GMAT Club Forumhttps://gmatclub.com:443/forum/ At a particular store, candy bars are normally priced at $1.https://gmatclub.com/forum/at-a-particular-store-candy-bars-are-normally-priced-at-145675.html Page 1 of 1  Author: trex16864 [ 13 Jan 2013, 03:16 ] Post subject: At a particular store, candy bars are normally priced at$1. At a particular store, candy bars are normally priced at $1.00 each. Last week, the store offered a promotion under which customers purchasing one candy bar at full price could purchase a second candy bar for$0.50. A third candy bar would cost $1.00, a fourth would cost$0.50, and so on.If, in a single transaction during the promotion, Rajiv spent D dollars on N candy bars, where D and N are integers, is N odd?(1) D is prime.(2) D is not divisible by 3.

 Author: stne [ 15 Jan 2013, 06:37 ] Post subject: Re: At a particular store, candy bars are normally priced at $1. trex16864 wrote:At a particular store, candy bars are normally priced at$1.00 each. Last week, the store offered a promotion under which customers purchasing one candy bar at full price could purchase a second candy bar for $0.50. A third candy bar would cost$1.00, a fourth would cost $0.50, and so on.If, in a single transaction during the promotion, Rajiv spent D dollars on N candy bars, where D and N are integers, is N odd?(1) D is prime.(2) D is not divisible by 3.1 candy cost 12 candies cost 1+.50=1.50 ( here D is not an integer, hence we cannot buy 2 candies . so we can reject all cases where D is non Integer)3 candies cost 1.50 +1 =2.504 candies cost 2.50+.50= 35 candies cost 3+1= 46 candies cost 4+.50= 4.507 candies cost 4.50+1=5.508 candies cost 5.50.+.50= 69 candies cost 6+1= 7 .....13 candies cost =10(i) D is prime D=3 and N=4 (N is even)D=7 N=9 (N is odd )not sufficient (ii) D is not Divisible by 3 D=1 N=1D=4 N =5 D=7 N=9D=10 N=13so we see if D is not divisible 3 then N is always odd.Hence B is sufficient Hope it's clear  Author: hfbamafan [ 27 Jun 2013, 21:26 ] Post subject: Re: At a particular store, candy bars are normally priced at$1. Can this problem be turned into an algebraic expression?

 Author: avinashrao9 [ 04 Jul 2013, 10:34 ] Post subject: Re: At a particular store, candy bars are normally priced at $1. stne wrote:trex16864 wrote:At a particular store, candy bars are normally priced at$1.00 each. Last week, the store offered a promotion under which customers purchasing one candy bar at full price could purchase a second candy bar for $0.50. A third candy bar would cost$1.00, a fourth would cost $0.50, and so on.If, in a single transaction during the promotion, Rajiv spent D dollars on N candy bars, where D and N are integers, is N odd?(1) D is prime.(2) D is not divisible by 3.1 candy cost 12 candies cost 1+.50=1.50 ( here D is not an integer, hence we cannot buy 2 candies . so we can reject all cases where D is non Integer)3 candies cost 1.50 +1 =2.504 candies cost 2.50+.50= 35 candies cost 3+1= 46 candies cost 4+.50= 4.507 candies cost 4.50+1=5.508 candies cost 5.50.+.50= 69 candies cost 6+1= 7 .....13 candies cost =10(i) D is prime D=3 and N=4 (N is even)D=7 N=9 (N is odd )not sufficient (ii) D is not Divisible by 3 D=1 N=1D=4 N =5 D=7 N=9D=10 N=13so we see if D is not divisible 3 then N is always odd.Hence B is sufficient Hope it's clearIs there any way to do this problem within 2 mins.Writing out all the values takes time and one is bound to make mistakes.It took almost 4 mins for me to complete  Author: mau5 [ 04 Jul 2013, 12:45 ] Post subject: Re: At a particular store, candy bars are normally priced at$1. avinashrao9 wrote:Is there any way to do this problem within 2 mins.Writing out all the values takes time and one is bound to make mistakes.It took almost 4 mins for me to complete  trex16864 wrote:At a particular store, candy bars are normally priced at $1.00 each. Last week, the store offered a promotion under which customers purchasing one candy bar at full price could purchase a second candy bar for$0.50. A third candy bar would cost $1.00, a fourth would cost$0.50, and so on.If, in a single transaction during the promotion, Rajiv spent D dollars on N candy bars, where D and N are integers, is N odd?(1) D is prime.(2) D is not divisible by 3.Any integer can only have 3 values for remainder when divided by 3, namely (0,1,2).Hence, any integer which is not a multiple of 3 can be represented as $$3*k+1$$ or $$3*k+2$$, for some positive integer k(k=0 for 1 and 2).Also,for D=1,N=1(odd),D=3,N=4(even).Hence,any spending which is a multiple of 3-->$$3*k$$ will always yield --> even # of candy bars(as it is a multiple of 4)Any spending in the form $$3*k+1$$--> # of bars is $$even+1 -->odd$$.From F.S 1, for D = 7 , we can represent 7 as $$3*2+1$$ --> # of bars is $$4*2+1$$= 9 bars(odd)Again, for D = 3 dollars, we anyways know that N=4(even). Thus, as we get both possibilities,this statement is Insufficient.From F.S 2: As we know that D is not divisible by 3, he would always get an odd no of bars as discussed above.Sufficient.Hope this helps.B.

 Author: AccipiterQ [ 03 Nov 2013, 10:02 ] Post subject: Re: At a particular store, candy bars are normally priced at $1. trex16864 wrote:At a particular store, candy bars are normally priced at$1.00 each. Last week, the store offered a promotion under which customers purchasing one candy bar at full price could purchase a second candy bar for $0.50. A third candy bar would cost$1.00, a fourth would cost $0.50, and so on.If, in a single transaction during the promotion, Rajiv spent D dollars on N candy bars, where D and N are integers, is N odd?(1) D is prime.(2) D is not divisible by 3.This is written incorrectly, in the actual question (2) states 'D IS divisible by 3'  Author: matthewpearse [ 10 Nov 2013, 20:19 ] Post subject: Re: At a particular store, candy bars are normally priced at$1. hfbamafan wrote:Can this problem be turned into an algebraic expression?Hey bamafan,You can turn this into a system of equations as follows:$$D=\frac{3}{4}N$$ (when N is even)$$D=\frac{3}{4}N + \frac{1}{4}$$ (when N is odd)The nice thing about this is you can easily see for N to be an even integer, D must be divisible by three:$$\frac{4D}{3} = N$$ (when N is even)So that shows that the second case is sufficient. For the first case the odd formula can be rearranged as follows:$$\frac{4D-1}{3} = N$$ (when N is odd)From the first equation, D must be divisible by three to be even. D = 3 is prime and fits this rule, so an even N can be created.From the second equation, N is whole number if D = 7, 13, etc., so N can also be odd when D is prime. Therefore, the first case is insufficient.Matthew

 Author: Rohan_Kanungo [ 25 Jan 2014, 11:18 ] Post subject: Re: At a particular store, candy bars are normally priced at $1. Quote:D=\frac{3}{4}N + \frac{1}{4} (when N is odd)Hi Can someone please help me understand how we arrived at this expression for N = oddAccording to my understanding it should be D=\frac{3(N-1)}{4}+ 1  Author: Bunuel [ 27 Jan 2014, 02:10 ] Post subject: Re: At a particular store, candy bars are normally priced at$1. Rohan_Kanungo wrote:Quote:$$D=\frac{3}{4}N + \frac{1}{4}$$ (when N is odd)Hi Can someone please help me understand how we arrived at this expression for N = oddAccording to my understanding it should be $$D=\frac{3(N-1)}{4}+ 1$$Both equations are the same: $$D=\frac{3(N-1)}{4}+ 1=\frac{3N}{4}-\frac{3}{4}+1=\frac{3N}{4}+\frac{1}{4}$$.Hope it's clear.

 Author: doordie13 [ 13 Aug 2014, 04:38 ] Post subject: Re: At a particular store, candy bars are normally priced at $1. We are given that each odd candy costs$1.00 and each even candy costs $0.50.We can have 2 conditions:Case1: N is evenSo the total cost of all candies would be (1)*(N/2) + (0.5)*(N/2) = 3N/4 = DCase 2: N is oddTotal cost is [(1)*{(N+1)/2} + (0.5)*{(N-1)/2}] = (3N+1)/4 = DSt 1:D is primeN=4 (in case 1 where N is even) gives D =3 N=9 (in case 2 where N is odd) gives D = 7So we get prime values for D from both conditions, hence INSUFFICIENT.St 2:D is not divisible by 3Case 1 clearly shows D should be divisible by 3. Thus we can reject this.Case 2 clearly shows, D is not divisible by 3. Hence, SUFFICIENT.Therefore, B.Thanks. Author: KarishmaB [ 07 Jul 2016, 01:18 ] Post subject: Re: At a particular store, candy bars are normally priced at$1. trex16864 wrote:At a particular store, candy bars are normally priced at $1.00 each. Last week, the store offered a promotion under which customers purchasing one candy bar at full price could purchase a second candy bar for$0.50. A third candy bar would cost $1.00, a fourth would cost$0.50, and so on.If, in a single transaction during the promotion, Rajiv spent D dollars on N candy bars, where D and N are integers, is N odd?(1) D is prime.(2) D is not divisible by 3.Responding to a pm:In the question stem, what does "D and N are integers" imply?This is how the total cost progresses with each new candy bought:$1 -$1.50 - $2.50 -$3$4 -$4.50 - $5.50 -$6$7 -$7.50 - $8.50 -$9...Note that we have integer cost whenever we buy candies in multiples of 4 or 1 more than a multiple of 4. The total cost is a multiple of 3 for every multiple of 4 total candies (N is even) bought. It is 1, 4, 7, 10, 13 ... etc for every 4a+1 (N is odd) candies bought.Question: Is N odd?If N is odd, D = 1 or 4 or 7 or 10 etcIf N is even, D = 3, or 6 or 9 ...(1) D is prime.D can be 3 or 7. In one case, N is even, in the other it is odd.Not sufficient. (2) D is not divisible by 3.D cannot be 3, 6, 9 etc. So N is not even. N must be odd. Sufficient.Answer (B)

 Author: minustark [ 07 May 2020, 20:09 ] Post subject: Re: At a particular store, candy bars are normally priced at $1. trex16864 wrote:At a particular store, candy bars are normally priced at$1.00 each. Last week, the store offered a promotion under which customers purchasing one candy bar at full price could purchase a second candy bar for $0.50. A third candy bar would cost$1.00, a fourth would cost $0.50, and so on.If, in a single transaction during the promotion, Rajiv spent D dollars on N candy bars, where D and N are integers, is N odd?(1) D is prime.(2) D is not divisible by 3.From the stem, we can say that the pattern of spending on candy bar will be like: 1+0.50+1+.05+1+....... Statement 1 says the amount spent by Rajiv is a prime. So he can spend$3(here N =4), or 7 (N=9). Not sufficientStmnt 2 says D is not divisible by 3. Rajiv can buy 2 bars or 3 bars. Not sufficientTogether, D is a prime number greater than 3. We can try to make a pattern of the spending. For every 2k bars, D will be 1.5k. So for every 4 bar D will be 3k. To make D prime and odd, N always has to be odd.C is the answer

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