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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.

1. INSUFFICIENT e.g. take y = 2, if n = 6, 6 x 2^2 = 24, sum of digits is 6 if n = 3, 3 x 2^2 = 12, sum of digits is 3

2. INSUFFICIENT e.g. y = 3 then y^4 = 81 (a 2 digit odd number) if y = 5^1/2 then y^4=25 (a 2 digit odd number)

1. & 2. SUFFICIENT y = 3 (the only prime number with an odd two digit result when raised to the 4th power) therefore n is a multiple of 9, and the sum of the digits of all multiples of 9 is 9.

1. & 2. SUFFICIENT y = 3 (the only prime number with an odd two digit result when raised to the 4th power) therefore n is a multiple of 9, and the sum of the digits of all multiples of 9 is 9.

The sum of the digits of all multiples of 9 is divisible by 9. It is not generally equal to 9. For example, 99 is divisible by 9, and the sum of its digits is 18.

That turns out not to affect the solution here, however, since we are only concerned with two-digit numbers less than 99, and the sum of the digits of every multiple of nine between 18 and 90 inclusive is always 9.
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Stmt 1 tells you that y is a small prime integer because the square has to be lower than 99 and has to have a multiple lower than 99. This leaves us 3 and 5 but we. So insuff

stmt 2 tells us that y is 3 or less but is insuff by itself because the stem of the question does not say anything about y

together y = 3 and stmt 1 tells us that it is divisible by 9 and for something to be divisible by 9 the digits add up to 9.

I think your conclusion is correct, harsh6239, though I'm pretty sure that this is an MGMAT question, and I found a slightly different version of it elsewhere:

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.

So if we go with the version that specifies that y is prime, then...

Statement 1: y could be any number of primes, which means that n must be divisible by 4, 9, 25, or 49. And that really doesn't narrow things down very much. -- not sufficient

Statement 2: tells us absolutely nothing about n. -- not sufficient

Together: since we know that y is prime, then y^4 could be 16 or 81... except that y^4 has to be odd. So y^4 must be 81. And if n < 99, then n must also be 81.

The answer is C, and you were completely correct.
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Re: What is the sum of the digits of the positive integer n [#permalink]

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23 Nov 2012, 21:05

GMATNinja wrote:

I think your conclusion is correct, harsh6239, though I'm pretty sure that this is an MGMAT question, and I found a slightly different version of it elsewhere:

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.

So if we go with the version that specifies that y is prime, then...

Statement 1: y could be any number of primes, which means that n must be divisible by 4, 9, 25, or 49. And that really doesn't narrow things down very much. -- not sufficient

Statement 2: tells us absolutely nothing about n. -- not sufficient

Together: since we know that y is prime, then y^4 could be 16 or 81... except that y^4 has to be odd. So y^4 must be 81. And if n < 99, then n must also be 81.

The answer is C, and you were completely correct.

Thank you sir for your reply. The slightly different version of question which you have given here states that y is a prime and hence an integer. My confusion in the original question which I had posted is that if we see statement 2 then y^4 is equal to a two digit odd integer and so y need not be an integer always. so going back to statement 1 then y^2 may be a fraction and not necessarily 9. In this case answer will be E. How to confirm this one mathematically ?

What is the sum of the digits of the positive integer n where n<99 ?

1) n is divisible by the square of y.

2) y*y*y*y (y raised to the power 4) is equal to a two digit positive odd integer.

Statement 1: We don't even know that y is an integer, so the square of y could be anything... which means that n could be anything, too. (The square of y could, for example, be 1.) -- incredibly insufficient

Statement 2: We don't know that y is an integer, so y^4 could be any two-digit positive odd integer... and that gives us 45 possible values for y^4, most of which are not integers. And the statement says absolutely nothing about n. -- still incredibly insufficient

Together: y^4 could be any two-digit odd integer, which means that we have tons of possible values for y^2, including 5, 7, and 9. N could then be any multiple of 5, 7, or 9. -- still not sufficient

So in the version without the phrase "prime number" in statement 1, the answer would definitely be E.
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Stmt 1 tells you that y is a small prime integer because the square has to be lower than 99 and has to have a multiple lower than 99. This leaves us 3 and 5 but we. So insuff

stmt 2 tells us that y is 3 or less but is insuff by itself because the stem of the question does not say anything about y

together y = 3 and stmt 1 tells us that it is divisible by 9 and for something to be divisible by 9 the digits add up to 9.

I think your conclusion to (1) is incomplete. y could also = 7, because (7^2)*2 = 98, which is < 99.

Re: What is the sum of the digits of the positive integer n [#permalink]

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04 Aug 2017, 01:34

n<99 or n can be any of the following no 98, 97......1

Statement 1- n is divisible by the square of the prime number y

Prime nos 2, 3, 5, 7, 11.....

Square of prime 4, 9, 25, 49, 121- n can be divisible by any of the above prime nos till 7-since n<99

Statement-2-y⁴ is a two-digit odd integer- what is y-No information hence statement is not sufficient

Statement 1 and 2. y is prime-y² divides n and y⁴ is a 2 digit odd no.

Test the nos 2-is even 3-3²=9 and 3⁴=81=odd. Hence the no is divisible by 9 and any no is divisible by 9 if its sum must be divisible by 9. since n<99 the n is a 2 digit no whose sum=9(18, 27, 36, 72, 81)