OG quant review problem 169

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nqhussain: Junior | Next Rank: 30 Posts; Posts: 14; Joined: Fri Feb 29, 2008 7:51 pm

by nqhussain » Sun Jun 01, 2008 8:41 am

If n is a positive integer and n^2 is divisible by 72, then the largest positive integer that must divide n is ??

answer is 12...please adivise.

Also I realize compare to MGMAT, the question in OG quant review are much much easier. I am not sure what this means in terms of the actual exams? comments please?

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Source: Beat The GMAT — Problem Solving | Sun Jun 01, 2008 8:41 am

kishore: Senior | Next Rank: 100 Posts; Posts: 62; Joined: Fri Dec 21, 2007 2:58 pm; Location: Fremont , California; Thanked: 3 times

by kishore » Sun Jun 01, 2008 10:51 am

If n is a positive integer and n^2 is divisible by 72, then the largest positive integer that must divide n is ??

Explanation:

dividend = divisor * quotient + remainder

=> since is n ^ 2 is divisble by 72

= > quotient is some x and remainder = 0

=> n ^ 2 = 72 * x + 0

=> n ^ 2 = 72*x

=> n = sqrt(72*x) = 6 * sqrt(2 *x)
=> since , he is asking for positive integer that must divide n

put x = 1

=> n = 6 * sqrt(2 *1)
=> n = 6 * sqrt(2) = 6* 1.414 = 8.4 (This is not integer)

put x = 2

=> n = 6*sqrt(2* 2)
=> n = 6* 2 = 12 (Which is integer)

so, 12 is the right answer

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jasonc: Master | Next Rank: 500 Posts; Posts: 127; Joined: Sat Apr 19, 2008 1:02 am; Thanked: 12 times

Re: OG quant review problem 169

by jasonc » Sun Jun 01, 2008 7:00 pm

nqhussain wrote:If n is a positive integer and n^2 is divisible by 72, then the largest positive integer that must divide n is ??

answer is 12...please adivise.

Also I realize compare to MGMAT, the question in OG quant review are much much easier. I am not sure what this means in terms of the actual exams? comments please?

Depends on what level/score you're at/aiming for.

OG questions are spread across difficulty levels, so if you're aiming for 700+/750+ then only a few of the questions are representative of the majority of the questions you'll likely see on the actual exam.

Although, many members have posted that MGMAT may have more difficult quant, but this is good for you if you've got the core concepts down because on the actual GMAT you'll be seeing easier questions than what you're used to.

I beat the GMAT! 760 (Q49/V44)

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Ian Stewart: GMAT Instructor; Posts: 2623; Joined: Mon Jun 02, 2008 3:17 amFollowed by: 355 members; Location: Montreal; Thanked: 1090 times; GMAT Score:780

Re: OG quant review problem 169

by Ian Stewart » Mon Jun 02, 2008 9:39 am

nqhussain wrote:If n is a positive integer and n^2 is divisible by 72, then the largest positive integer that must divide n is ??

answer is 12...please adivise.

Notice that when you square an integer, every exponent in the resulting prime factorization must be even. Take, for example, 108^2:

108^2 = (2^2 * 3^3)^2 = 2^4 * 3^6

(when you square an integer, you are just doubling every exponent in its prime factorization; when you cube an integer, you are tripling each exponent, etc.).

Now,
- n^2 is divisible by 72
- prime factorize 72:
- n^2 is divisible by 2^3 * 3^2
- but, n^2 is the square of an integer. All the exponents in its prime factorization must be even. If n^2 is divsible by 2^3, it must be divisible by 2^4, at least.
- Thus n^2 is divisible by 2^4 * 3^2, and n is divisible by 2^2 * 3 = 12.

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chipjet: Senior | Next Rank: 100 Posts; Posts: 35; Joined: Wed May 21, 2008 10:37 am; Location: Chicago; Thanked: 3 times; GMAT Score:700

by chipjet » Wed Jun 11, 2008 7:19 am

kishore wrote:If n is a positive integer and n^2 is divisible by 72, then the largest positive integer that must divide n is ??

Explanation:

dividend = divisor * quotient + remainder

=> since is n ^ 2 is divisble by 72

= > quotient is some x and remainder = 0

=> n ^ 2 = 72 * x + 0

=> n ^ 2 = 72*x

=> n = sqrt(72*x) = 6 * sqrt(2 *x)
=> since , he is asking for positive integer that must divide n

put x = 1

=> n = 6 * sqrt(2 *1)
=> n = 6 * sqrt(2) = 6* 1.414 = 8.4 (This is not integer)

put x = 2

=> n = 6*sqrt(2* 2)
=> n = 6* 2 = 12 (Which is integer)

so, 12 is the right answer

I think I'm confusing myself on this one. Isn't 12 the minimum that n could be to make this true? What about 36^2, which is also divisible by 72, or 72^2, etc. There are a lot of squares that can be divisible by 72. What am I missing?

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logitech: Legendary Member; Posts: 2134; Joined: Mon Oct 20, 2008 11:26 pmFollowed by: 25 members; Thanked: 237 times; GMAT Score:730

by logitech » Sat Oct 25, 2008 10:17 am

chipjet wrote:
kishore wrote:If n is a positive integer and n^2 is divisible by 72, then the largest positive integer that must divide n is ??

I think I'm confusing myself on this one. Isn't 12 the minimum that n could be to make this true? What about 36^2, which is also divisible by 72, or 72^2, etc. There are a lot of squares that can be divisible by 72. What am I missing?
That's true my friend. Why not 48 but 12 ?

LGTCH
---------------------
"DON'T LET ANYONE STEAL YOUR DREAM!"

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California4jx: Master | Next Rank: 500 Posts; Posts: 227; Joined: Thu Aug 14, 2008 10:43 amFollowed by: 1 member; Thanked: 7 times; GMAT Score:650

by California4jx » Mon Jun 15, 2009 10:35 am

in order to find the largest possible integer (not the value of n) but instead the largest possible interger that must divide n.

We all know that in order to find the largest integer - L - we must divide X with the smallest y in the following form of equation.

X = (y) * L

Example:

4 = (1) 4 => 4/1 = 4
4 = (2) 2 => 4/2 = 2 [L is going down with higher values of (y) ]

so, if we have any equation like this:

n / 6 * sqrt (2k) = L

where y = 6 * sqrt (2k) = as small as possible integer value

if k = 2, y = 12
k = 8, y = 24

so the higher the value of y, the lower the value of L, when divided by n

again: Maximize value of integer L = n / smallest denominator

Hopefully, this explains why we need the value of y = 12 and not 36, or 48, etc.

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Naruto: Master | Next Rank: 500 Posts; Posts: 121; Joined: Thu Jun 11, 2009 1:16 am; Thanked: 4 times; GMAT Score:720

by Naruto » Tue Jun 16, 2009 1:20 am

California4jx wrote:in order to find the largest possible integer (not the value of n) but instead the largest possible interger that must divide n.

We all know that in order to find the largest integer - L - we must divide X with the smallest y in the following form of equation.

X = (y) * L

Example:

4 = (1) 4 => 4/1 = 4
4 = (2) 2 => 4/2 = 2 [L is going down with higher values of (y) ]

so, if we have any equation like this:

n / 6 * sqrt (2k) = L

where y = 6 * sqrt (2k) = as small as possible integer value

if k = 2, y = 12
k = 8, y = 24

so the higher the value of y, the lower the value of L, when divided by n

again: Maximize value of integer L = n / smallest denominator

Hopefully, this explains why we need the value of y = 12 and not 36, or 48, etc.

Thats a good explanation, thanks i had the same doubt

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TryHarder: Senior | Next Rank: 100 Posts; Posts: 32; Joined: Fri Feb 27, 2009 8:38 pm

by TryHarder » Fri Jan 08, 2010 3:28 pm

What is wrong with this example:

n= 156 which is 12x13 (positive integer)

n sqared = 12x12x13x13 (n squared is divisible by 72)

Largest positive integer that must divide n is 13

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san2009: Master | Next Rank: 500 Posts; Posts: 171; Joined: Fri Apr 16, 2010 1:02 am; Thanked: 1 times

by san2009 » Tue Apr 27, 2010 11:33 pm

nqhussain wrote:
If n is a positive integer and n^2 is divisible by 72, then the largest positive integer that must divide n is ??

answer is 12...please adivise.
Notice that when you square an integer, every exponent in the resulting prime factorization must be even. Take, for example, 108^2:

108^2 = (2^2 * 3^3)^2 = 2^4 * 3^6

(when you square an integer, you are just doubling every exponent in its prime factorization; when you cube an integer, you are tripling each exponent, etc.).

Now,
- n^2 is divisible by 72
- prime factorize 72:
- n^2 is divisible by 2^3 * 3^2
- but, n^2 is the square of an integer. All the exponents in its prime factorization must be even. If n^2 is divsible by 2^3, it must be divisible by 2^4, at least.
- Thus n^2 is divisible by 2^4 * 3^2, and n is divisible by 2^2 * 3 = 12.

Ian: the question is asking for the LARGEST positive integer. Given what you're saying, wouldn't 12 be the SMALLEST positive integer that divides n? At the very least the PF of N must contain AT LEAST two 2s and one 3. Please advise.

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