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by pemdas » Mon Dec 26, 2011 2:13 am
first find N as the product of all multiples of 3 between 1 and 100
3(1*2*3...33) or N=3*33!

then find the greatest integer m for which (N/10^m) is an integer
3*33!/10^m
look above plz, to the parentheses (1*2*3...33) how many tens we have here? And how many 2*5 we have here? we have as following 0s
2*5*(2*5)*(3*5)*(2*8)*(2*10)(5*5)(2*13)(2*14)*(3*10)
2*5*10*15*16*20*(25)*26*28*30

in total 7 0s i guess

c
karthikpandian19 wrote:If N is the product of all multiples of 3 between 1 and 100, what is the greatest integer m for which (N/10^m) is an integer?

3
6
7
8
10
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by gurman.lidder » Mon Dec 26, 2011 2:28 am
Well i think it is 7. How did i reach here ??

well to get a multiple of 10 you need either X * 10 .. or 2 * 5 where X is any no.

So Coming back we need to find multiples of 3 that have the unit digit as 2 or 5 ( case 1) or multiples of 10( case 2)

Case 1:


12 * 15 = 180
42 * 45 = 1890
72 * 75 = 5400


4 Zeros here

Case 2 :
30 , 60 ,90

Hence we will have 7 Zeros hence we can say it is divided by 10^7


What is the OA ??
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by arpan20690 » Mon Dec 26, 2011 2:59 am
this question is very easy for those who have good grip on number system.
N = 3^33 * 33! (try to understand it)
Now we have to find no of 10s in 33!
No of 10 in 33! will be equal to no of 5 in the same.

5 ) 33 ( 6 is quotient 5 ) 6 ( 1 is quotient
30
......
3 so, no of five is equal to (6 + 1) = 7.
Hence No of 10 is equal to also 7. so the answer is 7.
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by karthikpandian19 » Mon Dec 26, 2011 4:04 am
OA is 7
gurman.lidder wrote:Well i think it is 7. How did i reach here ??

well to get a multiple of 10 you need either X * 10 .. or 2 * 5 where X is any no.

So Coming back we need to find multiples of 3 that have the unit digit as 2 or 5 ( case 1) or multiples of 10( case 2)

Case 1:


12 * 15 = 180
42 * 45 = 1890
72 * 75 = 5400


4 Zeros here

Case 2 :
30 , 60 ,90

Hence we will have 7 Zeros hence we can say it is divided by 10^7


What is the OA ??
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by pemdas » Mon Dec 26, 2011 6:36 am
@karthikpandian19, in my solution please read 3^33*33! (forgot while copying from my scratch rec.) 3^33 doesn't contain factors 2 and 5
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by ronnie1985 » Mon Dec 26, 2011 8:49 pm
N = 3*6*9*12*...*99
We have to find the max power of 10 or max power of 2 and 5. We shsll seek to find the power of 5 as 5?2
3&5 LCM =15 and 3and 25 LCM = 75
No of 5's = No of 15 + No of 25 = {99/15}+{99/75} = 7
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