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.