If x = 1^1*2^2*3^3*4^4*...*98^98*99^99*100^100, how many trailing zero

Question

If  x = 1^1*2^2*3^3*4^4*...*98^{98}*99^{99}*100^{100} , how many trailing zeros does x have ? (Trailing zeros are a sequence of 0's of a number, after which no other digits follow.)

(A) 1300
(B) 1200
(C) 1150
(D) 1050
(E) 1000

Are You Up For the Challenge: 700 Level Questions

Find the number of zeros in the product:
1^1*2*2*3^3*4^4............98^98*99^99*100^100
A)1200
B)1300
C)1050
D)None of these

CrackverbalGMAT · Accepted Answer

A zero is the result of the product of a 5 and a 0.

In the question above, all the 5's come from the multiples of 5, like 5, 10, 15, 20 and so on . The 2's come from even numbers. We need to count only the fives as there are more 2's than 5's.

5^5  contributes 5 zeroes

10^{10}  contributes 10 zeroes

15^{15}  contributes 15 zeroes

20^{20}  contributes 20 zeroes

25^{25} = (5^2)^{25} = 5^{50}  contributes 50 zeroes

So all the numbers where we have squares of 5 will give us the double number of zeroes. These numbers are 25, 50, 75 and 100...All other multiples will give us the same number of zeroes as the number itself.

So we have 5 + 10 + 15 + 20 + 50 + 30 + 35 + 40 + 45 + 100 + 55 + 60 + 65 + 70 + 150 + 80 + 85 + 90 + 95 + 200 = 1300

Option A

Arun Kumar

samarpan.g28 · Answer

Is there any other shortcut method to get the answer? I counted the power of 5s for every multiple of 5 present in x and added them but it took me some time to get the answer. Thank you in advance.

Oppenheimer1945 · Answer

Count powers of 5 and double count powers of 25

5(1+2…20)+25+50+75+100=1050+250=1300

Posted from my mobile device

pappal · Answer

trailing zeroes result from the product of 2 & 5
here no. of 5's availability will be limited in relation to the no. of 2's
so we have to find the highest power of 5 in the given value.
Each of the values 5^5,10^10,15^15........80^80,85^85,90^90,95^95 will have power of 5's =5,10,15,........80,85,90,95-------i
except for the values 25^25,50^50,75^75 & 100^100 which will have double power of these no.'s
=50,100,150,200------ii
for eg--75^75=(25*3)^75 =(5*5*3)^75 ,so power of 5 is 75*2=150
adding up powers of 5 will result in 1300 
A

If x = 1^12^23^34^4...98^9899^99*100^100, how many trailing zero

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