2002 AMC 10B Problem 14

P11-15~2 minPrint

Exponents, Logarithms & Radicals Bases & Digits Set up the equation/formula and compute; no special trick needed

The number $25^{64}\cdot 64^{25}$ is the square of a positive integer $N$ . In decimal representation, the sum of the digits of $N$ is

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Key insight

N is 5 to the 64th times 2 to the 75th. Pair each 5 with a 2 to make 64 tens; 2048 is left over.

Solution

The digit sum of a giant number is only easy if the number turns out to be simple, like a small number followed by zeros. A $2$ times a $5$ makes a $10$ , which just adds a zero. So break everything into $2$ s and $5$ s.

1. Write each base as a prime power: $25 = 5^2$ and $64 = 2^6$ . So
$25^{64} = 5^{128}, \qquad 64^{25} = 2^{150}.$
2. $N$ is the square root. Taking a square root halves each exponent (because $5^{64} \cdot 5^{64} = 5^{128}$ ):
$N = 5^{64} \cdot 2^{75}.$
3. There are $64$ fives and $75$ twos. Pair every five with a two to make $64$ tens. That uses $64$ twos, leaving $75 - 64 = 11$ twos:
$N = 10^{64} \cdot 2^{11}.$
4. $2^{11} = 2048$ . Multiplying by $10^{64}$ writes $64$ zeros after it, so $N$ is $2048$ followed by $64$ zeros.
5. Digit sum: $2 + 0 + 4 + 8 = 14$ .

The answer is $\boxed{\textbf{(B)}\ 14}$ .

Why this works

Powers of $10$ only add zeros, and zeros add nothing to a digit sum. So whenever a huge number is made of $2$ s and $5$ s, pair them up into tens and look only at what is left over. That small leftover number decides the digits.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

Write everything in primes: $25^{64} = 5^{128}$ and $64^{25} = (2^6)^{25} = 2^{150}$ . Taking the square root halves each exponent:
$N = 5^{64}\cdot 2^{75}.$

Every $5$ can be matched with a $2$ to make a $10$ ; there are $64$ fives and $75$ twos, so $64$ pairs form and $11$ twos are left over:
$N = (2\cdot 5)^{64}\cdot 2^{11} = 2048 \cdot 10^{64}.$

Multiplying by $10^{64}$ just appends $64$ zeros, so the nonzero digits of $N$ are those of $2048$ , with sum $2 + 0 + 4 + 8 = 14$ .

The answer is $\boxed{\textbf{(B)}\ 14}$ .

Review in the AoPS books

Introduction to Algebra, Chapter 1: Follow the Rules (§1.6, §1.7, §1.8)
Introduction to Number Theory, Chapter 8: Base Numbers
Introduction to Algebra, Chapter 11: Special Factorizations (§11.4)
Introduction to Number Theory, Chapter 9: Base Number Arithmetic