Find the number of zero at the end of the product of first 250 natural numbers
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GMAT Club Forum Index Problem Solving (PS)
Find the number of zeros at the end of the product : Problem Solving (PS)
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Out of Scope - Too Hard
Topic Discussion
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jrymbei Sep 15, 2014
00:00 ABCDE
DIFFICULTY: (N/A) QUESTION STATS: based on 7 sessions
100% (00:09) correct
0% (00:00) wrong
Find the number of zeros at the end of the product
A. Of first 100 multiples of 10
B. 5*10*15*20*25*30*35*40*40*45
C. 100!
I would also like to know whether this kind of question is possible in GMAT and what would be the complexity level.
1. 124
2. 10
3. 24
Kudos
Narenn
EXPERT'S
POST
Sep 15, 2014
jrymbei wrote:
Q : Find the number of zeros at the end of the product
a. Of first 100 multiples of 10
b. 5X10X15X20X25X30X35X40X40X45
C. 100!
I would also like to know whether this kind of question is possible in GMAT and what would be the complexity level.
OA-
1. 124
2. 10
3. 24
Not as difficult as it appears.
Theory :- To obtain a zero you need to multiply 2 by 5. Each pair of 2 and 5 will give you one zero. so we just have to look how many pairs of 2 and 5 exist in the multiplication.
A) First 100 multiples of 10. (Note that we need one 2 and one 5 to get one 0. In this multiplication you will get many more 2s than 5s, so looking for 2s and then fpr 5s will be waste of time. Just look for 5s and you are done.)
Multiplication of First 100 multiples of 10 can be written as (10*1)(10*2)(10*3)...................(10*100)
{10 * 100times} * {1*2*3*.................99*100}
Factors of 10 are 2 and 5. That means each ten will give you one 5. So {10 * 100times} will give you hundred 5s
Now consider {1*2*3*.................99*100}. This is actually 100!
Count all the multiples of 5 in here ----> 5, 10, 15, 20, 25, ......., 100
If you observe these integers closely, you will see each of those integers is divisible by 5151
However there are some integers which are also divisible by 5252 e.g. 25, 50, 75, and 100
You will also see that none of the integers is divisible by 5353 because 5^3 is greater than 100
So to get all the fives out of 100! we will first divide it by 5151 and then by 5252
100/5151 = 20
100/5252 = 4
So total count of fives is 100 + 20 + 4 = 124, which is also the count of number of zeros in the multiplication.
tep-by-step explanation: