If N is the product of all multiples of 3 between 1 and 100, what is

Question

If N is the product of all multiples of 3 between 1 and 100, what is the greatest integer m for which  \frac{N}{10^m}  is an integer?

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

Bunuel · Accepted Answer

If N is the product of all multiples of 3 between 1 and 100, what is the greatest integer m for which \(\frac{N}{10^m}\) is an integer?

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

We should determine # of trailing zeros of N=3*6*9*12*15*...*99 (a sequence of 0's of a number, after which no other digits follow).

Since there are at least as many factors 2 in N as factors of 5, then we should count the number of factors of 5 in N and this will be equivalent to the number of factors 10, each of which gives one more trailing zero.

Factors of 5 in N:
once in 15;
once in 30;
once in 45;
once in 60;
twice in 75 (5*5*3);
once in 90;

1+1+1+1+2+1=7 --> N has 7 trailing zeros, so greatest integer \(m\) for which \(\frac{N}{10^m}\) is an integer is 7.

Answer: C.

Check this for more:
https://gmatclub.com/forum/everything-ab ... 85592.html

Hope it helps.

abmyers · Answer

N = The product of the sequence of 3*6*9*12....*99

N therefore is also equal to 3* (1*2*3*.....*33)

Therefore N = 3* 33!

From here we want to find the exponent number of prime factors, specifically the factors of 10.

10 = 5*2 so we want to find which factors is the restrictive factor

We can ignore the 3, since a factor that is not divisible by 5 or 2 is still not divisible if that number is multiplied by 3.

Therefore:

33/ 2 + 33/4 + 33/8 = 16+8+4 = 28

33/ 5 + 33/25 = 6 + 1 = 7

5 is the restrictive factor.

Here is a similar problem: number-properties-from-gmatprep-84770.html

KarishmaB · Answer

Responding to a pm:

Once you are done, note that this question can be easily broken down into the factorial form.

3*6*9*...*99 = 3^{33} * (1*2*3*4*...*32*33) = 3^{33} * 33!

We need to find the number of 5s in 33! because you need a 2 and a 5 to make a 10. The number of 5s will certainly be fewer than the number of 2s.

33/5 = 6
6/5 = 1

So you will have a total of 6+1 = 7 5s and hence can make 7 10s.
So maximum power of 10 must be 7.

Answer C

Note that we ignore  3^{33}  because it has no 5s in it.

nonameee · Answer

Bunuel, is it necessary to count the number of trailing zeros? I have solved the problem by counting the number of 5's in N.

Bunuel · Answer

It's basically the same. Since there are at least as many factors 2 as factors of 5 in N, then finding the number of factors of 5 in N would be equivalent to the number of factors 10, each of which gives one more trailing zero.

Bunuel · Answer

abmyers · Answer

N = The product of the sequence of 3*6*9*12....*99

N therefore is also equal to 3* (1*2*3*.....*33)

Therefore N = 3* 33!

From here we want to find the exponent number of prime factors, specifically the factors of 10.

10 = 5*2 so we want to find which factors is the restrictive factor

We can ignore the 3, since a factor that is not divisible by 5 or 2 is still not divisible if that number is multiplied by 3.

Therefore:

33/ 2 + 33/4 + 33/8 = 16+8+4 = 28

33/ 5 + 33/25 = 6 + 1 = 7

5 is the restrictive factor.

Here is a similar problem: number-properties-from-gmatprep-84770.html

Amateur · Answer

How did you know that 2 factors and 5 factors in N are same?

Bunuel · Answer

No, that's not what I'm saying (see the red part). The power of 2 in N is  at least  as high as the power of 5 in N.

We are told that N=3*6*9*12*15*18*21*...*90*93*96*99 --> as you can observe, the power of 2 in N will be higher than the power of 5 (there are more even numbers than multiples of 5).

Hope it's clear.

Amateur · Answer

I did it in a different way..... since it is multiplication of all 3 multiples....
3*6*9*..... *99=(3^33)(1*2*3*4*5*......33)=(3^33)*33! (3 power 33 because a 3 can be extracted from each number inside)
(3^33) doesn't have any multiples between 1-9 which can contribute a 0.....
so number of trailing 0's should be number of trailing 0's of 33! which is 7.
So C is the answer... we don't need to count 5's and 2's and complicate things in this case!
Let me know if you think this approach of mine has loop holes.

lunar255 · Answer

I am not convinced by the answer of Bunuel, so I used excel to calculate the product.

The answer is 48,271,088,561,614,000,000,000,000,000,000,000,000,000,000,000,000,000, which means the maximum of m will be 39.

This is not a good question

Bunuel · Answer

1. There is nothing wrong with the question.

2. Solution is correct, answer is C.

3. Excel rounds big numbers. Actual result is 48,271,088,561,613,960,642,858,365,853,327,381,832,862,269,440,000,000.

mumbijoh · Answer

Dear Bunuel
I came across this question and i really do not understand it.I read the "Everything about factorial " link but i cant seem to apply what i have read there to this question.
How did you come up with this?Please help
"
once in 15; 
once in 30;
once in 45;
once in 60;
twice in 75 (5*5*3); 
once in 90;

Bunuel · Answer

15=5*3 30=5*6 45=5*9 60=5*12 75=5^ 2 *3 90=5*18 Similar questions to practice: if-n-is-the-greatest-positive-integer-for-which-2n-is-a-fact-144694.html what-is-the-largest-power-of-3-contained-in-103525.html if-n-is-the-product-of-all-positive-integers-less-than-103218.html if-n-is-the-product-of-integers-from-1-to-20-inclusive-106289.html if-n-is-the-product-of-all-multiples-of-3-between-1-and-101187.html if-p-is-the-product-of-integers-from-1-to-30-inclusive-137721.html what-is-the-greatest-value-of-m-such-that-4-m-is-a-factor-of-105746.html if-6-y-is-a-factor-of-10-2-what-is-the-greatest-possible-129353.html if-m-is-the-product-of-all-integers-from-1-to-40-inclusive-108971.html if-p-is-a-natural-number-and-p-ends-with-y-trailing-zeros-108251.html if-73-has-16-zeroes-at-the-end-how-many-zeroes-will-147353.html find-the-number-of-trailing-zeros-in-the-expansion-of-108249.html how-many-zeros-are-the-end-of-142479.html how-many-zeros-does-100-end-with-100599.html find-the-number-of-trailing-zeros-in-the-product-of-108248.html if-60-is-written-out-as-an-integer-with-how-many-consecuti-97597.html if-n-is-a-positive-integer-and-10-n-is-a-factor-of-m-what-153375.html if-d-is-a-positive-integer-and-f-is-the-product-of-the-first-126692.html Hope it helps.

TAL010 · Answer

Finding the powers of a prime number p, in the n!
The formula is:
Example:
What is the power of 2 in 25!?

^^ Taken from the GMAT Club book...what is the logic behind this question? What are they really asking?

Narenn · Answer

It means calculating number of instances of P in n! Consider the simple example ---> what is the power of 3 in 10! We can find four instances of three in 10! -----> 1 * 2 * 3 * 4 * 5 * (2* 3 ) * 7 * 8 * ( 3 * 3 ) * 10 You can see above we can get four 3s in the expression. Calculating the number of instances in this way could be tedious in the long expressions. but there is a simple formula to calculate the powers of a particular prime. the powers of Prime P in n! can be given by \frac{n}{p} + \frac{n}{p^2} + \frac{n}{p^3} + ................. till the denominator equal to or less than the numerator. what is the power of 3 in 10! ------> \frac{10}{3} + \frac{10}{3^2} = 3 + 1 = 4 Analyze how the process works........ We first divided 10 by 1st power of 3 i.e. by 3^1 in order to get all red 3s Later we divided 10 by 2nd power of 3 i.e. by 3^2 in order to get the leftover 3 (blue) we can continue in this way by increasing power of P as long as it does not greater than n Back to the original question.............. What is the power of 2 in 25!? ---------> 25/2 + 25/4 + 25/8 + 25/16 = 12 + 6 + 3 + 1 = 22 Hope that helps!

KarishmaB · Answer

Check out this post: https://anaprep.com/number-properties-h ... actorials/ It answers this question in detail explaining the logic behind it.

kshitij90 · Answer

We know that for a number to be divisible by 10 must have at least one zero. Let's break the 10 into its prime factors, ie. 5 and 2. Now, we need to find pairs of 2 and 5 in the numerator. Here, 5 is our limiting factor, as it appears less than 2 does. therefore two cont the number of 5s, we must count the 5s in all multiples of 3 between 1 and 100.

15= One 5
30= One 5
45= One 5
60= One 5
75 = Two 5s (5 x 5 x3=75)
90= One 5.\

Answer is C.

15= One 5
30= One 5
45= One 5
60= One 5
75 = Two 5s (5 x 5 x3=75)
90= One 5.\

Answer is C.

hamzakb · Answer

I found my answer by finding the number of multiples of 3 between 1 and 100 i.e 100/3 = 33.

Then I found the number of trailing zeroes in 33! = 7

so 10^7 can be the maximum for N/10^m to remain an integer.

Am I just lucky or can this also be a method of solving?

If N is the product of all multiples of 3 between 1 and 100, what is

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GMAT Problem Solving (PS) Questions

If N is the product of all multiples of 3 between 1 and 100, what is

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