AMC 10 · 2019 · #14

Grade 6 arithmetic

Archetype Small-Domain Constrained Search

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

📏 Long solution 💡 4 insights

Problem

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 $T+M+H$ ?

$\textbf{(A) }3 \qquad\textbf{(B) }8 \qquad\textbf{(C) }12 \qquad\textbf{(D) }14 \qquad\textbf{(E) } 17$

Pick an answer.

(A)

(B)

(C)

(D)

(E)

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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 $T + M + H$.

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

#7 Identify Subproblems 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.

$$\#(\text{trailing zeros of }19!) = \lfloor 19/5 \rfloor = 3$$

💡 Grade 6 GCF/LCM: every factor of $10$ requires one $5$, and there are exactly three $5$s up to $19$.

#7 Identify Subproblems 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 $H = 0$.

$$H00 = 000 \;\Rightarrow\; H = 0$$

💡 Grade 5 place value: the last three digits are the zeros from the trailing-zero count.

#8 Analyze the Units 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 $T + M + H = T + M + 0$.

$$\text{digit sum} = 33 + T + M \;\equiv\; 0 \pmod{9}$$

💡 Grade 4 factors and multiples: the digit-sum rule turns divisibility into a one-line check.

#8 Analyze the Units 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 $T + M = 12$.

$$T + M \in \{3, \, 12\}$$

💡 Grade 4: two single-digit numbers sum to at most $18$, so only two candidate sums remain.

#8 Analyze the Units 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 $0, 2, 8, 0, 0, 1, T, 1, 1$.

$$\text{odd-pos sum} = H + M + 20, \;\; \text{even-pos sum} = T + 13$$

💡 Grade 4 factors: the $11$-rule alternates digits — exactly what is needed here.

#8 Analyze the Units 6.EE.B.7 Step 6

Form the alternating sum with $H = 0$ and require divisibility by $11$.

$$(0 + M + 20) - (T + 13) = 7 + M - T \equiv 0 \pmod{11}$$

💡 Grade 6: a one-step mod relation between $T$ and $M$.

#6 Guess and Check 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 $M - T = -7$.

$$M - T \in \{4, \, -7\}$$

💡 Grade 6: only two integer gaps fit the single-digit range.

#6 Guess and Check 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.)

$$T = 4, \; M = 8, \; H = 0$$

💡 Grade 6: combining two linear constraints uniquely identifies the digit pair.

#7 Identify Subproblems 4.NBT.B.4 Step 9

Add the three digits.

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

💡 Grade 4 addition: a quick sum of three single digits.

#3 Eliminate Possibilities 4.NBT.A.2 Step 10

Match $12$ to the choices.

$$12 \;\Rightarrow\; \textbf{(C)}$$

💡 Grade 4: pick the matching whole number.

[1] #7 6.NS.B.4 Find the number of trailing zeros of $19!$ to pin down $H$. Trailing zeros come

[2] #7 5.NBT.A.1 The displayed last three digits are $H00$. Since $19!$ ends in exactly three zer

[3] #8 4.OA.B.4 Apply the divisibility-by-$9$ rule: a number is divisible by $9$ iff its digit s

[4] #8 4.OA.B.4 $33 \equiv 6 \pmod 9$, so $T + M \equiv 3 \pmod 9$. With $0 \le T, M \le 9$, the

[5] #8 4.OA.B.4 Apply the divisibility-by-$11$ rule: the alternating sum of digits (right-to-lef

[6] #8 6.EE.B.7 Form the alternating sum with $H = 0$ and require divisibility by $11$.

[7] #6 6.EE.B.7 Solve $M - T \equiv -7 \equiv 4 \pmod{11}$. With $0 \le T, M \le 9$, the gap $M

[8] #6 6.EE.B.7 Combine the two constraints by guess-and-check on the four pairs $(T+M, M-T)$. O

[9] #7 4.NBT.B.4 Add the three digits.

[10] #3 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.2 Read and write multi-digit whole numbers and compare using symbols (Matching $12$ to the multiple-choice value.)
4.NBT.B.4 Fluently add and subtract multi-digit whole numbers (Summing $T + M + H = 4 + 8 + 0 = 12$.)
4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite (Using divisibility-by-$9$ and divisibility-by-$11$ rules on $19!$.)
5.NBT.A.1 Recognize 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.4 Find greatest common factor and least common multiple of two numbers (Counting factors of $5$ in $19!$ to get three trailing zeros.)
6.EE.B.7 Solve 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 $(C)$.

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: C

Review

CCSS standards used (min grade 6)

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