Competition · AMC preparation · step 4 of 4

AMC 10 · 2019B · #14

Grade 6 arithmetic

Archetype Small-Domain Constrained Search └ Sub-archetype Factor-Pair Enumeration

factorialdivisibility-rulesdigit-sumlegendre-formulamodular-arithmetic identify-subproblemscasework ↑ Prerequisites: factorialdivisibility-rules

📏 Long solution 💡 4 insights

Problem

The decimal expansion of 19! is 121,6T5,100,40M,832,H00, where T, M, H are three single-digit blanks. Find T + M + H.

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 10 2019 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Identify Subproblems

Tool #7 (Subproblems): peel the unknown digits off in three independent attacks — (i) trailing zeros pin H via the count of factors of 5 in 19!, (ii) divisibility by 9 constrains T+M via the digit-sum rule, (iii) divisibility by 11 constrains M-T via the alternating-digit-sum rule. Tool #8 supplies the divisibility rules. Tool #6 finishes by checking the two integer pairs that survive. Tool #3 matches T+M+H = 12 to choice (C). Direct multiplication of 19! is possible but enormous; the divisibility route is dramatically cleaner.

1STEP 1

Count the trailing zeros

Twos are plentiful, so trailing zeros of 19! come only from factors of 5 — just 5, 10, 15 — giving 3 trailing zeros.

#(trailing zeros of 19!) = ⌊ 19/5 ⌋ = 3

2STEP 2

Read off the digit H

19! ends in exactly three zeros, so the shown H00 is really 000 — forcing H = 0.

H00 = 000 → H = 0

3STEP 3

Apply the divisible-by-9 rule

By the 9-rule the whole digit sum is a multiple of 9; the known digits add to 33, so 33 + T + M must be a multiple of 9 too.

digit sum = 33 + T + M ≡ 0 (mod 9)

4STEP 4

Reduce the digit sum

Since 33 leaves remainder 6 mod 9, T + M ≡ 3 (mod 9); with single digits that means T + M is 3 or 12.

T + M ∈ {3, 12}

5STEP 5

Apply the divisible-by-11 rule

For the 11-rule take the alternating digit sum: odd positions total H + M + 20 and even positions total T + 13.

odd-pos sum = H + M + 20, even-pos sum = T + 13

6STEP 6

Form the alternating sum

Plug H = 0 into the alternating sum: (M + 20) - (T + 13) = 7 + M - T must be divisible by 11.

(0 + M + 20) - (T + 13) = 7 + M - T ≡ 0 (mod 11)

7STEP 7

Solve for the digit difference

So M - T ≡ 4 (mod 11); since the gap M - T runs from -9 to 9, it must be 4 or -7.

M - T ∈ {4, -7}

8STEP 8

Test the leftover pairs

Test the four pairings of (T + M, M - T): only (12, 4) yields digits in 0-9, giving T = 4, M = 8.

T = 4, M = 8, H = 0

9STEP 9

Add the three digits

Add the three digits: T + M + H = 4 + 8 + 0 = 12.

T + M + H = 4 + 8 + 0 = 12

10STEP 10

Match against the choices

The sum 12 matches choice (C).

12 → (C)

Answer

Plug T=4, M=8, H=0 back into the number: 121,645,100,408,832,000. Digit sum = 1+2+1+6+4+5+1+0+0+4+0+8+8+3+2+0+0+0 = 45, a multiple of 9 ✓. Alternating sum (right to left) = 0-0+0-0+0-1+8-0+0-4+0-1+5-6+1-2+1 = 8 - 4 - 1 + 5 - 6 + 1 - 2 + 1 - 1 + 0 - 0 + 0 … — easier: (0+0+3+8+4+0+5+6+2) - (0+2+8+0+0+1+4+1+1) = 28 - 17 = 11 ✓ multiple of 11. Both checks pass.

💡Key takeaway

This AMC 10 problem only needs Grade 6 divisibility reasoning you already know! 19! has exactly 3 trailing zeros (one per factor of 5 in 5, 10, 15), so H = 0. Divisibility by 9 forces T + M ∈ {3, 12} and divisibility by 11 forces M - T ∈ {4, -7}; only T = 4, M = 8 fits both. Sum = 4 + 8 + 0 = 12, answer (C).

Count the trailing zeros
Read off the digit H
Apply the divisible-by-9 rule
Reduce the digit sum
Apply the divisible-by-11 rule
Form the alternating sum
Solve for the digit difference
Test the leftover pairs
Add the three digits
Match against the choices

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