2002 AMC 10B Problem 14

Below is the professionally curated solution for Problem 14 of the 2002 AMC 10B, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2002 AMC 10B solutions, or check the answer key.

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Concepts:exponentdigits

Difficulty rating: 1420

14.

The number 2564642525^{64} \cdot 64^{25} is the square of a positive integer N.N. In decimal representation, what is the sum of the digits of N?N?

77

1414

2121

2828

3535

Solution:

Since 25=5225 = 5^2 and 64=26,64 = 2^6, we have 25646425=51282150,25^{64}\cdot 64^{25} = 5^{128}\cdot 2^{150}, so N=51282150=564275.N = \sqrt{5^{128}\cdot 2^{150}} = 5^{64}\cdot 2^{75}.

Writing 275=264211,2^{75} = 2^{64}\cdot 2^{11}, we get N=(52)64211=10642048.N = (5\cdot 2)^{64}\cdot 2^{11} = 10^{64}\cdot 2048.

So NN is 20482048 followed by 6464 zeros, and its digit sum is 2+0+4+8=14.2 + 0 + 4 + 8 = 14.

Thus, the correct answer is B.

Problem 14 in Other Years