Stmt I INSUFF
20 not divisible by 15
60 divisible by 15
Stmt II INSUFF
3+6 = 9 is divisible by 3 where n is 3 is not divisible by 15
9+6 = 15 is divisible by 3 where n is 9 is divisible by 15
Stmt I and II SUFF
any number multiple of 20 ends with 0
any number to be multiple of 3 count of individual digits should be 3
if we add 6 to a number (since 6 is divisible by 3) original number should be divisible by 3
so n divisible by 5 and 3 which means divisible by 15.
This holds good for negative integers as well (-60+6=-54)
Multiples
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4meonly
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I like to solve this things by prime factorisation
Is the integer n a multiple of 15?
This means that n after prime factorisation should n have minimum 3 and 5 (3*5=15)
1. n is a multiple of 20.
20=2*2*5. Here is no 3, but n is a multiple of 20 means that it could be 300 (it is divided by 20 and 15)
Insufficient
2. n+6 is a multiple of 3.
means that n has 3 in prime factorisation.
But is it devided by 5? It could be 15 (15+6=21. deveide by 3, 21+6=27)
insufficient
1 and 2.
sufficient
N has 5 and 3 it is devided by 15
Answer C
Is the integer n a multiple of 15?
This means that n after prime factorisation should n have minimum 3 and 5 (3*5=15)
1. n is a multiple of 20.
20=2*2*5. Here is no 3, but n is a multiple of 20 means that it could be 300 (it is divided by 20 and 15)
Insufficient
2. n+6 is a multiple of 3.
means that n has 3 in prime factorisation.
But is it devided by 5? It could be 15 (15+6=21. deveide by 3, 21+6=27)
insufficient
1 and 2.
sufficient
N has 5 and 3 it is devided by 15
Answer C












