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X= 2^20*5^13
= 10^13 *2^7
= 128*10^13

Sum of digits = 11
Option B

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Since we are looking for the sum of the digits we can simplify this problem by removing the trailing zeros (zeros at the end of the number). Trailing zeros are added to a number when multiplied by 10.

Remember 10 = 2 * 5
Thus, in our problem 4^10*5^13 or:
2^20 * 5^13

As you can see we have the pair 2*5 thirteen times. We can remove 2^13 and 5^13 and work with the rest which is 2^7 = 128

1+2+8 = 11
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we can use the concept of cyclicity to find out the units digit.
units digit of 4^10 is 6 and units digit of 5^13 is 5
6+5 = 11
units digit 1.
(This method works because there is no repetition of units digit in the options.)
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Bunuel
What is the sum of the digits of integer x, where \(x = 4^{10} * 5^{13}\)?

(A) 13

(B) 11

(C) 10

(D) 8

(E) 5


We can rewrite the expression as:

2^20 x 5^13

2^7 x 2^13 x 5^13

2^7 x 10^13

128 x 10^13


This is the number 128 followed by 13 zeros, so the sum of the digits is 1 + 2 + 8 = 11 (note: the 13 zeros will not contribute anything to the sum).

Answer: B
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B)

2^7 is 128

Sum of digits is equal to 11.

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