Finding the number of powers of a prime number k, in the n!.

The formula is:
\(\frac{n}{k}+\frac{n}{k^2}+\frac{n}{k^3}\) ... till \(n>k^x\)

For example: what is the power of 2 in 25! (the highest value of m for which 2^m is a factor of 25!)
\(\frac{25}{2}+\frac{25}{4}+\frac{25}{8}+\frac{25}{16}=12+6+3+1=22\). So the highest power of 2 in 25! is 22: \(2^{22}*k=25!\), where k is the product of other multiple of 25!.

Check for more: http://gmatclub.com/forum/everything-ab ... 85592.html and http://gmatclub.com/forum/math-number-theory-88376.html

Back to the original question:
If p is the product of integers from 1 to 30, inclusive, what is the greatest integer k for which \(3^k\) is a factor of p?

A. 10
B. 12
C. 14
D. 16
E. 18

Given \(p=30!\).

Now, we should check the highest power of 3 in 30!: \(\frac{30}{3}+\frac{30}{3^2}+\frac{30}{3^3}=10+3+1=14\). So the highest power of 3 in 30! is 14.

Answer: C.

Hope it's clear.
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For more on this check below:
Everything about Factorials on the GMAT
Power of a Number in a Factorial Problems
Trailing Zeros Problems

For other subjects check Ultimate GMAT Quantitative Megathread

Hope it helps.
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Simplest and the easy way
Plz find the attached picture

15. No need to actually find the product of 1 to 30. Just look at the numbers that have factors of 3.

3 = 3
6 = 3 * 2
9 = 3 * 3
12 = 3 * 2 * 2
15 = 3 * 5
18 = 3 * 3 * 2
21 = 3 * 7
24 = 3 * 2 * 2 * 2
27 = 3 * 3 * 3
30 = 3 * 2 * 5

There are 14 3's, so the largest factor of p for 3k is 3*14.

Check out this post for an explanation of the method you can use to solve such questions:
http://www.veritasprep.com/blog/2011/06 ... actorials/
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If p is the product of integers from 1 to 30, inclusive, what is the greatest integer k for which 3^k is a factor of p?

a) 10
b) 12
c) 14
d) 16
e) 18

Values which we are looking for are 3,6,9,12,.. all multiples till 30
now everything will give you atleast 1 power of 3 but there are values which will give more than 1 power of 3 and those values will be multiples of 9
9 -> will give 2 powers
18-> will give 2 powers
27 -> will give 3 powers
NUmber of single powers of 3 = (30-3)/3 +1 - 1(for 9) - 1(for 18) -1(for 27) = 7
so total powers = 2(for 9) + 2(for 18) + 3(for 27) + 7 = 14

Hope it helps!
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Hi Bunuel,

This explanation makes complete sense but i'm curious as to why another method doesn't work.

I first found the number of integers (30-1+1) = 30
Average of 30+1/2 = 15.5

30 * 15.5 = 465

Factoring 465 only gave me one "3" and therefore I knew it was wrong but I had wasted valuable time. I realized that what I had done doesn't account for the product but rather the sumof the consecutive integers. Correct?

If the question was changed just a bit to say that "p" is the sum of the integers from 1 - 30 then my method would've worked and the answer would've been 1. Is that correct?

Yes, you are correct. If it were "p is the SUM of integers from 1 to 30, inclusive", then the answer would be 1.
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Just wrote down the multiples of 3 & the count of 3's in them

3 >> 1

6 >> 1

9 >> 2

12 > 1

15 > 1

18 > 2

21 > 1

24 > 1

27 > 3

30 > 1

Total = 14

Answer = C

You need to find out the number of 3's in 30!

There is a direct formula for this:

\([30/3] + [30/3^2] + [30/3^3]\) = 10 + 3 + 1 = 14
Option C

Here [x] = integral part of the number

Solution:

We are given that p is the product of the integers from 1 to 30 inclusive, which is the same as 30!.

We know that 30! must be divisible by all the numbers from 30 to 1, inclusive. In this question, however, we only care about the numbers in the factorial that are divisible by 3. All of the other numbers are irrelevant. For example, 20 is not a multiple of 3, and because 3 does not evenly divide into 20, there’s no need to consider it. Only the multiples of 3 matter for this problem and they are: 30, 27, 24, 21, 18, 15, 12, 9, 6, and 3.

What we’re really being asked is how many prime factors of 3 are contained in the product of 30, 27, 24, 21, 18, 15, 12, 9, 6, and 3.

However, instead of actually calculating the product and then doing the prime factorization, we can simply count the number of prime factors of 3 in each number from our set of multiples of 3: 30, 27, 24, 21, 18, 15, 12, 9, 6, and 3.

30 has 1 prime factor of 3.

27 has 3 prime factors of 3.

24 has 1 prime factor of 3.

21 has 1 prime factor of 3.

18 has 2 prime factor of 3.

15 has 1 prime factor of 3.

12 has 1 prime factor of 3.

9 has 2 prime factors of 3.

6 has 1 prime factor of 3.

3 has 1 prime factor of 3.

When we sum the total number of prime factors of 3 in our list we get 14. It follows that 14 is the greatest value K for which 3^k divides evenly into 30!, and thus k = 14.

Answer: C

Alternate Solution

There’s actually a shortcut that can be used. The question is what is the largest possible value of k such that 30!/3^k is an integer. To use the shortcut, divide the divisor 3 into 30, which is simply the numerator without the factorial. This division is 30/3 = 10. We then divide 3 into the resulting quotient of 10, and ignore any remainder. This division is 10/3 = 3. We again divide 3 into the resulting quotient of 3. This division is 3/3 = 1. Since our quotient of 1 is smaller than the divisor of 3, we can stop here. The final step is to add together all the quotients: 10 + 3 + 1 = 14.

Thus, the largest value of k is 14. One caveat to this shortcut is that it only works when the denominator is a prime number. For example, had the denominator been 9, this rule could not have been used.
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Attached is a visual that should help.

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This question is really asking us to determine the number of 3's "hiding" in the prime factorization of p.

p = (1)(2)(3)(4)(5)(6)(7)(8)(9) . . . (27)(28)(29)(30)
= (1)(2)(3)(4)(5)(2)(3)(7)(8)(3)(3)(10)(11)(3)(4)(13)(14)(3)(5)(16)(17)(3)(3)(2)(19)(20)(3)(7)(22)(23)(3)(8)(25)(26). . . (3)(3)(3)(28)(29)(3)(10)
= (3)^14(other non-3 stuff)

Answer:

RELATED VIDEO

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Hi All,

These types of questions are based on a math concept called "prime factorization", which basically means that any integer greater than 1 is either prime OR the product of a bunch of primes.

Here's a simple example:

24 = (2)(2)(2)(3)

Now, when it comes to this question, we're asked to multiply all the integers from 1 to 30, inclusive and find the greatest integer K for which 3^K is a factor of this really big number.

Here's a simple example with a smaller product:
1 to 6, inclusive…
(1)(2)(3)(4)(5)(6)

Then numbers 1, 2, 4 and 5 do NOT have any 3's in them, so we can essentially ignore them:
3 = one 3
6 = (2)(3) = one 3
Total = two 3's

So 3^2 is the biggest "power of 3" that goes into the product of 1 to 6, inclusive.

Using that same idea, we need to find all of the 3's in the product of 1 to 30, inclusive. Here though, you have to be careful, since there are probably MORE 3's than immediately realize:

3 = one 3
6 = one 3
9 = (3)(3) = two 3s
12 = one 3
15 = one 3
18 = (2)(3)(3) = two 3s
21 = one 3
24 = one 3
27 = (3)(3)(3) = three 3s
30 = one 3

Total = 14 3's

Final Answer:

GMAT assassins aren't born, they're made,
Rich
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Could you further explain how you arrived at "10+3+1=14" from the equation which you created before it?

You should take only the quotient of the division, that is 30/9 = 3 and 30/27 = 1.
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Bunuel chetan2u niks18 VeritasPrepKarishma JeffTargetTestPrep



Can experts please advise what I am missing in my below approach:
I need to find a factor of 3^k such that when 30! is divided by 3^k the remainder is zero.
Since k is a positive integer we need to see max value of 3^k ; now 3^k can take
3,9, 27.. I was stuck beyond this point to co-relate 30! and max power of 3^k such that remainder is zero.



Are not we supposed to find no in factorial divisible by 3^k and not 3 ?
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Hi..
You have to find how many 3s are there in 30!..
30/3=10 gives us the number which have 3 in it.. 3,6,9...,30
30/9=3.xy~3 gives us the number which are div by 9 as there is extra 3 in these numbers... 9,18,27
30/27=1.. numbers containing 3*3s..27
So our Ans 10+3+1=14

Check out this post for a detailed discussion on this concept:
https://www.veritasprep.com/blog/2011/0 ... actorials/
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