What is the sum of the digits of the positive...

[boiarski]

I believe the answer to the following question is incorrect [UPDATE: I misread something when I reviewed the question; the answer is actually correct]:

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) y^4 is a two-digit odd integer.

The answer to this question is listed as C for the following reason:

(1) INSUFFICIENT: n could be divisible by any square of a prime number, e.g. 4 (22), 9 (32), 25 (52), etc.

(2) INSUFFICIENT: This gives us no information about n. It is not established that y is an integer, so n could be many different values.

(1) AND (2) SUFFICIENT: We know that y is a prime number. We also know that y^4 is a two-digit odd number. The only prime number that yields a two-digit odd integer when raised to the fourth power is 3: 34 = 81. Thus y = 3.

We also know that n is divisible by the square of y or 9. So n is divisible by 9 and is less than 99, so n could be 18, 27, 36, 45, 54, 63, 72, 81, or 90. We do not know which number n is but we do know that all of these two-digit numbers have digits that sum to 9.

The correct answer is C.


The problem with the question is that the answer fails to consider that y is 2, since 2^4 is 16. So, since we can't tell whether y is 2 or 3 we can't tell what the two digits of n add up to. The answer then must be E. Interestingly, I actually answered C by making the exact same assumptions as in the explanation, while taking the CAT.

[frigorificos]

nope

it says y^4 is a two-digit odd integer. so y cannot be 2 as 2^4 = 16 which is even which leaves us with 3 only

Cheers
Buitre

[boiarski]

frigorificos, you're completely right. I seemed to have spotted it when I answered the question correctly, but when I was reviewing the CAT I must have just skipped over it. I need to get more sleep.

Thanks for the correction!

[JonathanSchneider]

: )

[malikrulzz]

Yup B. Smart one.

[JonathanSchneider]

Seems we're all clear here : )

[Beatrice Michael]

are you sure the ansvver is C

vvat if

y = √3, √5, √7, 3, etc. So y4 = 9 or 25 or 49 and y2 = 3 or 5 or 7 or 9 ... so n may be divisible
by 3 or 5 or 7 or 9 ... we are not sure what the sum of the digits of n will be ... suppose n = 21 or 25 ...
we get different answers for the sum of digits as 3 or 7... Not sufficient. Answer E.

[tim]

y has to be a prime number..

BTW there is a w key between q and e on most keyboards, thereby eliminating the need to write the letter v twice.. :)