Problem Solving

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?

A. 3
B. 6
C. 7
D. 8
E. 10

Is the Answer [spoiler]{C}[/spoiler]

Multiple of 3 from 1 to 100 means
3(1.2.3.4....33) ==> 3.33!

So, the question is basically asking for MAXIMUM power of 10 in N

33/5 = 6
6/5 = 1

6+1 = 7

N = 3 x 6 x 9 x ... 99

N when divided by 10^m should yield an integer result.

We just need to get hold of all the multiples of 10 & 5
5 --> Because 5*2 = 10 so each 5 can form a pair with an even number (plenty in the list of multiples of 3)


Special mention of 75. 75 has two 5s in it (5 * 5 * 2) so it can cancel out 1 additional 10 when grouped with a even number. Thus we can form 7 10s out of the denominator 6 + 1 ( additional 5 from 75)
Answer : [spoiler]m = 7[/spoiler]

Regards,
Vivek

rakeshd347 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?

A. 3
B. 6
C. 7
D. 8
E. 10

We have to find the multiples of 3 which are also multiple of 2 or 5 or both:
3*(2,4,5,6,8,10,12,14,15,16,18,20,22,24,25,26,28,30,32)
Power of 5 = 5^(1 + 1 + 1 + 1 + 2 + 1) = 5^7
So we can have a maximum of 7th power of 10.

Choose C

Cheers

@Rahul
Can you pls elaborate on the solution that you provided. This is not clear to me

ani781 wrote:@Rahul
Can you pls elaborate on the solution that you provided. This is not clear to me

According to question, "N is the product of all multiples of 3 between 1 and 100,"

STEP 1:
we can re-write this as

3x6x9x12x15x18x....99

So, 3 (1x2x3x4x5.....33) --> 3.33!

STEP 2:
We need greatest integer "m" for N/10^m

that is, we can we-write this question as, we need the find the maximum power of "10" in "N" (3.33!)

IMP: There's a very simple and helpful logic, whenever you need to find the power of any number in any factorial.. divide the number(removing factorial) until you cannot divide further; For power of 10, divide by "5"

SO,
33/5 = 6 [we are using "33" and not "33!"]
6/5 = 1 [re-divide]
1/5 = 0 [no further division possible; terminate]

So, Maximum power = 6+1 = 7

I hope the solution is now much more understandable

Recently, many similar questions were posted, you may refer and try to apply this method on those question as well..

rakeshd347 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?

A. 3
B. 6
C. 7
D. 8
E. 10

N = 3*6*9*...*93*96*99 = (3³³)(1*2*3*...*31*32*33) = 3³³ * 33!.
10^m = the number of 10's that can divide into N.

We need to count how many 10's can divide into 33!.
Since 10=2*5, EVERY COMBINATION OF 2*5 contained within the prime-factorization of 33! will provide a 10 than can divide into N.
The prime-factorization of 33! will be composed of FAR MORE 2'S than 5's.
Thus, to determine HOW MANY 10'S can divide into 33!, we must count THE NUMBER OF 5'S contained within the prime-factorization of 33!.

To determine the number of 5's, count how many times EACH POWER OF 5 can divide into 33.

Every multiple of 5 that can divide into 33 provides at least one 5:
33/5¹ = 6.

Every multiple of 5² that can divide into 33 adds a second 5:
33/5² = 33/25 = 1.

Thus, the greatest number of 10's than can divide into N = 6+1 = 7.

The correct answer is C.

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Thanks Rich, you explanation is crystal clear.
I dont know why , but maybe because my medium of study was non-english, I could not at all understand the problem... untill I saw your solution.

Thanks !