chetan2u wrote:

What are the number of zeroes in the end of integer x, if x = 123!+124! ?

(1) 28

(2) 29

(3) 30

(4) 31

(5) 56

To determine the number of trailing zeros in a number, we need to determine the number of 5-and-2 pairs within the prime factorization of that number. Let’s start by simplifying 123!+124!.

123!+124! = 123!(1 + 124) = 123!(125)

Since we know there are fewer 5s in 123!(125) than 2s, we can find the number of 5s and thus be able to determine the number of 5-and-2 pairs.

To determine the number of 5s within 123!, we can use the following shortcut in which we divide 123 by 5, then divide the quotient of 123/5 by 5 and continue this process until we no longer get a nonzero quotient.

123/5 = 24

24/5 = 4 (we can ignore the remainder)

Since 4/5 does not produce a nonzero quotient, we can stop.

The final step is to add up our quotients; that sum represents the number of factors of 5 within 123!.

Thus, there are 24 +4 = 28 factors of 5 within 123!

Finally, we see that there are three factors of 5 within 125, since 125 = 5^3.

Since there are a total of 31 factors of 5 within 123!(125), there are 31 5-and-2 pairs and, thus, 31 trailing zeros.

Answer: D

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