Place consecutive whole numbers to sum decimals

5.NBT.B.7 · take · grade 5

Archetype: Get Closest to a Target Value · step in a 4-type progression

Five consecutive one-digit whole numbers, written in order from smallest to largest, are $A$ , $B$ , $C$ , $D$ , $E$ . The decimals $A.BC$ and $C.DE$ are formed from these digits. If the sum $A.BC + C.DE$ is greater than $6$ and less than $7$ , find $100$ times $A.BC$ .

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Understand

Five one-digit whole numbers in a row (each one bigger than the last) are named A, B, C, D, E. Build the decimals A.BC and C.DE from those digits. Their sum is somewhere between 6 and 7. Find 100 times A.BC.

Givens

A, B, C, D, E are five consecutive whole numbers in increasing order, so B=A+1, C=A+2, D=A+3, E=A+4.
A.BC has whole part A and decimal digits B (tenths) and C (hundredths).
C.DE has whole part C and decimal digits D (tenths) and E (hundredths).
6 < A.BC + C.DE < 7.

Unknowns

The value of 100 x A.BC.

Constraints

All five digits are single digits (0-9) and consecutive.
The sum of the two decimals lies strictly between 6 and 7.

Plan

#6 Guess and Check · also uses: #9 Solve an Easier Related Problem

Because the digits are consecutive, the whole choice is fixed by a single starting digit A. The whole-number part of the sum is A + C = A + (A+2) = 2A+2, which already tells us roughly how big the sum is, so we test the few possible A values.

Execute

#9 Solve an Easier Related Problem 5.NBT.A.1

Since the numbers are consecutive, once A is chosen the digits are A, A+1, A+2, A+3, A+4. The whole parts of A.BC and C.DE are A and C = A+2, so the whole part of the sum is about A + (A+2) = 2A+2.

A+C = A+(A+2) = 2A+2

Reducing five unknown digits to one starting number makes the problem easy to test.

#6 Guess and Check 5.NBT.A.3

We need the sum near 6, so 2A+2 should be about 6, giving A = 2. Test A=2: the digits are 2,3,4,5,6, so the decimals are A.BC = 2.34 and C.DE = 4.56.

2A+2=6 \Rightarrow A=2 \Rightarrow A.BC=2.34,\ C.DE=4.56

A small guess pinned by the whole-number size is quick to verify.

#6 Guess and Check 5.NBT.B.7

Add the two decimals: 2.34 + 4.56 = 6.90, which is greater than 6 and less than 7. (A=1 gives 1.23 + 3.45 = 4.68, too small; A=3 gives 3.45 + 5.67 = 9.12, too big.) So A=2 is the only fit and A.BC = 2.34.

2.34+4.56=6.90,\quad 6<6.90<7

Only one starting digit makes the sum sit in the 6-to-7 window.

#6 Guess and Check 5.NBT.B.7

Multiplying a hundredths decimal by 100 shifts the decimal point two places right, turning 2.34 into 234.

2.34 \times 100 = 234

Times 100 moves the point two spots, a place-value shift.

Answer: 234

Review

The digits 2,3,4,5,6 are consecutive and single-digit. The sum 6.90 is indeed between 6 and 7. Multiplying 2.34 by 100 gives 234, a whole number, which makes sense because two decimal places cleared exactly.

Systematic List: tabulate the sum for each possible A (1 -> 4.68, 2 -> 6.90, 3 -> 9.12) and keep only the row inside (6, 7); only A=2 survives, giving 100 x 2.34 = 234.

Standards · min grade 5

5.NBT.B.7 Add, subtract, multiply, and divide decimals to hundredths — Adding 2.34 + 4.56 and multiplying 2.34 x 100.
5.NBT.A.1 Recognize that a digit in one place represents ten times as much as to its right — Reducing the five consecutive digits to a single starting digit A.
5.NBT.A.3 Read, write, and compare decimals to thousandths — Forming the decimals 2.34 and 4.56 from the digit list.

💡 When the digits are consecutive, one starting number fixes them all - guess it from the whole-number size, then check!