One of the toughest I encountered:Algebra

[harsh.champ]

find the number of all rational numbers m/n such that
(1) 0 < m/n < 1
(2) m and n are relatively prime
(3) m * n = 25!

(1)255
(2)256
(3)128
(4)64
(5)None of these

_________________

[shashank.ism]

find the number of all rational numbers m/n such that
(1) 0 < m/n < 1
(2) m and n are relatively prime
(3) m * n = 25!

(1)255
(2)256
(3)128
(4)64
(5)None of these

_

no of primes less than 25=2,3,5,7,11,13,17,19,23=9 so number is 2^(9−1) = 2^8 = 256

[ldoolitt]

no of primes less than 25=2,3,5,7,11,13,17,19,23=9 so number is 2^(9−1) = 2^8 = 256

Hey Shashank,

Could you expand on this a little bit? I didn't totally understand your solution.

[subgeeth]

Even I dont understand:-(

[Stuart@KaplanGMAT]

"relatively prime" is not a term you'll ever see on the GMAT - what does it mean?

[ldoolitt]

"relatively prime" is not a term you'll ever see on the GMAT - what does it mean?

Hey Stuart,

Yeah I've never seen it on a GMAT question either. For reference though two numbers who are "relatively prime" have no common factors.

[Stuart@KaplanGMAT]

find the number of all rational numbers m/n such that
(1) 0 < m/n < 1
(2) m and n are relatively prime
(3) m * n = 25!

(1)255
(2)256
(3)128
(4)64
(5)None of these

Really don't think this can reasonably be solved in under 5 minutes, but hey, let's try.

Ok, given that relatively prime means no primes in common:

If m*n = 25!, then m and n must share all the primes between 2 and 23, namely:

2, 3, 5, 7, 11, 13, 17, 19 and 23.

However, there are multiples of many of those primes, so we really need to think about all the other numbers as well:

4, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20, 21, 24 and 25.

Now, since m and n can't have any primes in common, certain of these numbers must be paired together.

For example, whichever one of them has 2 also has to have 3, since 6, 12, 18 and 24 are multiples of both of those numbers. Similarly, 10 is a multiple of 2 and 5, so that also must be included. 14 is 2*7, 18 is 2*9, 22 is 2*11, so:

2, 3, 5, 7, 9 and 11 must all be factors of the same one of m or n.

This means that 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 14, 15, 16, 18, 20, 21, 22, 24 and 25 must all be factors of one of m or n, leaving only:

13, 17, 19 and 23 up for grabs.

Given that m/n is a proper fraction, we know that m < n, so:

n MUST be a multiple of 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 14, 15, 16, 18, 20, 21, 22, 24 and 25.

With 4 numbers left, each one can be assigned to either m or n (after all, we could have m=1 and n = 25!).

There are 16 ways to distribute 4 objects to 2 different groups, so 16 is our final answer... choose (5).

I'd love to see the official explanation - where is this from?

[bhumika.k.shah]

what are the chances of such problems coming in GMAT ??

Above what level is this problem ? above 700? 750 ?

It was indeed a quite difficult qs

find the number of all rational numbers m/n such that
(1) 0 < m/n < 1
(2) m and n are relatively prime
(3) m * n = 25!

(1)255
(2)256
(3)128
(4)64
(5)None of these

Really don't think this can reasonably be solved in under 5 minutes, but hey, let's try.

Ok, given that relatively prime means no primes in common:

If m*n = 25!, then m and n must share all the primes between 2 and 23, namely:

2, 3, 5, 7, 11, 13, 17, 19 and 23.

However, there are multiples of many of those primes, so we really need to think about all the other numbers as well:

4, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20, 21, 24 and 25.

Now, since m and n can't have any primes in common, certain of these numbers must be paired together.

For example, whichever one of them has 2 also has to have 3, since 6, 12, 18 and 24 are multiples of both of those numbers. Similarly, 10 is a multiple of 2 and 5, so that also must be included. 14 is 2*7, 18 is 2*9, 22 is 2*11, so:

2, 3, 5, 7, 9 and 11 must all be factors of the same one of m or n.

This means that 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 14, 15, 16, 18, 20, 21, 22, 24 and 25 must all be factors of one of m or n, leaving only:

13, 17, 19 and 23 up for grabs.

Given that m/n is a proper fraction, we know that m < n, so:

n MUST be a multiple of 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 14, 15, 16, 18, 20, 21, 22, 24 and 25.

With 4 numbers left, each one can be assigned to either m or n (after all, we could have m=1 and n = 25!).

There are 16 ways to distribute 4 objects to 2 different groups, so 16 is our final answer... choose (5).

I'd love to see the official explanation - where is this from?

[Stuart@KaplanGMAT]

what are the chances of such problems coming in GMAT ??

Above what level is this problem ? above 700? 750 ?

It was indeed a quite difficult qs

I'd say chance is 0!

Still not sure what the source is (or what the official answer is).

[Ian Stewart]

find the number of all rational numbers m/n such that
(1) 0 < m/n < 1
(2) m and n are relatively prime
(3) m * n = 25!

(1)255
(2)256
(3)128
(4)64
(5)None of these

_________________

There is no chance whatsoever of encountering a question like this on the GMAT (you don't even need to know what 'rational number' or 'relatively prime' mean on the test), so anyone reading this forum only for GMAT help can move on to other problems without giving this question another thought. If anyone wants a solution for interest's sake, I can try to provide one (shashank's solution above is correct, though I'll try to go into more detail) --

* we are looking for two numbers m and n for which m*n = 25!. While it's not necessary, it will be easier to explain the solution if we prime factorize 25!:

25! = (2^22)(3^10)(5^6)(7^3)(11^2)(13)(17)(19)(23)

* Now, m and n are relatively prime, which means they have no common prime divisors. That means that if m is divisible by, say, 2, then n cannot be. So if m*n will be equal to 25!, and if m is divisible by 2, then m actually must be divisible by 2^22, since those twenty-two 2's must be in m*n, and only one of m or n can be divisible by 2. The same is true for each of the remaining primes in the prime factorization above: all of the 3's are in either m or n, all of the 5's are in either m or n, and so on. So for each of the nine different primes in the prime factorization of 25!, we have two choices: give the prime to m or give the prime to n. Since there are nine primes in total, we could assign these primes to m and n in 2^9 ways. To give some examples of the values we might get for m and n doing this:

m = (2^22)(3^10)(5^6)(7^3)(11^2) and n = (13)(17)(19)(23)

m = (3^10) and n = (2^22)(5^6)(7^3)(11^2)(13)(17)(19)(23)

m = (13)(17)(19)(23) and n = (2^22)(3^10)(5^6)(7^3)(11^2)

m = 1 and n = (2^22)(3^10)(5^6)(7^3)(11^2)(13)(17)(19)(23)

etc.

* In this way, we can be sure that m*n = 25!, and that m and n are relatively prime (i.e. they have no prime divisors in common). We've ignored one condition, however: m must be less than n. Since half the time m will be smaller than n, and half the time m will be larger than n, we must divide by 2; the answer is 2^9/2 = 2^8 = 256.

[laxmanrr]

Amazing explanation!!

Thanks a ton