2011 AMC 8 Problem 15

Attempt Problem 15 of the 2011 AMC 8 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2011 AMC 8 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

15.

How many digits are in the product 45⋅510?4^5 \cdot 5^{10}?
88
99
1010
1111
1515
Answer: D
Concepts:exponentdigits
Small hint:
Pair 2102^{10} with 5105^{10} to make 101010^{10}.
Big hint:
Rewrite 454^5 as a power of 22.
Solution:
To find the number of digits, we can try to express this number in terms of powers of 10.10.
We get that 45⋅510=210⋅510=1010.\begin{align*} 4^5 \cdot 5^{10} &= 2^{10} \cdot 5^{10} \\ &= 10^{10}. \end{align*}
This shows that the desired number is 11 followed by 1010 zeros, for a total of 1111 digits.
Thus, D is the correct answer.
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