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What is the sum of the digits of integer \(k,\) if \(k = 10^{40}- 46.\)

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Re: What is the sum of the digits of integer \(k,\) if \(k = 10^{40}- 46.\)

by alishagmat » Sat Oct 10, 2020 7:54 am
expression =10^{40}-46= 999.....38 times ....54 (last two digits is 54)
So, sum = 9 X 38 +5+4 = 351 Answer
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Re: What is the sum of the digits of integer \(k,\) if \(k = 10^{40}- 46.\)

by Scott@TargetTestPrep » Sun Oct 11, 2020 5:41 am
M7MBA wrote: ↑
Wed Oct 07, 2020 10:41 am
 What is the sum of the digits of integer \(k,\) if \(k = 10^{40}- 46.\)

(A) 351
(B) 360
(C) 363
(D) 369
(E) 378

Answer: A

Solution:

Note that 10^40 is 1 followed by 40 zeros. When we subtract 46 from this number, we have to borrow from the first digit of 1 (making it 0), leaving 38 digits of 9 before the final 2 digits of 54 in the answer. Therefore, the sum of the digits of k is 38 x 9 + 5 + 4 = 351.

Answer: A

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