if a and b are distinct prime numbers, Find the remainder when a+b is

Question

if a and b are distinct prime numbers, Find the remainder when a+b is divided by 2? 1) a < b 2) Sum of the distinct positive factors of a^b is ODD Source : https://www.GMATinsight.com

chetan2u · Answer

Divison by 2 will lead to only two remainders, either 1 or 0.

Therefore, if a and b are odd, ans is 0, as a+b=even
But if a or b is 2, then a+b=Odd, and remainder will be 1.

REQUIREMENT :- If any of a or b is 2.

1) a < b
So a=2, and b=3...a+b=5 and remainder is 1
a=3 and b=5...a+b=8 and remainder is 0.

2) Sum of the distinct positive factors of  a^b  is ODD
 Now there can be two cases.. 
a)  a is even ....All the factors will be even except factor 1, and sum of these factors will be ODD, so  b could be anything but different from 2 . Thus a=2, and b=odd, and remainder =1.
b)  a is odd ....Since a is odd prime, all its factors will be odd. So, for the SUM to be odd, we will have to have odd number of factors, that is b+1 is odd, so  b is even or 2 .
 Also ODD number of factors means the number is a SQUARE of an integer. 
Hence a+b=O+E=O, and the remainder will be 1.
 Thus, in both the cases, the remainder is 1. 
Suff

B

The statement II could have been as below( That is what I read it as initially) and still the answer would be B. 
2) Number of the distinct positive factors of  a^b  is ODD
Since a and b are prime numbers, the number of distinct factors of  a^b  is b+1. It is given that b+1 is odd, so b is even or 2.
 Also ODD number of factors means the number is a SQUARE of an integer. 
Hence a+b=O+E=O, and the remainder will be 1.
Suff

preetamsaha · Answer

if a and b are distinct prime numbers, Find the remainder when a+b is divided by 2?

1) a < b
say,a=2,b=3 ; the remainder when a+b is divided by 2 = 1
again,a=3,b=5 ; the remainder when a+b is divided by 2 = 0  insufficient

2) Number of the distinct positive factors of a^b is ODD
say,a=2,b=3 ;Number of the distinct positive factors of 2^3 is =3+1=4 ( NOT ODD ) so,it can't be the case.
so, b can't be an odd number.
so,  b must be an even number  so that we can find the Number of the distinct positive factors as odd. (EVEN+1=ODD)
only even prime number = 2 (b=2)
so, a^2 (a may be any prime number except 2 ,it wouldn't effect the result ) is the general term. 
now,  a must be odd  ( all prime numbers are odd except 2)
so,a+b = odd 
so, the remainder when a+b is divided by 2 = 1  sufficient

correct answer B

GMATinsight · Answer

Question: Remainder when a+b is divided by 2 = ?

Question REPHRASED: Is a+b Odd?

a+b will be odd when one of a and b is 2

Statement 1 : a < b 
Case 1 : a = 2 and b = 3 i.e. a+b = odd
Case 2 : a = 3 and b = 5 i.e. a+b = even hence

NOT SUFFICIENT

STatement 2: Sum of the distinct positive factors of a^b is ODD

Case 1: a = 2 and b = 3, a^b = 2^3 = 8 and sum of the factors is odd and a+b = odd

Case 2: a = 3 and b = 2, a^b = 3^2 = 9 and sum of the factors is odd and a+b = odd

Case 3: a = 3 and b = 5, a^b = 3^5 = 243 and 6 odd factors sum of the factors is Even  so the values a and b both ODD can NOT be accepted

But a+b is always Odd hence 
 SUFFICIENT

Answer: Option B

GMATinsight · Answer

chetan2u . preetamsaha Why should b be even? (Check highlighted part) a^b could be 2^3 = 8 factors {1, 2, 4, 8} i.e sum of factors = odd and NO CONTRADICTION Let me know if I misunderstood something... Always ready to improvise.

chetan2u · Answer

Hi, 
Apology. I read it as 'number of distinct factors', and NOT 'sum of distinct factors', and my solution takes it as number of distinct factors.
By the way, this can be another question, where statement gives number of distinct factors is ODD, and we still get the answer as B.
I'll edit the solution in light of the error.

preetamsaha · Answer

GMATinsight 
yes, you are right. i have misread the question and treated the problem as "number " instead of sum. anyways, with that case ,
we have solved two problems in one shot. 
thanks.

hiranmay · Answer

if a and b are distinct prime numbers, Find the remainder when a+b is divided by 2?

1) a < b --> insuff: if a=2<b=3, then a+b=5, remainder=1, but if a=3 <b=5, then a+b=8, remainder=0
2) Sum of the distinct positive factors of a^b is ODD --> suff: a^b has (b+1) factors i.e. b+1=odd=> b=even & there is only one even prime number & other prime numbers are all odd, so b=2 and a not =b, b = odd prime number, so a+b= odd+even=odd, remainder=1
Answer: B

Archit3110 · Answer

For prime no a+b would be either even or odd if bothe are odd or either of one is 2 ; so we would get 0 or 1 as remainder when divided by 2;
#1 a<b ; we can have 2<3 or 3<5 insufficient 
#2 a^b sum of +ve factors is odd possible when odd^even eg 3^2.. so remainder is 1 always when divided by 2 hence sufficient option B

sampad · Answer

total number of factors and sum of the distinct positive factors are different. Here all the solutions are based on the total number of factor. But in the problem it stated that we need to check for sum of the distinct positive factors. I think we need to revisit the solution.

if a and b are distinct prime numbers, Find the remainder when a+b is

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if a and b are distinct prime numbers, Find the remainder when a+b is

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