m is a positive integer. m! is ending with k zeros, (m+2)! is ending

Question

m is a positive integer. m! is ending with k zeros, (m+2)! is ending with k+2 zeros. Find the number of possible values of m, if 90 ≤ m ≤ 190.

(A) 2
(B) 3
(C) 4
(D) 6
(E) 8

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CEO2021 · Answer

chetan2u KarishmaB maam i Am getting m= 98 , 123, 148 & 173 (4 nos.) What are the two other values Please help!!!!

KarishmaB · Answer

If m! has k zeroes, (m+2)! will have k+2 zeroes only if (m+1) or (m+2) is a multiple of 25. This will bring in two more 5s and hence we will be able to make 2 more 10s. 
Also, note that (m+1) or (m+2) should not be a multiple of 125 because that will increase the number of 0s by 3.

Between 90 and 190, we have 3 multiples of 25 (but not of 125). They are 100, 150 and 175. So either (m+1) or (m+2) could take any of these three values. 
So m could be 98 or 99 or 148 or 149 or 173 or 174. 
Total 6 values possible.

Answer (D)

rdrdrd1201 · Answer

Hi  KarishmaB , Would you mind explaining how can we gettot he conclusion that If m! has k zeroes, (m+2)!  will have k+2 zeroes only if (m+1) or (m+2) is a multiple of 25?

Sanyami19 · Answer

Hi Karishma mam,

If m! has k zeroes, (m+2)! will have k+2 zeroes only if (m+1) or (m+2) is a multiple of 25. This will bring in two more 5s and hence we will be able to make 2 more 10s.

To make 2 more 10s we need two 2's and two 5's. As per your explanation, as 150 is multiple of 25, but it gives only one of the 2. And since it is even, the consecutive no. will be odd (+1/-1). So if 150 is either value of (m+1) or (m+2), it will be a pair of odd even number which will add only one 2 and two 5s. So 150 is not the option.

Similarly, 175 must be (m+1) and 176 must be (m+2) since it is divisible by 4 so it will add two 2s. If we consider 175 as (m+2), then (m+1) will be 174 which will add only 1 of the 2s. Hence, 175 gives only 1 value of m.

As for 100 - 100 itself adds 2 pairs of (2*5) so it can give 2 values of M.

So final answer would be 3. Do you agree?

KarishmaB · Answer

Starting from 1, a factor of 5 appears in every fifth term. 1, 2, 3, 4,  5 , 6, 7, 8, 9,  10 , 11, ...

Consider this: 
1! = 1
2! = 2
3! = 6
4! = 24
5! = 120
6! = 720
7! = 5040
8! = 40320
9! = 362880
10! = 3628800
...

To make a 10 (i.e. a 0 at the end), you need a 2 and a 5. Till 4!, we have no 5s so we can make no 10s. 
The moment we get 5! = 5*4*3*2*1, we get a 5 in the mix. Now 5*2 will combine to give us a 0.

Now the next 4 numbers 6, 7, 8 and 9 when multiplied by the previous terms will not give any extra 0s because they don't have 5 as a factor (we have more than enough 2s). But when we bring in 10 in 10!, we get another 5 and hence another 0. 
So 10! ends with two 0s. Then 11!, 12!, 13! and 14! will end with two 0s only because we don't have any new 5s. The moment, we come to 15!, there is another 5 in the mix and now we will have three 0s at the end. From 16! to 19! we will continue having three 0s at the end.

Then 20! will have four 0s at the end and so will 21! to 24!

What happens at 25!? 25 has two 5s so it will bring in two new 0s and 25! will end with six 0s.

This is the logic.

Now, if m! has k zeroes and (m+2)! has k+2 zeroes, it means a multiple of 25 appeared either at (m+1) or at (m+2) which made the two extra 0s. 
It is like saying that 23! has four 0s while 25! has six 0s (Here, m = 23 and m+2 is a multiple of 25)
OR 24! has four 0s while 26! has six 0s (Here m = 24 and m+1 is a multiple of 25)
So, considering a multiple of 25, m can take 2 previous values (here they are 24 and 23).

In our question, there are 3 multiples of 25 (that will add two 0s i.e. 100, 150 and 175). 
125 will add three 0s so we should ignore it.

So m can take 2 previous values: 98, 99, 148, 149, 173, 174
In all, 6 values.

KarishmaB · Answer

When we are considering 150!, we already have plenty of 2s to go around. The new numbers needn't bring in the 2s. Every second number (2, 4, 6, 8, 10, 12...) already brought in a 2. We are far far short of 5s. e.g. in 5! = 1*2*3*4*5, we already have three 2s but we have only one 5 so I can make only one 10. If you give me another two 5s, I can easily make another two 10s. 25 * 5! will have three 0s at the end. 
Hope this makes sense now.

Sanyami19 · Answer

But why are we considering 150! in the first place. Since we have simplified the Q. to the form (m+2)*(m+1)*m! such that the product of (m+2) * (m+1) must give us 2 trailing zeros . So 150 = 25*2*3 and now 150 gives only 1 two so only 1 trailing zero. Can you pls reconsider your response and let me where my approach is incorrect.

ThatDudeKnows · Answer

Adding a zero at the end means adding another multiple of 10. 
In order to go from k zeros to k+1 zeros, we'd need to add one multiple of 10, which can be achieved by adding one multiple of 2 and one multiple of 5. 
In order to go from k zeros to k+2 zeros, we need to add TWO multiples of 10, so we need TWO 2s and TWO 5s. 
We have plenty of 2s, so the 5s are the limiter. Let's think about how we can add two 5s (but not more than two 5s or we will be adding three zeros and we don't want that).

As we slide m up the number line, there will never be a time when m and m+2 are separated by two different numbers each of which is a multiple of 5, so that means the only way to get exactly TWO 5s between m and m+2 is to get them both at the same time. That only happens at multiples of 25 that aren't also multiples of 125 (multiples of 125 will have three 5s, so they will add three zeros). What are the multiples of 25 between 90 and 190 that aren't also multiples of 125? 100, 150, 175.

We want to count the ways that m is less than those cutoffs and m+2 is equal to or greater than those cutoffs.
That occurs when m = 98, 99, 148, 149, 173, and 174.
Six ways.

Answer choice D.

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m is a positive integer. m! is ending with k zeros, (m+2)! is ending

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