Deduce hidden digits from the ones place up

4.NBT.B.5 · take · grade 4

Archetype: Recover Hidden Digits from Carries · step in a 5-type progression

The long-multiplication below shows a three-digit number multiplied by a two-digit number. Find the digit that belongs in each of $A$ , $B$ , $C$ , $D$ , and $E$ .

\begin{array}{r} 4\,A\,2 \\ \times \quad 3\,B \\ \hline 2\,C\,1\,0 \\ 1\,D\,8\,6 \phantom{0} \\ \hline E\,6\,1\,7\,0 \end{array}

(Layout) The top three-digit number is $4\,A\,2$ and the two-digit multiplier is $3\,B$ . The first partial product is $2\,C\,1\,0$ , the second partial product is $1\,D\,8\,6$ written shifted one place to the left, and adding the two partial products gives the final product $E\,6\,1\,7\,0$ .

Show solution

Understand

A three-digit number 4A2 is multiplied by a two-digit number 3B. The two partial products are 2C10 and 1D86 (shifted one place left), and the final product is E6170. We must find each hidden digit A, B, C, D, and E.

Givens

Top factor is 4A2 (a three-digit number)
Bottom factor is 3B (a two-digit number)
First partial product (top factor times ones digit B) is 2C10
Second partial product (top factor times tens digit 3) is 1D86, shifted one place left
Final product is E6170

Unknowns

The digits A, B, C, D, and E

Constraints

Each letter is a single digit 0-9
The second partial product equals 4A2 times 3
The first partial product equals 4A2 times B

Plan

#6 Guess and Check · also uses: #7 Identify Subproblems#5 Look for a Pattern

The second partial product 1D86 must equal 4A2 x 3, which depends only on A; testing digits for A in a bounded range pins it down. With A known, the first partial product 4A2 x B = 2C10 fixes B, and the addition fixes E. Breaking the multiplication into its known partial products is the key subproblem split.

Execute

#6 Guess and Check 4.NBT.B.5

The second partial product 1D86 is the top number multiplied by the tens digit 3. So 4A2 x 3 must equal 1D86. Test digits: 462 x 3 = 1386, which matches the pattern 1D86 with the tens digit 8 and ones digit 6. So A = 6 and D = 3.

462 \times 3 = 1386 \Rightarrow A=6,\ D=3

Multiplying a 3-digit number by a single digit is a Grade 4 skill, and only one value of A makes the tens digit come out as 8.

#6 Guess and Check 4.NBT.B.5

The first partial product 2C10 is 462 multiplied by the ones digit B. Since it ends in 0, B must make 2 x B end in 0, so B = 5. Then 462 x 5 = 2310, which matches 2C10 with C = 3.

462 \times 5 = 2310 \Rightarrow B=5,\ C=3

Looking only at the ones place tells you B must be 5; one more multiplication confirms C.

#7 Identify Subproblems 4.NBT.B.5

The full multiplier is 35. Compute 462 x 35 = 2310 + 13860 = 16170, which matches E6170 with E = 1.

462 \times 35 = 2310 + 13860 = 16170 \Rightarrow E=1

Adding the two partial products is just multi-digit addition, and the leading digit of the sum gives E.

Answer: A = 6, B = 5, C = 3, D = 3, E = 1

Review

Check the whole multiplication: 462 x 35 = 16170, a five-digit number matching E6170 = 16170. Both partial products (2310 and 1386 shifted) add correctly, so all five digits are consistent.

Instead of guess-and-check, use Look for a Pattern (tool 5) on the ones digits: the second partial ends in 6, and since 2 x 3 = 6 the ones place is automatic, then read each column of the standard algorithm to deduce digits one place at a time.

Standards · min grade 4

4.NBT.B.5 Multiply a whole number of up to four digits by a one-digit whole number — Computing each partial product (462 x 3 and 462 x 5) and the full product 462 x 35 to identify the hidden digits.

💡 This only needs the Grade 4 multiplication you already know -- just match each partial product one place at a time!