AMC 10 · 2019 · #14
Grade 6 arithmeticProblem
The base-ten representation for 19! is 121,6T5,100,40M,832,H00, where T, M, and H denote digits that are not given. What is T+M+H?
Pick an answer.
AMC 10 2019 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: The decimal expansion of $19!$ is $121{,}6T5{,}100{,}40M{,}832{,}H00$, where $T$, $M$, $H$ are three single-digit blanks. Find $T + M + H$.
Givens: $19! = 121{,}6T5{,}100{,}40M{,}832{,}H00$ — an $18$-digit number; $T, M, H$ are each one of the digits $0,1,\dots,9$; $19!$ is the product of $1 \cdot 2 \cdot 3 \cdots 19$; Answer choices: $3,\, 8,\, 12,\, 14,\, 17$
Unknowns: $T + M + H$
Understand
Restated: The decimal expansion of $19!$ is $121{,}6T5{,}100{,}40M{,}832{,}H00$, where $T$, $M$, $H$ are three single-digit blanks. Find $T + M + H$.
Givens: $19! = 121{,}6T5{,}100{,}40M{,}832{,}H00$ — an $18$-digit number; $T, M, H$ are each one of the digits $0,1,\dots,9$; $19!$ is the product of $1 \cdot 2 \cdot 3 \cdots 19$; Answer choices: $3,\, 8,\, 12,\, 14,\, 17$
Plan
Primary tool: #7 Identify Subproblems
Secondary: #8 Analyze the Units, #6 Guess and Check, #3 Eliminate Possibilities
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.
Execute — Answer: C
6.NS.B.4 Step 1 - Find the number of trailing zeros of $19!$ to pin down $H$.
- Trailing zeros come from pairs of factors $2 \cdot 5$, and twos are plentiful, so count factors of $5$ in $\{1,2,\dots,19\}$: only $5, 10, 15$ each contribute one factor of $5$.
- So $19!$ has exactly $3$ trailing zeros.
💡 Grade 6 GCF/LCM: every factor of $10$ requires one $5$, and there are exactly three $5$s up to $19$.
5.NBT.A.1 Step 2 - The displayed last three digits are $H00$.
- Since $19!$ ends in exactly three zeros, those three digits are $000$, forcing $H = 0$.
💡 Grade 5 place value: the last three digits are the zeros from the trailing-zero count.
4.OA.B.4 Step 3 - Apply the divisibility-by-$9$ rule: a number is divisible by $9$ iff its digit sum is.
- Since $9 \mid 19!$, the digit sum is a multiple of $9$.
- Add the known digits ($1+2+1+6+5+1+0+0+4+0+8+3+2+0+0+0 = 33$) plus the unknowns $T + M + H = T + M + 0$.
💡 Grade 4 factors and multiples: the digit-sum rule turns divisibility into a one-line check.
4.OA.B.4 Step 4 - $33 \equiv 6 \pmod 9$, so $T + M \equiv 3 \pmod 9$.
- With $0 \le T, M \le 9$, the only possibilities are $T + M = 3$ or $T + M = 12$.
💡 Grade 4: two single-digit numbers sum to at most $18$, so only two candidate sums remain.
4.OA.B.4 Step 5 - Apply the divisibility-by-$11$ rule: the alternating sum of digits (right-to-left) is a multiple of $11$.
- List positions $1$ (rightmost) through $18$.
- The displayed digits from left to right are $1,2,1,6,T,5,1,0,0,4,0,M,8,3,2,H,0,0$.
- From the right, odd positions hold $0, H, 3, M, 4, 0, 5, 6, 2$ and even positions hold $0, 2, 8, 0, 0, 1, T, 1, 1$.
💡 Grade 4 factors: the $11$-rule alternates digits — exactly what is needed here.
6.EE.B.7 Step 6 Form the alternating sum with $H = 0$ and require divisibility by $11$.
💡 Grade 6: a one-step mod relation between $T$ and $M$.
6.EE.B.7 Step 7 - Solve $M - T \equiv -7 \equiv 4 \pmod{11}$.
- With $0 \le T, M \le 9$, the gap $M - T$ lies in $[-9, 9]$, so the only options are $M - T = 4$ or $M - T = -7$.
💡 Grade 6: only two integer gaps fit the single-digit range.
6.EE.B.7 Step 8 - Combine the two constraints by guess-and-check on the four pairs $(T+M, M-T)$.
- Only $(T+M, M-T) = (12, 4)$ gives integer digits in $0$–$9$: $M = 8, T = 4$.
- (The other three pairings give half-integers or negatives.)
💡 Grade 6: combining two linear constraints uniquely identifies the digit pair.
4.NBT.B.4 Step 9 Add the three digits.
💡 Grade 4 addition: a quick sum of three single digits.
4.NBT.A.2 Step 10 Match $12$ to the choices.
💡 Grade 4: pick the matching whole number.
6.NS.B.4 Find the number of trailing zeros of $19!$ to pin down $H$. Trailing zeros come 5.NBT.A.1 The displayed last three digits are $H00$. Since $19!$ ends in exactly three zer 4.OA.B.4 Apply the divisibility-by-$9$ rule: a number is divisible by $9$ iff its digit s 4.OA.B.4 $33 \equiv 6 \pmod 9$, so $T + M \equiv 3 \pmod 9$. With $0 \le T, M \le 9$, the 4.OA.B.4 Apply the divisibility-by-$11$ rule: the alternating sum of digits (right-to-lef 6.EE.B.7 Form the alternating sum with $H = 0$ and require divisibility by $11$. 6.EE.B.7 Solve $M - T \equiv -7 \equiv 4 \pmod{11}$. With $0 \le T, M \le 9$, the gap $M 6.EE.B.7 Combine the two constraints by guess-and-check on the four pairs $(T+M, M-T)$. O 4.NBT.B.4 Add the three digits. 4.NBT.A.2 Match $12$ to the choices. Review
Reasonableness: 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) $= 0-0+0-0+0-1+8-0+0-4+0-1+5-6+1-2+1 = 8 - 4 - 1 + 5 - 6 + 1 - 2 + 1 - 1 + 0 - 0 + 0 \dots$ — 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.
Alternative: Tool #4 (Matrix Logic) — write a $2 \times 2$ table of the four candidate pairs $(T+M, M-T) \in \{3,12\} \times \{4,-7\}$ and mark each cell "valid digits" or "reject". Only the cell $(12, 4)$ is marked valid because the other three yield $T$ or $M$ outside $\{0,1,\dots,9\}$ or non-integer. Same conclusion, same answer $(C)$.
CCSS standards used (min grade 6)
4.NBT.A.2Read and write multi-digit whole numbers and compare using symbols (Matching $12$ to the multiple-choice value.)4.NBT.B.4Fluently add and subtract multi-digit whole numbers (Summing $T + M + H = 4 + 8 + 0 = 12$.)4.OA.B.4Find all factor pairs and recognize multiples; determine prime or composite (Using divisibility-by-$9$ and divisibility-by-$11$ rules on $19!$.)5.NBT.A.1Recognize that a digit in one place represents ten times as much as to its right (Reading the last three digits $H00$ as the trailing zeros of $19!$.)6.NS.B.4Find greatest common factor and least common multiple of two numbers (Counting factors of $5$ in $19!$ to get three trailing zeros.)6.EE.B.7Solve real-world problems by writing and solving equations of the form px = q (Combining $T + M \equiv 3 \pmod 9$ and $M - T \equiv 4 \pmod{11}$ to get $T = 4, M = 8$.)
⭐ 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 $T + M \in \{3, 12\}$ and divisibility by $11$ forces $M - T \in \{4, -7\}$; only $T = 4, M = 8$ fits both. Sum $= 4 + 8 + 0 = \mathbf{12}$, answer $(C)$.
⭐ 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 $T + M \in \{3, 12\}$ and divisibility by $11$ forces $M - T \in \{4, -7\}$; only $T = 4, M = 8$ fits both. Sum $= 4 + 8 + 0 = \mathbf{12}$, answer $(C)$.
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