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15 Sep 2014, 12:00
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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
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
15 Sep 2014, 12:36
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jrymbei
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 \(5^1\)
However there are some integers which are also divisible by \(5^2\) e.g. 25, 50, 75, and 100
You will also see that none of the integers is divisible by \(5^3\) because 5^3 is greater than 100
So to get all the fives out of 100! we will first divide it by \(5^1\) and then by \(5^2\)
100/\(5^1\) = 20
100/\(5^2\) = 4
So total count of fives is 100 + 20 + 4 = 124, which is also the count of number of zeros in the multiplication.
Solve remaining two examples using this principle.
15 Sep 2014, 20:58
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jrymbei
These questions are just a take on "the maximum power of m in n!" type of questions. Check out this post for details on the concept: https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2011/06 ... actorials/
First let's do part (C)
(c) 100!
To find the number of 0s in 100!, we need the number of 5s in 100!.
100/5 = 20
20/5 = 4
Total = 20+4 = 24
So 100! will have 24 zeroes.
Now part (a) is simple.
(a) Of first 100 multiples of 10
\(10*20*30* ... *1000\)
\(= 10^{100}*(1*2*3*... *100)\)
\(= 10^{100}*100!\)
Now we know that \(10^{100}\) will have 100 zeroes. We need to find the zeroes at the of 100!. As found above, 100! has 24 zeroes.
So total we have 100+24 = 124 zeroes.
(b) 5X10X15X20X25X30X35X40X45X40 (first 9 positive multiples of 5 and an extra 40)
\(5^9*(1*2*3*...*9) * 40\)
\(5^{10}*(1*2*3*...*9) * 8\)
\(5^{10}*(9!) * 8\)
There is only one 5 in 1 to 9 so total we have eleven 5s. Note that we have a lot of extra 5s in this case. We are not sure whether we have enough 2s. So let's find out the number of 2s too. First, let's find the number of 2s in 9!
9/2 = 4
4/2 = 2
2/2 = 1
Total number of 2s in 9! = 4+2+1 = 7
We get another three 2s from 8 so total number of 2s = 7+3 = 10
So even though we have eleven 5s, we have only ten 2s so we can make only ten 10s.
