AMC 8 · 2024 · #15

Easy mode Grade 4

Problem

Imagine the six letters $F$ , $L$ , $Y$ , $B$ , $U$ , $G$ each stand for a different digit (0 through 9). No two letters share the same digit.

When you read $\underline{F}~\underline{L}~\underline{Y}~\underline{F}~\underline{L}~\underline{Y}$ , treat it as a six-digit number whose digits are $F$ , $L$ , $Y$ , $F$ , $L$ , $Y$ in that order. The number $\underline{B}~\underline{U}~\underline{G}~\underline{B}~\underline{U}~\underline{G}$ works the same way.

We want these two six-digit numbers to satisfy the equation

$8\cdot\underline{F}~\underline{L}~\underline{Y}~\underline{F}~\underline{L}~\underline{Y}=\underline{B}~\underline{U}~\underline{G}~\underline{B}~\underline{U}~\underline{G}.$

Among all digit choices that make this equation true, pick the one where $\underline{F}~\underline{L}~\underline{Y}~\underline{F}~\underline{L}~\underline{Y}$ is as large as possible.

For that choice, what is $\underline{F}~\underline{L}~\underline{Y}+\underline{B}~\underline{U}~\underline{G}$ ?

Pick an answer.

(A)

1089

(B)

1098

(C)

1107

(D)

1116

(E)

1125

AMC 8 2024 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.

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Toolkit + CCSS Solution

Understand

Restated: Six different letters $F, L, Y, B, U, G$ each stand for a different digit (0-9). The six-digit number $\underline{F}\underline{L}\underline{Y}\underline{F}\underline{L}\underline{Y}$ (the three-digit block $FLY$ written twice) is **the greatest** number for which $8 \cdot \underline{F}\underline{L}\underline{Y}\underline{F}\underline{L}\underline{Y} = \underline{B}\underline{U}\underline{G}\underline{B}\underline{U}\underline{G}$. Find $FLY + BUG$ (treating $FLY$ and $BUG$ as three-digit numbers).

Givens: $F, L, Y, B, U, G$ are six **distinct** digits 0-9; $\underline{F}\underline{L}\underline{Y}\underline{F}\underline{L}\underline{Y}$ is the three-digit block $FLY$ repeated, and likewise for $\underline{B}\underline{U}\underline{G}\underline{B}\underline{U}\underline{G}$; $8 \cdot \underline{F}\underline{L}\underline{Y}\underline{F}\underline{L}\underline{Y} = \underline{B}\underline{U}\underline{G}\underline{B}\underline{U}\underline{G}$; We want the **greatest** such six-digit number; Answer choices: (A) 1089, (B) 1098, (C) 1107, (D) 1116, (E) 1125

Unknowns: The value of $FLY + BUG$

Understand

Plan

Primary tool: #6 Guess and Check

Secondary: #9 Solve an Easier Related Problem, #2 Make a Systematic List, #3 Eliminate Possibilities

The six-digit number $\underline{F}\underline{L}\underline{Y}\underline{F}\underline{L}\underline{Y}$ looks scary, but Tool #9 (**Easier Related Problem**) cracks it: try a small three-digit number first, say $123$, and compute $123{,}123$. Notice $123{,}123 = 1000 \times 123 + 123 = 1001 \times 123$ — the same trick works for every three-digit block. So the messy six-digit equation simplifies to $8 \cdot FLY = BUG$. From there the task becomes: find the **greatest** three-digit $FLY$ such that $8 \cdot FLY$ is also a three-digit number and all six digits are different. That is a perfect job for Tool #6 (**Guess and Check**) — start with the biggest legal $FLY$ and step down one at a time. Tool #2 keeps the descending search organized, and Tool #3 confirms the final sum sits in the answer list.

Execute — Answer: C

#9 Solve an Easier Related Problem 4.NBT.A.2 Step 1

Crack the structure with a small example.
Take any 3-digit number, say $254$, and write it twice: $254{,}254$.
By place value, $254{,}254 = 254 \times 1000 + 254 = 254 \times 1001$.
The same identity holds for every 3-digit block, so $\underline{F}\underline{L}\underline{Y}\underline{F}\underline{L}\underline{Y} = 1001 \times FLY$ and $\underline{B}\underline{U}\underline{G}\underline{B}\underline{U}\underline{G} = 1001 \times BUG$.
The given equation becomes $8 \cdot (1001 \cdot FLY) = 1001 \cdot BUG$, and dividing both sides by $1001$ leaves a clean equation.

$$254{,}254 = 254 \times 1001 \;\Rightarrow\; 8 \cdot FLY = BUG$$

💡 Writing a 3-digit number in its thousands-and-ones places to see the $1001$ factor is exactly Grade 4 multi-digit place-value reading.

#9 Solve an Easier Related Problem 4.NBT.B.6 Step 2

Now figure out **how big $FLY$ can be**.
Since $BUG = 8 \cdot FLY$ must also be a 3-digit number, we need $BUG \le 999$, so $8 \cdot FLY \le 999$, which means $FLY \le 999 \div 8 = 124.875$.
The largest integer is $FLY = 124$.
(The problem asks for the *greatest* six-digit answer, and the six-digit value $1001 \cdot FLY$ grows with $FLY$, so we want the largest $FLY$ that works.)

$$FLY \le \dfrac{999}{8} = 124.875 \;\Rightarrow\; FLY \le 124$$

💡 Dividing $999 \div 8$ to find the largest 3-digit number whose octuple still fits in 3 digits is Grade 4 multi-digit division.

#6 Guess and Check 4.NBT.B.5 Step 3

Now **guess and check from the top**, listing values $124, 123, 122, \ldots$ in order until all six digits come out distinct.
Start with $FLY = 124$: $F=1, L=2, Y=4$ (distinct so far).
Compute $BUG = 8 \times 124 = 992$, giving $B=9, U=9, G=2$.
But $B = U = 9$ and $L = G = 2$, so the six digits $\{1,2,4,9,9,2\}$ are **not** all distinct.
Reject $124$.

$$FLY = 124 \;\Rightarrow\; BUG = 8 \times 124 = 992,\ \text{digits} = \{1,2,4,9,9,2\}\ \text{(repeats)}$$

💡 Computing $8 \times 124 = 992$ is a Grade 4 multi-digit-by-one-digit multiplication.

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

Step down to the next guess.
Try $FLY = 123$: $F=1, L=2, Y=3$ (distinct).
Compute $BUG = 8 \times 123 = 984$, giving $B=9, U=8, G=4$.
The six digits are $\{1, 2, 3, 9, 8, 4\}$ — **all six are different**.
Since we are walking downward and this is the first hit, $FLY = 123$ is the greatest valid value.

$$FLY = 123 \;\Rightarrow\; BUG = 8 \times 123 = 984,\ \text{digits} = \{1,2,3,9,8,4\}\ \text{(all distinct)} \;\checkmark$$

💡 Another Grade 4 one-digit-times-three-digit multiplication, plus a simple check that six listed digits are different.

#3 Eliminate Possibilities 4.NBT.B.4 Step 5

Add the two three-digit numbers to get the final answer: $FLY + BUG = 123 + 984$.
Lining up the place values: $3 + 4 = 7$ (ones), $2 + 8 = 10$ (write 0, carry 1), $1 + 9 + 1 = 11$ (write 1, carry 1), giving $1{,}107$.
Matching $1107$ to the answer choices: (A) 1089, (B) 1098, **(C) 1107**, (D) 1116, (E) 1125 — so the answer is choice (C).
The other choices can be eliminated since they are not equal to $123 + 984$.

$$123 + 984 = 1107 \;\Rightarrow\; \textbf{(C)}$$

💡 Adding two three-digit numbers with carrying and comparing the result to a short list of options is Grade 4 multi-digit addition and comparison.

[1] #9 4.NBT.A.2 Crack the structure with a small example. Take any 3-digit number, say $254$, an

[2] #9 4.NBT.B.6 Now figure out **how big $FLY$ can be**. Since $BUG = 8 \cdot FLY$ must also be

[3] #6 4.NBT.B.5 Now **guess and check from the top**, listing values $124, 123, 122, \ldots$ in

[4] #6 4.NBT.B.5 Step down to the next guess. Try $FLY = 123$: $F=1, L=2, Y=3$ (distinct). Comput

[5] #3 4.NBT.B.4 Add the two three-digit numbers to get the final answer: $FLY + BUG = 123 + 984$

Review

Reasonableness: Double-check by plugging $FLY = 123$ back into the **original six-digit equation**: the left side is $8 \cdot 123{,}123 = 984{,}984$, and the right side $\underline{B}\underline{U}\underline{G}\underline{B}\underline{U}\underline{G} = 984{,}984$ — exact match. The six digits $\{F, L, Y, B, U, G\} = \{1, 2, 3, 9, 8, 4\}$ are all different. And $FLY = 123$ being one short of the arithmetic ceiling $124$ is reasonable, since exactly one tiny case ($124$) had to be skipped for the distinctness rule. The sum $1107$ sits squarely in the middle of the answer choices, which is also where guessing-and-checking with a 'first failure, then success' pattern tends to land.

Alternative: An alternative is Tool #3 (**Eliminate Possibilities**) directly on the answer choices: since $FLY + BUG = FLY + 8 \cdot FLY = 9 \cdot FLY$, the answer must be a multiple of $9$. Test each choice: $1089 = 9 \times 121$, $1098 = 9 \times 122$, $1107 = 9 \times 123$, $1116 = 9 \times 124$, $1125 = 9 \times 125$ — all five are multiples of $9$, so divisibility alone does not pick a winner. But each choice tells us a candidate $FLY$, and we just check distinctness from the **largest** $FLY = 125$ downward: $125 \Rightarrow 8\times125=1000$ (4 digits, illegal); $124 \Rightarrow 992$ (repeated digits); $123 \Rightarrow 984$ (all six distinct, win) — answer (C). Same answer with no guessing needed once the multiple-of-9 trick is seen.

CCSS standards used (min grade 4)

4.NBT.A.2 Read and write multi-digit whole numbers and compare using symbols (Reading $\underline{F}\underline{L}\underline{Y}\underline{F}\underline{L}\underline{Y}$ by place value as $FLY \times 1000 + FLY = 1001 \times FLY$ to simplify the six-digit equation.)
4.NBT.B.4 Fluently add and subtract multi-digit whole numbers (Adding $123 + 984 = 1107$ with carrying to obtain the final sum, and comparing $1107$ to the five answer choices.)
4.NBT.B.5 Multiply a whole number of up to four digits by a one-digit whole number (Computing $8 \times 124 = 992$ and $8 \times 123 = 984$ during the guess-and-check search.)
4.NBT.B.6 Find whole-number quotients and remainders with up to four-digit dividends (Dividing $999 \div 8 = 124.875$ to bound $FLY$ from above so $BUG$ stays a three-digit number.)

⭐ This AMC 8 problem only needs Grade 4 multi-digit place value, multiplication, division, and addition you already know!

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: C

Review

CCSS standards used (min grade 4)

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