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2023 AMC 10A Problem 5

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2023 AMC 10A Problem 5AMC 10 Step by Step
Name:                             Date: September 13, 2026
Exponents, Logarithms & Radicals · difficulty 1 of 5 · about 2 min · position P1-10

How many digits are in the base-ten representation of 85⋅510⋅1558^5 \cdot 5^{10} \cdot 15^5 ?

  1. A)

    1414

  2. B)

    1515

  3. C)

    1616

  4. D)

    1717

  5. E)

    1818

Answer

E

Key insight

Rewrite in primes: 2^15 times 5^15 times 3^5 equals 243 followed by fifteen zeros.

Solution

Break every base into primes:
85=215,155=35⋅55.8^5 = 2^{15}, \qquad 15^5 = 3^5 \cdot 5^5.
So the product is
215⋅510⋅35⋅55=215⋅515⋅35=1015⋅243.2^{15} \cdot 5^{10} \cdot 3^5 \cdot 5^5 = 2^{15} \cdot 5^{15} \cdot 3^5 = 10^{15} \cdot 243.

Multiplying 243243 by 101510^{15} appends fifteen zeros to the three-digit number 243243, so the result has 3+15=183 + 15 = 18 digits.

The answer is (E) 18\boxed{\textbf{(E)}\ 18}.

Why this works

Digit-count questions about big products are almost always about pairing factors of 22 with factors of 55 to form powers of 1010. Whatever is left over is a small number whose digit count you can read off directly. Always factor into primes before trying to multiply anything.

The trap

Counting only the 15 zeros and forgetting the three digits of 243, giving 15.

Common mistakes

  • Counting only the 15 zeros and forgetting the three digits of 243, giving 15.
  • Writing 85=288^5 = 2^{8} or 155=3⋅5515^5 = 3 \cdot 5^5 and losing track of exponents.

Techniques

Apply an identity: SFFT, sum of squares, difference of cubes, Vieta · Set up the equation/formula and compute; no special trick needed