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## Events & Promotions

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13 Jan 2013, 03:16
3
22
00:00

Difficulty:

95% (hard)

Question Stats:

35% (02:58) correct 65% (03:05) wrong based on 826 sessions

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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?

(1) D is prime.

(2) D is not divisible by 3.
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27 Jun 2013, 21:26
1
Can this problem be turned into an algebraic expression?
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04 Jul 2013, 12:45
2
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.
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10 Nov 2013, 20:19
2
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.

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27 Jan 2014, 02:10
1
Rohan_Kanungo wrote:
Quote:
$$D=\frac{3}{4}N + \frac{1}{4}$$ (when N is odd)

Hi

According to my understanding it should be $$D=\frac{3(N-1)}{4}+ 1$$

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

Hope it's clear.
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07 Jul 2016, 01:18
2
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 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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Re: At a particular store, candy bars are normally priced at $1. [#permalink] ### Show Tags 29 Sep 2018, 09:11 Hello from the GMAT Club BumpBot! Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos). Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________ Re: At a particular store, candy bars are normally priced at$1.   [#permalink] 29 Sep 2018, 09:11
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