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

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

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.

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

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

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: \(D=\frac{3(N-1)}{4}+ 1=\frac{3N}{4}-\frac{3}{4}+1=\frac{3N}{4}+\frac{1}{4}\).

Hope it's clear.

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.

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)

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

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