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28 Mar 2020, 05:25
Kudos
Bookmarks
AKG1593
If n is an odd natural number and n! ends with 32 zeros, then how many values of n are possible?
A. 2
B. 3
C. 1
D. 4
E. none
A. 2
B. 3
C. 1
D. 4
E. none
Show SpoilerMy doubt & OA
I think the answer should be B(135,137,139).But the OA is A.Can anyone explain?
Asked: If n is an odd natural number and n! ends with 32 zeros, then how many values of n are possible?
Number of zeros in 130! = 26+5+ 1= 32
Number of zeros in 135! = 27+5+1 = 33
n = {131,133} since 130,132 & 134 are even.
IMO A
24 May 2020, 19:41
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kelvind13
AKG1593
If n is an odd natural number and n! ends with 32 zeros, then how many values of n are possible?
A. 2
B. 3
C. 1
D. 4
E. none
A. 2
B. 3
C. 1
D. 4
E. none
Show SpoilerMy doubt & OA
I think the answer should be B(135,137,139).But the OA is A.Can anyone explain?
I got the right answer but it took me close to 4 minutes to get to it.
by trial and error i plugged in first 100 and realized i was too low, then went to 125 and realized i was still too low, Finally went to 131 and hit jackpot.
I got the answer by using the below formula
(n/5) +(n/5^2) + (n/5^3) = 32
the final numbers are 131 and 133 (since they are odd Numbers only)
Is there a faster solution to this?
I tried using the following way
[n/5] + [n/25] + [n/125] = 32
Now the max value of number when only 5 is contributing for zeroes will be = 5*32 = 160
However, In actuality, even 25 and 125 will contribute towards trailing zeroes.
Now [160/25] = 6 and [160/125] = 1
Therefore to accomodate these contributions you have to adjust our earlier calculation of just considering 5 which gave number 160.
Therefore our answer might be close to 160 - (6+1)*5 = 125
Hence you need to check around this number to get the required number of trailing zeroes.
Hope it helps
Posted from my mobile device
14 Jul 2020, 18:52
Kudos
Bookmarks
AKG1593
If n is an odd natural number and n! ends with 32 zeros, then how many values of n are possible?
A. 2
B. 3
C. 1
D. 4
E. none
A. 2
B. 3
C. 1
D. 4
E. none
Show SpoilerMy doubt & OA
I think the answer should be B(135,137,139).But the OA is A.Can anyone explain?
Bunuel, you can confirm whether my understanding is correct.
One does not need to find actual number for which n! has 32 zeroes.
For examples between 30! to 35! - 30! has n trailing zeroes and 35! has n+1 trailing zeroes.
so between 30 to 35 - 4 digits.. likewise true for 32 zeroes number.
