AMC 10 · 2002 · #14
Grade 8 number-theoryThe number 2564⋅6425 is the square of a positive integer N. In decimal representation, the sum of the digits of N is
Pick an answer.
AMC 10 2002 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: A gigantic number, $25^{64}\cdot 64^{25}$, happens to be a perfect square; call its square root $N$. Write $N$ in ordinary decimal digits and add up all of those digits.
Givens: The number is $25^{64}\cdot 64^{25}$; This number equals $N^2$ for some positive integer $N$; Answer choices for the digit sum: (A) $7$, (B) $14$, (C) $21$, (D) $28$, (E) $35$
Unknowns: The sum of the decimal digits of $N=\sqrt{25^{64}\cdot 64^{25}}$
Understand
Restated: A gigantic number, $25^{64}\cdot 64^{25}$, happens to be a perfect square; call its square root $N$. Write $N$ in ordinary decimal digits and add up all of those digits.
Givens: The number is $25^{64}\cdot 64^{25}$; This number equals $N^2$ for some positive integer $N$; Answer choices for the digit sum: (A) $7$, (B) $14$, (C) $21$, (D) $28$, (E) $35$
Plan
Primary tool: #9 Solve an Easier Related Problem
Secondary: #15 Organize Information in More Ways, #5 Look for a Pattern
The number is astronomically large, so Tool #9 (Solve an Easier Related Problem) replaces the impossible arithmetic with an easier one: break $25$ and $64$ into powers of the primes $5$ and $2$, where taking a square root just means halving the exponents. Tool #15 (Organize Information in More Ways) then regroups the leftover $2$s and $5$s so that equal numbers of them pair into $10$s. Tool #5 (Look for a Pattern) finishes it: multiplying by a power of $10$ only tacks zeros onto the end, so the digit sum comes from the few nonzero digits in front.
Execute — Answer: B
8.EE.A.1 Step 1 Break the bases into primes
- Both bases are powers of a single prime.
- Since $25=5^2$ and $64=2^6$, use the power-of-a-power rule (multiply the exponents): $25^{64}=(5^2)^{64}=5^{128}$ and $64^{25}=(2^6)^{25}=2^{150}$.
- The whole product becomes $5^{128}\cdot 2^{150}$ — the same number, now written entirely in the primes $2$ and $5$.
💡 A power raised to a power just multiplies the exponents, turning messy bases into clean prime powers.
8.EE.A.1 Step 2 Take the root by halving exponents
- Taking a square root of a prime power halves its exponent, because $\sqrt{p^{2k}}=p^{k}$.
- Both exponents here are even, so $N=\sqrt{5^{128}\cdot 2^{150}}=5^{128/2}\cdot 2^{150/2}=5^{64}\cdot 2^{75}$.
- This is the exact integer $N$, still in prime form.
💡 Squaring doubles an exponent, so undoing it — the square root — cuts the exponent in half.
8.EE.A.1 Step 3 Pair twos and fives into tens
- A $2$ and a $5$ multiply to $10$, so match up as many of them as possible.
- Split the $2$s to expose $64$ of them: $2^{75}=2^{64}\cdot 2^{11}$.
- Now pair the $64$ twos with the $64$ fives: $5^{64}\cdot 2^{64}=(2\cdot 5)^{64}=10^{64}$.
- That leaves $N=2^{11}\cdot 10^{64}$.
💡 Every $2$ paired with a $5$ becomes a $10$, and $10$s are exactly what create trailing zeros.
6.EE.A.1 Step 4 Evaluate the front number
- Only the leftover $2^{11}$ carries the actual digits.
- Compute it: $2^{11}=2048$.
- So $N=2048\cdot 10^{64}$.
💡 The paired-off $10$s only add zeros, so the meaningful digits all live in the small factor out front.
5.NBT.A.2 Step 5 Append the zeros and add the digits
- Multiplying $2048$ by $10^{64}$ writes $2048$ and then $64$ zeros after it.
- Zeros add nothing to a digit sum, so the sum of the digits of $N$ is just $2+0+4+8=14$.
- That is choice (B).
💡 Multiplying by a power of $10$ only slides digits left and fills in zeros, which never change the digit sum.
8.EE.A.1 Both bases are powers of a single prime. Since $25=5^2$ and $64=2^6$, use the po 8.EE.A.1 Taking a square root of a prime power halves its exponent, because $\sqrt{p^{2k} 8.EE.A.1 A $2$ and a $5$ multiply to $10$, so match up as many of them as possible. Split 6.EE.A.1 Only the leftover $2^{11}$ carries the actual digits. Compute it: $2^{11}=2048$. 5.NBT.A.2 Multiplying $2048$ by $10^{64}$ writes $2048$ and then $64$ zeros after it. Zero Review
Reasonableness: Sanity-check the exponents by squaring back: $(2^{11}\cdot 10^{64})^2=2^{22}\cdot 10^{128}=2^{22}\cdot 2^{128}\cdot 5^{128}=2^{150}\cdot 5^{128}$, which matches $64^{25}\cdot 25^{64}$ exactly — so $N=2^{11}\cdot 10^{64}$ is right. The digit sum $14$ also sits comfortably among the offered choices, and it makes sense that a number that is mostly trailing zeros has a small digit sum.
Alternative: Instead of halving exponents, take the square root first as $\sqrt{25^{64}}\cdot\sqrt{64^{25}}$. The first is $25^{32}=5^{64}$; for the second, $\sqrt{64^{25}}=8^{25}=2^{75}$ since $8=\sqrt{64}$. That again gives $N=5^{64}\cdot 2^{75}=2^{11}\cdot 10^{64}$, leading to the same digits $2048$ and digit sum $14$.
CCSS standards used (min grade 8)
8.EE.A.1Know and apply the properties of integer exponents (Rewriting $(5^2)^{64}=5^{128}$ and $(2^6)^{25}=2^{150}$, halving exponents to take the square root ($5^{64}\cdot 2^{75}$), and regrouping $5^{64}\cdot 2^{64}=10^{64}$.)6.EE.A.1Write and evaluate numerical expressions involving whole-number exponents (Evaluating the leftover factor $2^{11}=2048$, the only part of $N$ that carries nonzero digits.)5.NBT.A.2Explain patterns in number of zeros and placement of decimal point (Recognizing that multiplying $2048$ by $10^{64}$ appends $64$ zeros, so those zeros contribute nothing to the digit sum.)
⭐ Turn scary bases into powers of $2$ and $5$, pair every $2$ with a $5$ to make $10$s (which are just trailing zeros), and the tiny leftover $2^{11}=2048$ hands you the digit sum $2+0+4+8=14$.
⭐ Turn scary bases into powers of $2$ and $5$, pair every $2$ with a $5$ to make $10$s (which are just trailing zeros), and the tiny leftover $2^{11}=2048$ hands you the digit sum $2+0+4+8=14$.
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