What is the sum of the digits of the positive integer n

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.

+1 C

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.

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 ?

OK, so if we're going with your version...



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.

Statement 1 : Square of prime number less than 99 leaves us with 3,5,7,9 (Not sufficient)

Statement 2 : y^4 is a two digit odd integer (Not sufficient)

But when we combine both the we are left only with y=3 and the n is divisible by 9, so we get a unique answer which also 9

Answer : C

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

Fun Property of first ten multiples 9 ->
09
18
27
36
45
54
63
72
81
90

The units digit is increasing by 1 and units digit is decreasing by 1=> Sum of digits remain the same =9

From 1 and 2 => y must be 3
Hence x must be a multiple of 9 between 0 and 99
Hence sum of its digits must be 9


Hence C

From Statement 1, n is divisible by square of y .
y can be 1 , 4 , 9, 16 ...
Clearly Insufficient .

From Statement 2 , Y^4 is a two digit positive odd integer
y^4 = 81 => y = 3
No information on n .
So, statement is Insufficient .

Combine , we know that sum of digits of n is equal to 9 .
Answer is C .

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)

Statement 1. N = ay2 (y is a prime number)
N has different value for y = 3, 5, 7 etc. Hence, Insufficient.
Statement 2. y4 is a two digit odd integer.
y= 3.
But we don’t know the relation between ‘n’ and ‘y’ in statement 2. Hence, Insufficient.
Statement 1 & 2 together. Using the results of statement 1 and 2, we get,
n = ay2 and y = 3
n = 9a
So, n is a no divisible by 9. Any two digit no divisible by 9 has sum of its digits = 9.
Hence, Sufficient.

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

(1) says: n is divisible by the square of the prime number y

[color=#9e0039]How can it be inferred that n has to 81. S2 only confirms that y=3 , and if we apply it in S1 then we can say denom. has to be 9 but nothing is clear about numer. can be 9,18,27 ?? please guide

mohammad Ali Khan

You are right that combining both n can take all multiples values of square of number 3 which are 9,18, 27, 36, 45, 54, 63, 72, 81, 90.

But the question is sum of digits of n which is ALWAYS 9 for any of the above possible values. That’s why the answer is C

Hope it’s clear.

Posted from my mobile device

good question
#1
n is divisible by the square of the prime number y
y can be 2,3,5,7 many possiblities
#2
y^4 is a two-digit odd integer
y=2,3 again we have many values of n
from 1 &2
y=3 ; y^2 = 9 and sum of digits divisible by 9 has sum of digits =9
IMO C

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