GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 25 Apr 2019, 17:03

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# If x is the product of the positive integers from 1 to 9, inclusive,

Author Message
TAGS:

### Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 54544
If x is the product of the positive integers from 1 to 9, inclusive,  [#permalink]

### Show Tags

25 May 2016, 04:27
1
00:00

Difficulty:

25% (medium)

Question Stats:

79% (01:26) correct 21% (01:44) wrong based on 111 sessions

### HideShow timer Statistics

If x is the product of the positive integers from 1 to 9, inclusive, and if i, k, m, and p are positive integers such that $$x = 2^i3^k5^m7^p$$, then i + k + m + p =

A. 4
B. 7
C. 8
D. 11
E. 13

_________________
Verbal Forum Moderator
Status: Greatness begins beyond your comfort zone
Joined: 08 Dec 2013
Posts: 2260
Location: India
Concentration: General Management, Strategy
Schools: Kelley '20, ISB '19
GPA: 3.2
WE: Information Technology (Consulting)
Re: If x is the product of the positive integers from 1 to 9, inclusive,  [#permalink]

### Show Tags

25 May 2016, 04:54
x= 9!
= 9* 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1
= (3^2) * (2^3) * 7 * (2*3) * 5 * 2^2 * 3 * 2 * 1
= 2^7 * 3^4 * 5 * 7
= 2^i * 3^k * 5^m * 7^p

i+k+m+p = 7+4+1+1
= 13
_________________
When everything seems to be going against you, remember that the airplane takes off against the wind, not with it. - Henry Ford
The Moment You Think About Giving Up, Think Of The Reason Why You Held On So Long
+1 Kudos if you find this post helpful
Manager
Joined: 18 May 2016
Posts: 67
GMAT 1: 720 Q49 V39
GPA: 3.7
WE: Analyst (Investment Banking)
If x is the product of the positive integers from 1 to 9, inclusive,  [#permalink]

### Show Tags

31 May 2016, 15:27
1
Given x = 9! = $$2^i * 3^k * 5^m * 7^p$$

Find the number of powers of prime numbers: $$n / i^1 + n / i^2 ...$$ until i^z <n, z any positive number z=>1

Number of 2s in expression 9!: i = $$\frac{9}{2} + \frac{9}{8}≈ 5 + 1 = 6$$

Number of 3s in expression 9!: k = $$\frac{9}{3} + \frac{9}{9} ≈ 3 + 1 = 4$$

Number of 5s in expression 9!: m = $$\frac{9}{5} ≈ 2$$

Number of 7s in expression 9!: p = $$\frac{9}{7} ≈ 1$$

i + k + m + p = 6 + 4 + 2 + 1 = 13

_________________

My debrief: Self-study: How to improve from 620(Q39,V36) to 720(Q49,V39) in 25 days!
Verbal Forum Moderator
Status: Greatness begins beyond your comfort zone
Joined: 08 Dec 2013
Posts: 2260
Location: India
Concentration: General Management, Strategy
Schools: Kelley '20, ISB '19
GPA: 3.2
WE: Information Technology (Consulting)
Re: If x is the product of the positive integers from 1 to 9, inclusive,  [#permalink]

### Show Tags

31 May 2016, 20:42
1
fantaisie wrote:
Given x = 9! = $$2^i * 3^k * 5^m * 7^p$$

Find the number of powers of prime numbers: $$n / i^1 + n / i^2 ...$$ until i^z <n, z any positive number z=>1

Number of 2s in expression 9!: i = $$\frac{9}{2} + \frac{9}{8}≈ 5 + 1 = 6$$

Number of 3s in expression 9!: k = $$\frac{9}{3} + \frac{9}{9} ≈ 3 + 1 = 4$$

Number of 5s in expression 9!: m = $$\frac{9}{5} ≈ 2$$

Number of 7s in expression 9!: p = $$\frac{9}{7} ≈ 1$$

i + k + m + p = 6 + 4 + 2 + 1 = 13

Hi fantaisie ,
It seems that you have missed 2^2 while trying to find powers of 2 in 9! . Also you have rounded up in a few cases . Hope this helps !!

$$\frac{9}{2}+\frac{9}{4}+\frac{9}{8}= 4+2+1 = 7$$

$$\frac{9}{3}+\frac{9}{9}= 3+1 = 4$$

$$\frac{9}{5} = 1$$

$$\frac{9}{7} = 1$$

Finding the powers of a prime number p, in the n!

The formula is:
$$\frac{n}{p}+\frac{n}{p^2}+\frac{n}{p^3}+...+\frac{n}{p^k}$$, where $$k$$ must be chosen such that $$p^k\leq{n}$$
_________________
When everything seems to be going against you, remember that the airplane takes off against the wind, not with it. - Henry Ford
The Moment You Think About Giving Up, Think Of The Reason Why You Held On So Long
+1 Kudos if you find this post helpful
Manager
Joined: 18 May 2016
Posts: 67
GMAT 1: 720 Q49 V39
GPA: 3.7
WE: Analyst (Investment Banking)
Re: If x is the product of the positive integers from 1 to 9, inclusive,  [#permalink]

### Show Tags

01 Jun 2016, 02:56
Skywalker18 wrote:
fantaisie wrote:
Given x = 9! = $$2^i * 3^k * 5^m * 7^p$$

Find the number of powers of prime numbers: $$n / i^1 + n / i^2 ...$$ until i^z <n, z any positive number z=>1

Number of 2s in expression 9!: i = $$\frac{9}{2} + \frac{9}{8}≈ 5 + 1 = 6$$

Number of 3s in expression 9!: k = $$\frac{9}{3} + \frac{9}{9} ≈ 3 + 1 = 4$$

Number of 5s in expression 9!: m = $$\frac{9}{5} ≈ 2$$

Number of 7s in expression 9!: p = $$\frac{9}{7} ≈ 1$$

i + k + m + p = 6 + 4 + 2 + 1 = 13

Hi fantaisie ,
It seems that you have missed 2^2 while trying to find powers of 2 in 9! . Also you have rounded up in a few cases . Hope this helps !!

$$\frac{9}{2}+\frac{9}{4}+\frac{9}{8}= 4+2+1 = 7$$

$$\frac{9}{3}+\frac{9}{9}= 3+1 = 4$$

$$\frac{9}{5} = 1$$

$$\frac{9}{7} = 1$$

Finding the powers of a prime number p, in the n!

The formula is:
$$\frac{n}{p}+\frac{n}{p^2}+\frac{n}{p^3}+...+\frac{n}{p^k}$$, where $$k$$ must be chosen such that $$p^k\leq{n}$$

Thank you very much for pointing out my mistake! I have one question: what's the rule when it comes to rounding in this formula? $$\frac{9}{5}$$ = 1.8, do we only take into account actual integers? Thank you!
_________________

My debrief: Self-study: How to improve from 620(Q39,V36) to 720(Q49,V39) in 25 days!
Math Expert
Joined: 02 Aug 2009
Posts: 7589
Re: If x is the product of the positive integers from 1 to 9, inclusive,  [#permalink]

### Show Tags

01 Jun 2016, 03:03
fantaisie wrote:

Thank you very much for pointing out my mistake! I have one question: what's the rule when it comes to rounding in this formula? $$\frac{9}{5}$$ = 1.8, do we only take into account actual integers? Thank you!

Hi,

rounding of is always done to the lower integer...
Reason --
say we are looking for number of 5 in 9! -
9/5 + 9/5^2 = 1.8 + 0.3 = 1+0 = 1..
so 1 in 1.8 talks of integer 5 in the product....
_________________
Non-Human User
Joined: 09 Sep 2013
Posts: 10637
Re: If x is the product of the positive integers from 1 to 9, inclusive,  [#permalink]

### Show Tags

09 Sep 2017, 05:06
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
Re: If x is the product of the positive integers from 1 to 9, inclusive,   [#permalink] 09 Sep 2017, 05:06
Display posts from previous: Sort by