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If p is the product of integers from 1 to 30, inclusive

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If p is the product of integers from 1 to 30, inclusive [#permalink] New post 23 Aug 2012, 10:18
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If p is the product of the 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
[Reveal] Spoiler: OA

Last edited by Bunuel on 12 Dec 2012, 04:36, edited 2 times in total.
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Re: 3^k is a factor of p [#permalink] New post 23 Aug 2012, 19:27
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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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Re: 3^k is a factor of p [#permalink] New post 23 Aug 2012, 21:07
Hi nkdotgupta,

Thanks for your reply but this is the explaation as provided in the OG. I wanted to know if there are other ways of approaching the problem since this method might be cumbersome we encounter larger numbers.

nktdotgupta wrote:
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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If p is the product of integers from 1 to 30, inclusive [#permalink] New post 24 Aug 2012, 00:55
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ajju2688 wrote:
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

[Reveal] Spoiler:
c


This is an OG12 question. Can someone tell me if there is a quick way to solve these kinds of questions since the OG explanation seems to be very time consuming?


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: everything-about-factorials-on-the-gmat-85592.html and 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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Re: 3^k is a factor of p [#permalink] New post 24 Aug 2012, 01:26
Bunuel wrote:
ajju2688 wrote:
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

[Reveal] Spoiler:
c


This is an OG12 question. Can someone tell me if there is a quick way to solve these kinds of questions since the OG explanation seems to be very time consuming?


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: everything-about-factorials-on-the-gmat-85592.html and 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 1.

Answer: C.

Hope it's clear.


Thanks Bunuel for the crystal clear explanation! :)
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Re: If p is the product of integers from 1 to 30, inclusive [#permalink] New post 22 Oct 2013, 12:20
Correct me if I am mistaken but understanding number properties seems to be in line with these type of problems.

Going back to basics - these type of problems seem to relate to the factor foundation rule - if a is a factor of b and b is a factor of c then a is a factor of c.

In these problems you are trying to find the greatest degree of a common prime- in this case- of 3 in 30!.

Is this correct?
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Re: 3^k is a factor of p [#permalink] New post 25 May 2014, 08:54
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BOOKMARKED
Bunuel wrote:
ajju2688 wrote:
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

[Reveal] Spoiler:
c


This is an OG12 question. Can someone tell me if there is a quick way to solve these kinds of questions since the OG explanation seems to be very time consuming?


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: everything-about-factorials-on-the-gmat-85592.html and 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 1.

Answer: C.

Hope it's clear.


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?
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Re: 3^k is a factor of p [#permalink] New post 25 May 2014, 10:00
Expert's post
1
This post was
BOOKMARKED
russ9 wrote:
Bunuel wrote:
ajju2688 wrote:
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

[Reveal] Spoiler:
c


This is an OG12 question. Can someone tell me if there is a quick way to solve these kinds of questions since the OG explanation seems to be very time consuming?


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: everything-about-factorials-on-the-gmat-85592.html and 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 1.

Answer: C.

Hope it's clear.


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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Re: If p is the product of integers from 1 to 30, inclusive [#permalink] New post 24 Jun 2014, 02:16
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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
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Re: If p is the product of integers from 1 to 30, inclusive [#permalink] New post 11 Sep 2014, 11:17
ajju2688 wrote:
If p is the product of the 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



product of the integers from 1 to 30 = 30!

to find the greatest integer K i.e 3^k

30/3 +30/9 +30/27 = 10+3+1 =14
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Re: If p is the product of integers from 1 to 30, inclusive [#permalink] New post 08 Mar 2015, 19:50
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Bunuel wrote:
ajju2688 wrote:
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

[Reveal] Spoiler:
c


This is an OG12 question. Can someone tell me if there is a quick way to solve these kinds of questions since the OG explanation seems to be very time consuming?


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: everything-about-factorials-on-the-gmat-85592.html and 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 1.

Answer: C.

Hope it's clear.


Br


Bunuel..."So the highest power of 3 in 30! is 1."...You mean 14 right?
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Re: If p is the product of integers from 1 to 30, inclusive [#permalink] New post 08 Mar 2015, 19:53
Expert's post
Avinashs87 wrote:
Bunuel wrote:
ajju2688 wrote:
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

[Reveal] Spoiler:
c


This is an OG12 question. Can someone tell me if there is a quick way to solve these kinds of questions since the OG explanation seems to be very time consuming?


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: everything-about-factorials-on-the-gmat-85592.html and 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 1.

Answer: C.

Hope it's clear.


Br


Bunuel..."So the highest power of 3 in 30! is 1."...You mean 14 right?


Yes, it's a typo. Edited. Thank you.
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COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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Re: If p is the product of integers from 1 to 30, inclusive [#permalink] New post 09 Mar 2015, 00:29
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ajju2688 wrote:
If p is the product of the 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



Simplest and the easy way
Plz find the attached picture

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Re: If p is the product of integers from 1 to 30, inclusive   [#permalink] 09 Mar 2015, 00:29
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