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At a particular store, candy bars are normally priced at $1. [#permalink]

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13 Jan 2013, 03:16

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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?

Re: At a particular store, candy bars are normally priced at $1. [#permalink]

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15 Jan 2013, 06:37

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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 1 2 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 =2.50 4 candies cost 2.50+.50= 3 5 candies cost 3+1= 4 6 candies cost 4+.50= 4.50 7 candies cost 4.50+1=5.50 8 candies cost 5.50.+.50= 6 9 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=1 D=4 N =5 D=7 N=9 D=10 N=13

so we see if D is not divisible 3 then N is always odd.

Re: At a particular store, candy bars are normally priced at $1. [#permalink]

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04 Jul 2013, 12:45

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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.

Re: At a particular store, candy bars are normally priced at $1. [#permalink]

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10 Nov 2013, 20:19

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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.

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:

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 etc If 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.

Re: At a particular store, candy bars are normally priced at $1. [#permalink]

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04 Jul 2013, 10:34

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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 1 2 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 =2.50 4 candies cost 2.50+.50= 3 5 candies cost 3+1= 4 6 candies cost 4+.50= 4.50 7 candies cost 4.50+1=5.50 8 candies cost 5.50.+.50= 6 9 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=1 D=4 N =5 D=7 N=9 D=10 N=13

so we see if D is not divisible 3 then N is always odd.

Hence B is sufficient

Hope it's clear

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.

Re: At a particular store, candy bars are normally priced at $1. [#permalink]

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03 Nov 2013, 10:02

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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'

Can someone please help me understand how we arrived at this expression for N = odd According to my understanding it should be \(D=\frac{3(N-1)}{4}+ 1\)

Both equations are the same: \(D=\frac{3(N-1)}{4}+ 1=\frac{3N}{4}-\frac{3}{4}+1=\frac{3N}{4}+\frac{1}{4}\).

Re: At a particular store, candy bars are normally priced at $1. [#permalink]

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03 Sep 2015, 17:40

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Re: At a particular store, candy bars are normally priced at $1. [#permalink]

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29 Aug 2017, 03:19

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