Problem

Deduce hidden digits from the ones place up

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.

Base-ten numbers

Your answer

How to solve

Strategy Guess and Check — 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.

1STEP 1

Use the second partial product to find A and D

462 × 3 = 1386 fits the pattern 1D86 (tens digit 8), so A = 6 and D = 3.

462 × 3 = 1386 → 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.

4.NBT.B.5Guess And Check

The second partial-product row 1D86 is the top number 4A2 multiplied by the tens digit 3, so its value is fixed by the single unknown digit A.

Why?

Long multiplication builds each partial-product row by multiplying the whole top number by one digit of the bottom number, and the lower row uses the bottom number's tens digit 3.

Why?

Multiplying by the two-digit number 3B means multiplying by 3 tens together with B ones, and a number times a sum can be handled one part at a time.

Why?

The 3 stands in the tens place, so it means 3 tens, that is 30, not 3 ones.

🧱Place-value groupingTen ones make one ten and ten tens make one hundred — each place bundles ten of the last.

Why?

A number times a sum equals that number times each part, added back together.

🧱Distributive propertya groups of (b and c) is a groups of b plus a groups of c — you can split the group.

Why?

Multiplying the top number by 3 tens gives the same digits as multiplying it by 3 and sliding the result one place to the left, so apart from that shift the row is exactly the top number times 3.

Why?

Multiplying by 3 tens is the same as multiplying by 3 and then by 10, because regrouping which factors you combine first does not change the product.

🧱Associativity of multiplicationMultiplying three numbers, grouping any two first gives the same product.

Why?

Multiplying by ten turns every one into a ten, sliding each digit one place to the left.

🧱Place-value groupingTen ones make one ten and ten tens make one hundred — each place bundles ten of the last.

Why?

The digits 4, 2, and 3 are all already known, so the whole row's value is decided by the one missing digit A.

Why?

Multiplying the top number by 3 is repeated equal-group adding that gives one definite total for each choice of A, so only one A can produce the shown row.

🧱Multiplication as equal groupsa times b means a equal groups of b — it is just repeated adding of the same amount.

2STEP 2

Use the first partial product to find B and C

462 × 5 = 2310 matches 2C10 (ends in 0), giving B = 5 and C = 3.

462 × 5 = 2310 → B=5, C=3

3STEP 3

Add the partial products to find E

The full multiplier is 35: 462 × 35 = 2310 + 13860 = 16170, matching E6170 with E = 1.

462 × 35 = 2310 + 13860 = 16170 → E=1

Answer

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

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

Takeaway

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

Use the second partial product to find A and D
Use the first partial product to find B and C
Add the partial products to find E

Where next?

Another one like thissuggested

11 practice problems

Step back and solidifysuggested

Recover missing digits from column carries Deduce missing digits in a multiplication

9 reasoning problems grow out of this type — the bridge toward competition →

▶ Practice — 11 problems