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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, 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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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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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'

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.

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

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27 Jan 2014, 02:10

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Expert's post

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 = 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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07 Jul 2016, 01:18

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Expert's post

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:

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.

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