At a particular store, candy bars are normally priced at $1.

Question

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.

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.

stne · Accepted Answer

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

KarishmaB · Answer

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.

Answer (B)

hfbamafan · Answer

Can this problem be turned into an algebraic expression?

avinashrao9 · Answer

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

mau5 · Answer

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.

AccipiterQ · Answer

This is written incorrectly, in the actual question (2) states 'D  IS  divisible by 3'

matthewpearse · Answer

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

Rohan_Kanungo · Answer

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

Bunuel · Answer

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.

doordie13 · Answer

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 even

So the total cost of all candies would be (1)*(N/2) + (0.5)*(N/2) = 3N/4 = D

Case 2: N is odd

Total cost is [(1)*{(N+1)/2} + (0.5)*{(N-1)/2}] = (3N+1)/4 = D

St 1:

D is prime

N=4 (in case 1 where N is even) gives D =3
N=9 (in case 2 where N is odd) gives D = 7

So we get prime values for D from both conditions, hence INSUFFICIENT.

St 2:

D is not divisible by 3

Case 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.
:-D

minustark · Answer

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 sufficient
Stmnt 2 says D is not divisible by 3. Rajiv can buy 2 bars or 3 bars. Not sufficient

Together, 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

bumpbot · Answer

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At a particular store, candy bars are normally priced at $1.

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