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

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05 Sep 2014, 21:34

statement 1 : n is divisible by the square of the prime number y.

y can be 2,3,5 or 7. so n can be a multiple of 4,9,25 or 49. No definite answer

statement 2 : y^4 is a two-digit odd integer

statement is clearly insufficient

1 + 2 combined,

only y=3 can satisfy both conditions. so n is a multiple of 9 and n<99 Therefore n could be 9,18,27,36,45.....90 sum of digits for all the possibilities is 9. so combined statements are sufficient.

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.

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