Bunuel wrote:
samark wrote:
What is the sum of the digits of the positive integer n where n < 99?
1) n is divisible by the square of the prime number y.
2) y4 is a two-digit odd integer.
What is the sum of the digits of the positive integer n where n < 99?(1) n is divisible by the square of the prime number y --> clearly insufficient, as no info about y.
2) y^4 is a two-digit odd integer --> also insufficient, as no info about n, but from this statement we know that if y is an integer then y=3 (y must be odd in order y^4 to be odd and it cannot be less than 3 or more than 3 since 1^4 and 5^4 are not two digit numbers).
(1)+(2) Since from (1) y=integer then from (2) y=3, so n is divisible by 3^2=9. Number to be divisible by 9 sum of its digits must be multiple of 9, as n is two-digit number <99 then the sum of its digits must be 9 (18, 27, 36, ..., 90.). Suffiicient.
Answer: C.
Hello,
Thanks for all your responses and help as it is a great help for preparations
I figured out a flaw here , may be I am wrong but I needed your opinion.
The question only says that n is a positive integer and doesnot says anything about y,
I marked the answer "E".
As per all comments, we can see that 9 is happening to be the sum.
considering y is 3 , but since there is no mentioning of Y, it can be a -ve , fraction or decimal.
I took y as √7 then y raised to 4th power becomes 49.
So, this made me mark E.
Please could you tell me where did I go wrong.
Always thankful to you for your explanations.
Thanks
Harshit