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The function f is defined for all positive integers n by the following rule. f(n) is the number of positive integers each of which is less than n and has no positive factor in common with n other than 1. If p is a prime number then f(p)= A. p-1 B. p-2 C. (p+1)/2 D. (p-1)/2 E. 2

If not the wording the question wouldn't be as tough as it is now. The GMAT often hides some simple concept in complicated way of delivering it.

This question for instance basically asks: how many positive integers are less than given prime number p which have no common factor with p except 1.

Well as p is a prime, all positive numbers less than p have no common factors with p (except common factor 1). So there would be p-1 such numbers (as we are looking number of integers less than p).

For example: if p=7 how many numbers are less than 7 having no common factors with 7: 1, 2, 3, 4, 5, 6 --> 7-1=6.

Re: NEW SET of good PS(3) [#permalink]
16 Oct 2012, 06:07

Bunuel wrote:

5. Mrs. Smith has been given film vouchers. Each voucher allows the holder to see a film without charge. She decides to distribute them among her four nephews so that each nephew gets at least two vouchers. How many vouchers has Mrs. Smith been given if there are 120 ways that she could distribute the vouchers? (A) 13 (B) 14 (C) 15 (D) 16 (E) more than 16

Answer: C.

Clearly there are more than 8 vouchers as each of four can get at least 2. So, basically 120 ways vouchers can the distributed are the ways to distribute \(x-8\) vouchers, so that each can get from zero to \(x-8\) as at "least 2", or 2*4=8, we already booked. Let \(x-8\) be \(k\).

In how many ways we can distribute \(k\) identical things among 4 persons? Well there is a formula for this but it's better to understand the concept.

Let \(k=5\). And imagine we want to distribute 5 vouchers among 4 persons and each can get from zero to 5, (no restrictions).

Consider:

\(ttttt|||\) We have 5 tickets (t) and 3 separators between them, to indicate who will get the tickets:

\(ttttt|||\) Means that first nephew will get all the tickets,

\(|t|ttt|t\) Means that first got 0, second 1, third 3, and fourth 1

And so on.

How many permutations (arrangements) of these symbols are possible? Total of 8 symbols (5+3=8), out of which 5 \(t\)'s and 3 \(|\)'s are identical, so \(\frac{8!}{5!3!}=56\). Basically it's the number of ways we can pick 3 separators out of 5+3=8: \(8C3\).

So, # of ways to distribute 5 tickets among 4 people is \((5+4-1)C(4-1)=8C3\).

For \(k\) it will be the same: # of ways to distribute \(k\) tickets among 4 persons (so that each can get from zero to \(k\)) would be \((K+4-1)C(4-1)=(k+3)C3=\frac{(k+3)!}{k!3!}=120\).

\((k+1)(k+2)(k+3)=3!*120=720\). --> \(k=7\). Plus the 8 tickets we booked earlier: \(x=k+8=7+8=15\).

Answer: C (15).

P.S. Direct formula:

The total number of ways of dividing n identical items among r persons, each one of whom, can receive 0,1,2 or more items is \(n+r-1C_{r-1}\).

The total number of ways of dividing n identical items among r persons, each one of whom receives at least one item is \(n-1C_{r-1}\).

Hope it helps.

I tried to backsolve, but even after I saw the formula I failed miserably.

Is it possible to be solving this through backsolving, or would that be too cumbersome? If it is possible, how would you then proceed?

Re: NEW SET of good PS(3) [#permalink]
18 Mar 2013, 13:24

Expert's post

annutalreja wrote:

Hi, still not clear why answer to the 2nd ques should be P-1 and not P-2. Shouldn't both P and 1 be deducted from the set?

p yes, but not 1.

Consider this, say p=7 how many numbers are less than 7 having no common factors with 7 other than 1: 1, 2, 3, 4, 5, 6 --> 7-1=6. (7 and 1 do not share any common factor other than 1.)

Re: NEW SET of good PS(3) [#permalink]
26 Mar 2013, 10:38

Bunuel are these all official? The first problem uses the word different which I think is, different from, the word 'unique'. Subtle difference I never noticed but huge implications _________________

Re: NEW SET of good PS(3) [#permalink]
06 Apr 2013, 09:04

Augustus wrote:

yangsta8 wrote:

Bunuel wrote:

1. ABCDE is a regular pentagon with F at its center. How many different triangles can be formed by joining 3 of the points A,B,C,D,E and F? (A) 10 (B) 15 (C) 20 (D) 25 (E) 30

6 points in total to make triangles. I think a combination of any 3 will make a unique triangle so: 6C3 = 20

Will anyone please explain what the C means in the notation?

Thanks in advance

That's a notation for Combinations. You have six points to choose from and you're choosing three so (6!)/(3!*3!).

Re: NEW SET of good PS(3) [#permalink]
04 Jul 2013, 07:40

Bunuel wrote:

5. Mrs. Smith has been given film vouchers. Each voucher allows the holder to see a film without charge. She decides to distribute them among her four nephews so that each nephew gets at least two vouchers. How many vouchers has Mrs. Smith been given if there are 120 ways that she could distribute the vouchers? (A) 13 (B) 14 (C) 15 (D) 16 (E) more than 16

Answer: C.

Clearly there are more than 8 vouchers as each of four can get at least 2. So, basically 120 ways vouchers can the distributed are the ways to distribute \(x-8\) vouchers, so that each can get from zero to \(x-8\) as at "least 2", or 2*4=8, we already booked. Let \(x-8\) be \(k\).

In how many ways we can distribute \(k\) identical things among 4 persons? Well there is a formula for this but it's better to understand the concept.

Let \(k=5\). And imagine we want to distribute 5 vouchers among 4 persons and each can get from zero to 5, (no restrictions).

Consider:

\(ttttt|||\) We have 5 tickets (t) and 3 separators between them, to indicate who will get the tickets:

\(ttttt|||\) Means that first nephew will get all the tickets,

\(|t|ttt|t\) Means that first got 0, second 1, third 3, and fourth 1

And so on.

How many permutations (arrangements) of these symbols are possible? Total of 8 symbols (5+3=8), out of which 5 \(t\)'s and 3 \(|\)'s are identical, so \(\frac{8!}{5!3!}=56\). Basically it's the number of ways we can pick 3 separators out of 5+3=8: \(8C3\).

So, # of ways to distribute 5 tickets among 4 people is \((5+4-1)C(4-1)=8C3\).

For \(k\) it will be the same: # of ways to distribute \(k\) tickets among 4 persons (so that each can get from zero to \(k\)) would be \((K+4-1)C(4-1)=(k+3)C3=\frac{(k+3)!}{k!3!}=120\).

\((k+1)(k+2)(k+3)=3!*120=720\). --> \(k=7\). Plus the 8 tickets we booked earlier: \(x=k+8=7+8=15\).

Answer: C (15).

P.S. Direct formula:

The total number of ways of dividing n identical items among r persons, each one of whom, can receive 0,1,2 or more items is \(n+r-1C_{r-1}\).

The total number of ways of dividing n identical items among r persons, each one of whom receives at least one item is \(n-1C_{r-1}\).

Re: NEW SET of good PS(3) [#permalink]
04 Jul 2013, 07:48

Expert's post

akijuneja wrote:

Bunuel wrote:

5. Mrs. Smith has been given film vouchers. Each voucher allows the holder to see a film without charge. She decides to distribute them among her four nephews so that each nephew gets at least two vouchers. How many vouchers has Mrs. Smith been given if there are 120 ways that she could distribute the vouchers? (A) 13 (B) 14 (C) 15 (D) 16 (E) more than 16

Answer: C.

Clearly there are more than 8 vouchers as each of four can get at least 2. So, basically 120 ways vouchers can the distributed are the ways to distribute \(x-8\) vouchers, so that each can get from zero to \(x-8\) as at "least 2", or 2*4=8, we already booked. Let \(x-8\) be \(k\).

In how many ways we can distribute \(k\) identical things among 4 persons? Well there is a formula for this but it's better to understand the concept.

Let \(k=5\). And imagine we want to distribute 5 vouchers among 4 persons and each can get from zero to 5, (no restrictions).

Consider:

\(ttttt|||\) We have 5 tickets (t) and 3 separators between them, to indicate who will get the tickets:

\(ttttt|||\) Means that first nephew will get all the tickets,

\(|t|ttt|t\) Means that first got 0, second 1, third 3, and fourth 1

And so on.

How many permutations (arrangements) of these symbols are possible? Total of 8 symbols (5+3=8), out of which 5 \(t\)'s and 3 \(|\)'s are identical, so \(\frac{8!}{5!3!}=56\). Basically it's the number of ways we can pick 3 separators out of 5+3=8: \(8C3\).

So, # of ways to distribute 5 tickets among 4 people is \((5+4-1)C(4-1)=8C3\).

For \(k\) it will be the same: # of ways to distribute \(k\) tickets among 4 persons (so that each can get from zero to \(k\)) would be \((K+4-1)C(4-1)=(k+3)C3=\frac{(k+3)!}{k!3!}=120\).

\((k+1)(k+2)(k+3)=3!*120=720\). --> \(k=7\). Plus the 8 tickets we booked earlier: \(x=k+8=7+8=15\).

Answer: C (15).

P.S. Direct formula:

The total number of ways of dividing n identical items among r persons, each one of whom, can receive 0,1,2 or more items is \(n+r-1C_{r-1}\).

The total number of ways of dividing n identical items among r persons, each one of whom receives at least one item is \(n-1C_{r-1}\).

Re: NEW SET of good PS(3) [#permalink]
01 Dec 2013, 00:28

yossarian84 wrote:

Bunuel wrote:

5. Mrs. Smith has been given film vouchers. Each voucher allows the holder to see a film without charge. She decides to distribute them among her four nephews so that each nephew gets at least two vouchers. How many vouchers has Mrs. Smith been given if there are 120 ways that she could distribute the vouchers? (A) 13 (B) 14 (C) 15 (D) 16 (E) more than 16

Answer: C.

Clearly there are more than 8 vouchers as each of four can get at least 2. So, basically 120 ways vouchers can the distributed are the ways to distribute \(x-8\) vouchers, so that each can get from zero to \(x-8\) as at "least 2", or 2*4=8, we already booked. Let \(x-8\) be \(k\).

In how many ways we can distribute \(k\) identical things among 4 persons? Well there is a formula for this but it's better to understand the concept.

Let \(k=5\). And imagine we want to distribute 5 vouchers among 4 persons and each can get from zero to 5, (no restrictions).

Consider:

\(ttttt|||\) We have 5 tickets (t) and 3 separators between them, to indicate who will get the tickets:

\(ttttt|||\) Means that first nephew will get all the tickets,

\(|t|ttt|t\) Means that first got 0, second 1, third 3, and fourth 1

And so on.

How many permutations (arrangements) of these symbols are possible? Total of 8 symbols (5+3=8), out of which 5 \(t\)'s and 3 \(|\)'s are identical, so \(\frac{8!}{5!3!}=56\). Basically it's the number of ways we can pick 3 separators out of 5+3=8: \(8C3\).

So, # of ways to distribute 5 tickets among 4 people is \((5+4-1)C(4-1)=8C3\).

For \(k\) it will be the same: # of ways to distribute \(k\) tickets among 4 persons (so that each can get from zero to \(k\)) would be \((K+4-1)C(4-1)=(k+3)C3=\frac{(k+3)!}{k!3!}=120\).

\((k+1)(k+2)(k+3)=3!*120=720\). --> \(k=7\). Plus the 8 tickets we booked earlier: \(x=k+8=7+8=15\).

Answer: C (15).

P.S. Direct formula:

The total number of ways of dividing n identical items among r persons, each one of whom, can receive 0,1,2 or more items is \(n+r-1C_{r-1}\).

The total number of ways of dividing n identical items among r persons, each one of whom receives at least one item is \(n-1C_{r-1}\).

Hope it helps.

Awesome...hats off...this is totally new to me...widens my realm..and strengthens my reasoning...thanks a lot

Karishma/Bunnel,

There one more formula to distribute n things in m, such that each m can receive 0 to n items = m^n. Thing to be distributed goes in the power.

Re: NEW SET of good PS(3) [#permalink]
01 Dec 2013, 20:35

Expert's post

cumulonimbus wrote:

VeritasPrepKarishma wrote:

yogesh1984 wrote:

Bunuel,

1- Collinear point issue will arise in case of overlapping values of x, y ? (as in here we have all the overlapping range for x & y). Also since range here is small for both x, y (ie.=3) we can manually calculate the collinear points but in case of large range how do we go about it ? should it be = # overlapping points on X + # overlapping points on Y + # diagonal points (which will essentially be min(# overlapping points on X , Y) -1 )-- Not so sure on this though ...

2- I see a similar Question in OG12 PS Q.229- The method explained here in the above example does not seems to fit too well there. basically in the question we have -4 <= X <=5, 6<= Y <=16. Can you please throw some light in the context of OG question....

To find out the possible number of right triangles I tried as below: No. of rectangles *4 (for each orientation)

For example: How many triangles with positive area can be drawn on the coordinate plane such that the vertices have integer coordinates (x,y) satisfying 2 ≤ x ≤ 4 and 5 ≤ y ≤ 7?

No. of rectangles = 9 (this by actual counting of rectangles) No. of right triangles = 4*9 = 36, Is this correct?

Secondly, I tried to calculate no. of rectangles via combinations but i count understand why no. of rectangles would be 3c2*3c2?

To me it should be = 3c2*(3-1)=6, but this is not correct.

The method is correct for this question. Number of right triangles will be 36. _________________

Re: NEW SET of good PS(3) [#permalink]
05 Dec 2013, 20:33

Expert's post

cumulonimbus wrote:

The method is correct for this question. Number of right triangles will be 36.

Hi Karishma, Do you mean I cannot use the above method for all questions for getting the number of right triangles?

Whenever an innovative method is used, we need to understand its assumptions. Here we are looking for right triangles with all integer coordinates lying within a certain range. If one of these conditions is not met, the method will change. Of course there is nothing special about these particular values: 2 ≤ x ≤ 4 and 5 ≤ y ≤ 7 and as discussed above, the method will work for any other such set of values e.g. 2 <= x <= 5 and 5<= y <= 7 or -1 <= x <= 7 and 2<= y <= 7 etc. There is nothing wrong with this approach - in fact it's great. You can certainly solve a similar question using this approach. But it is important for you to understand the generic method too (for more generic questions, say all triangles, not just right triangles or some other variation) so that you can twist it according to the question at hand, if required. _________________

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