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

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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.

Also check:
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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

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.

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.

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.

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!

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.

Dear Reto,
My friend, before you post anything else, please familiarize yourself with the protocols. This question has been posted many times before, for example, here:
if-n-is-the-product-of-all-multiples-of-3-between-1-and-101187.html
where there's already a long discussion. Always search for a question before you start a new thread from scratch. Presumably, Bunuel, the math genius moderator, will merge this post into one of the larger previous posts on the same topic. If you have any questions that are not already answered there, you are more than welcome to ask me.
Best of luck,
Mike

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.

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;

15=5*3
30=5*6
45=5*9
60=5*12
75=5^2*3
90=5*18

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

You don't have to count the multiples of 3. Just look at the pattern.

Multiples of 3:

3 * 6 * 9 * 12 * ... * 96 * 99

3 = 3*1
6 = 3*2
9 = 3*3
...
96 = 3*32
99 = 3*33

So in all, we have 33 multiples of 3.

(3*1) * (3*2) * (3*3) * (3*4) * ... * (3*32) * (3*33)

Now from each term, separate out the 3 and put all 3s together in the front. You have 33 terms so you will get 33 3s. Also you will be left with all second terms 1, 2, 3, 4 etc

= (3*3*3..*3) * (1 * 2 * 3 * 4 * ... * 32 * 33)

= 3^(33) * (1 * 2 * 3 * 4 * ... * 32 * 33)

But 33! = (1 * 2 * 3 * 4 * ... * 32 * 33)

So you get 3^(33) * 33!

­
The restrictive factor among 2 and 5 would be the one with the lowest power in 3*6*9*12*15*...*99. Since we need the power of 10, which is equal to 2*5, the factor with the lowest power in 3*6*9*12*15*...*99 will determine the power of 10 in 3*6*9*12*15*...*99. As 5 appears with a lower power than 2 in 3*6*9*12*15*...*99, then 5 will be the restrictive factor, also known as the limiting factor, and its power, 7, would be the power of 10.

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.

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