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Maximum value of m for the expression to yield integer value

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Maximum value of m for the expression to yield integer value

by rohitacmilan » Tue Oct 15, 2013 11:32 pm
N is the product of multiples of 3 lying between 1 and 100. What is the maximum value of m for which the following expression N/(10^m) is an integer?

I really don't remember the answer choices. Apologies for that.

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by theCodeToGMAT » Wed Oct 16, 2013 12:02 am
N = 3 x 6 x 9 .... 99
= 3 ( 1x2x3...33) ==> 3.33!


Basically question is asking for the power of 10 in 33! .. so .

33/5 = 6
6/5 = 1

Answer {7}

Is m = 7?
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by rohitacmilan » Wed Oct 16, 2013 12:20 am
I am not sure about the correct answer - you are right.
Now I recollect this approach that from one of Brent's post.

Thanks a ton :)
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by theCodeToGMAT » Wed Oct 16, 2013 12:22 am
rohitacmilan wrote:I am not sure about the correct answer - you are right.
Now I recollect this approach that from one of Brent's post.

Thanks a ton :)
you are welcome!
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by Premalatha Anand » Wed Oct 16, 2013 2:21 pm
theCodeToGMAT wrote:N = 3 x 6 x 9 .... 99
= 3 ( 1x2x3...33) ==> 3.33!


Basically question is asking for the power of 10 in 33! .. so .

33/5 = 6
6/5 = 1

Answer {7}

Is m = 7?
Didn't get it from 33/5 step.
Can you explain pls!
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by theCodeToGMAT » Wed Oct 16, 2013 6:09 pm
Premalatha Anand wrote: Didn't get it from 33/5 step.
Can you explain pls!
To find power of the smallest factor in a number divide the number (without factorial) with that number.

In this question we want to find the power of 10 in 33!... 10 = 2 x 5 .. so by finding power of "5" we can conclude..

To find power of "5" in 33!.. keep on dividing 33 untill further division is not possible.


33/5 = 6 [we took 33 and not 33!]
6/5 = 1
1/5 = 0 [division not possible]

So, 6+1 = 7
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by [email protected] » Wed Oct 16, 2013 7:15 pm
Hi rohitacmilan,

This questions asks us for the MAXIMUM M for which N/(10^M) is an integer, so we're looking for the "powers of 10" that can be found by multiplying all of the multiples of 3 from 3 to 99: 3 x 6 x 9 x ......96 x 99. Sometimes a bit of "brute force" math is the easiest way to answer a Quant question.

We need to find all the multiples of 10 in that multiplication string.

The immediate multiples of 10: 30, 60, 90 that's three 0s

When you multiply a number that ends in 5 by an even number, you get a number that ends in a 0, so we have to factor those in as well:
15, 45 and 75
15 x 2 = 30
45 x 2 = 90
75 x 2 x 2 = 300
Here we have another four 0s.

Total = 3 + 4 = seven 0s.

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