Use the fact on $a+b$ and $c$

#7 Identify Subproblems 7.EE.B.4

a+b

is also nonnegative, so apply the fact with

m=a+b

,

n=c

.

2^{a+b}+2^c\le1+2^{a+b+c}

Treat

a+b

as one block.

4Chain the inequalities

#13 Convert to Algebra 8.EE.A.1

Put the two steps together and use

a+b+c=8

.

2^a+2^b+2^c\le1+1+2^{a+b+c}=2+2^8=258

2^{a+b+c}=2^8=256

.

5Check the bound is reached

#6 Guess and Check 8.EE.A.1

Equality needs

a=0

or

b=0

, and

a+b=0

or

c=0

. Try

a=0

,

b=0

,

c=8

.

2^0+2^0+2^8=1+1+256=258

2^0=1

, so the bound

258

is actually attained.

Answer:

258

4 · Reviewdoes it hold up?

Other splits give less: $(7,1,0)$ gives $128+2+1=131$ , $(6,2,0)$ gives $64+4+1=69$ , $(6,1,1)$ gives $64+2+2=68$ , and $\left(\frac{8}{3},\frac{8}{3},\frac{8}{3}\right)$ gives about $19.0$ . Putting all of the $8$ into one exponent makes one term large, which makes sense for repeated multiplying.

Another way: The fact itself says $(2^m-1)(2^n-1)\ge0$ . Thinking of it this way, the sum is largest when the exponents are as unequal as possible: try whole-number splits $(8,0,0)$ , $(7,1,0)$ , $(6,2,0)$ , $(6,1,1)$ to get $258$ , $131$ , $69$ , $68$ ; the chained inequality shows no split, whole or not, beats $258$ .

Standardsmin grade 8

7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Adding to both sides of an inequality and chaining two inequalities.
8.EE.A.1 Know and apply the properties of integer exponents — Combining $2^{a+b}$ and $2^c$ and using $2^0=1$.

💡Takeaway. A fact about two numbers can handle three -- glue two of them into one block and use it twice.

Variant 11 hard

For nonnegative numbers $m$ and $n$ , it is known that
$2^m+2^n\le1+2^{m+n}$ , and equality holds when $m=0$ or $n=0$ .
If $a$ , $b$ , and $c$ are nonnegative numbers with $a+b+c=9$ , what is the largest possible value of $2^a+2^b+2^c$ ?

Your answer

Show solution

1 · Understandwhat's really being asked

We are given a fact: for nonnegative $m$ , $n$ , $2^m+2^n\le1+2^{m+n}$ , with equality when $m=0$ or $n=0$ . For nonnegative $a$ , $b$ , $c$ with $a+b+c=9$ , find the largest value of $2^a+2^b+2^c$ .

Givens

$2^m+2^n\le1+2^{m+n}$ for nonnegative $m$ , $n$ ; equality when $m=0$ or $n=0$ .
$a$ , $b$ , $c$ are nonnegative and $a+b+c=9$ .

Unknowns

The largest possible value of $2^a+2^b+2^c$ .

Constraints

An upper bound is the answer only if some allowed $a$ , $b$ , $c$ actually reach it.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #13 Convert to Algebra#6 Guess and Check

The given fact handles two powers at a time. Use it on $a$ and $b$ , then treat $a+b$ as a single exponent and use it again with $c$ . That leaves only $2^{a+b+c}=2^9$ .

3 · Execute5 carry out the plan

1Use the fact on $a$ and $b$

#7 Identify Subproblems 7.EE.B.4

a

and

b

are nonnegative, so the fact applies with

m=a

,

n=b

.

2^a+2^b\le1+2^{a+b}

This merges two powers into one, at the cost of a

1

.

2Add $2^c$ to both sides

#13 Convert to Algebra 7.EE.B.4

Adding the same number to both sides keeps the inequality.

2^a+2^b+2^c\le1+2^{a+b}+2^c

Now the right side has only two powers of

2

.

3Use the fact on $a+b$ and $c$

#7 Identify Subproblems 7.EE.B.4

a+b

is also nonnegative, so apply the fact with

m=a+b

,

n=c

.

2^{a+b}+2^c\le1+2^{a+b+c}

Treat

a+b

as one block.

4Chain the inequalities

#13 Convert to Algebra 8.EE.A.1

Put the two steps together and use

a+b+c=9

.

2^a+2^b+2^c\le1+1+2^{a+b+c}=2+2^9=514

2^{a+b+c}=2^9=512

.

5Check the bound is reached

#6 Guess and Check 8.EE.A.1

Equality needs

a=0

or

b=0

, and

a+b=0

or

c=0

. Try

a=0

,

b=0

,

c=9

.

2^0+2^0+2^9=1+1+512=514

2^0=1

, so the bound

514

is actually attained.

Answer:

514

4 · Reviewdoes it hold up?

Other splits give less: $(8,1,0)$ gives $256+2+1=259$ , $(7,2,0)$ gives $128+4+1=133$ , $(7,1,1)$ gives $128+2+2=132$ . Putting all of the $9$ into one exponent makes one term large, which makes sense for repeated multiplying.

Another way: The fact itself says $(2^m-1)(2^n-1)\ge0$ . Thinking of it this way, the sum is largest when the exponents are as unequal as possible: try whole-number splits $(9,0,0)$ , $(8,1,0)$ , $(7,2,0)$ , $(7,1,1)$ to get $514$ , $259$ , $133$ , $132$ ; the chained inequality shows no split, whole or not, beats $514$ .

Standardsmin grade 8

7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Adding to both sides of an inequality and chaining two inequalities.
8.EE.A.1 Know and apply the properties of integer exponents — Combining $2^{a+b}$ and $2^c$ and using $2^0=1$.

💡Takeaway. A fact about two numbers can handle three -- glue two of them into one block and use it twice.

Variant 12 hard

For nonnegative numbers $m$ and $n$ , it is known that
$2^m+2^n\le1+2^{m+n}$ , and equality holds when $m=0$ or $n=0$ .
If $a$ , $b$ , and $c$ are nonnegative numbers with $a+b+c=10$ , what is the largest possible value of $2^a+2^b+2^c$ ?

Your answer

Show solution

1 · Understandwhat's really being asked

We are given a fact: for nonnegative $m$ , $n$ , $2^m+2^n\le1+2^{m+n}$ , with equality when $m=0$ or $n=0$ . For nonnegative $a$ , $b$ , $c$ with $a+b+c=10$ , find the largest value of $2^a+2^b+2^c$ .

Givens

$2^m+2^n\le1+2^{m+n}$ for nonnegative $m$ , $n$ ; equality when $m=0$ or $n=0$ .
$a$ , $b$ , $c$ are nonnegative and $a+b+c=10$ .

Unknowns

The largest possible value of $2^a+2^b+2^c$ .

Constraints

An upper bound is the answer only if some allowed $a$ , $b$ , $c$ actually reach it.

2 · Planchoose the strategy

#7 Identify Subproblems · also uses: #13 Convert to Algebra#6 Guess and Check

The given fact handles two powers at a time. Use it on $a$ and $b$ , then treat $a+b$ as a single exponent and use it again with $c$ . That leaves only $2^{a+b+c}=2^{10}$ .

3 · Execute5 carry out the plan

1Use the fact on $a$ and $b$

#7 Identify Subproblems 7.EE.B.4

a

and

b

are nonnegative, so the fact applies with

m=a

,

n=b

.

2^a+2^b\le1+2^{a+b}

This merges two powers into one, at the cost of a

1

.

2Add $2^c$ to both sides

#13 Convert to Algebra 7.EE.B.4

Adding the same number to both sides keeps the inequality.

2^a+2^b+2^c\le1+2^{a+b}+2^c

Now the right side has only two powers of

2

.

3Use the fact on $a+b$ and $c$

#7 Identify Subproblems 7.EE.B.4

a+b

is also nonnegative, so apply the fact with

m=a+b

,

n=c

.

2^{a+b}+2^c\le1+2^{a+b+c}

Treat

a+b

as one block.

4Chain the inequalities

#13 Convert to Algebra 8.EE.A.1

Put the two steps together and use

a+b+c=10

.

2^a+2^b+2^c\le1+1+2^{a+b+c}=2+2^{10}=1026

2^{a+b+c}=2^{10}=1024

.

5Check the bound is reached

#6 Guess and Check 8.EE.A.1

Equality needs

a=0

or

b=0

, and

a+b=0

or

c=0

. Try

a=0

,

b=0

,

c=10

.

2^0+2^0+2^{10}=1+1+1024=1026

2^0=1

, so the bound

1026

is actually attained.

Answer:

1026

4 · Reviewdoes it hold up?

Other splits give less: $(9,1,0)$ gives $512+2+1=515$ , $(8,2,0)$ gives $256+4+1=261$ , $(8,1,1)$ gives $256+2+2=260$ , and $\left(\frac{10}{3},\frac{10}{3},\frac{10}{3}\right)$ gives about $30.2$ . Putting all of the $10$ into one exponent makes one term large, which makes sense for repeated multiplying.

Another way: The fact itself says $(2^m-1)(2^n-1)\ge0$ . Thinking of it this way, the sum is largest when the exponents are as unequal as possible: try whole-number splits $(10,0,0)$ , $(9,1,0)$ , $(8,2,0)$ , $(8,1,1)$ to get $1026$ , $515$ , $261$ , $260$ ; the chained inequality shows no split, whole or not, beats $1026$ .

Standardsmin grade 8

7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities — Adding to both sides of an inequality and chaining two inequalities.
8.EE.A.1 Know and apply the properties of integer exponents — Combining $2^{a+b}$ and $2^c$ and using $2^0=1$.

💡Takeaway. A fact about two numbers can handle three -- glue two of them into one block and use it twice.

Chain the given inequality twice · 12 practice problems

Generated variants — 12

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Constraints

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Constraints

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Unknowns

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Unknowns

Constraints

Givens

Unknowns

Constraints

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Constraints

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Where next?