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Bunuel
n is a positive integer. What is the remainder when n is divided by 6?

(1) n is a multiple of 3.
(2) When n is divided by 2, the remainder is 1.

Ans: C

Reminder when n divided by 6
Stat:1) n is multiple of 3: we know every even multiple of 3 will leave reminder 0 when divided by 6 and every odd multiple of 3 will leave reminder 3 when divided by 6. (Not Sufficient to ans)

Stat:2) when n is divided by 2 reminder is 1: so n = 2I+1
means n= 1,3,5,7,9,11,13,15.... in this case when n divided by 6 reminder is 1,3,5,1,3,5... respectively so (Not Suffi.)

Together Stat :1 & 2) n is a multiple of 3 and leaves 1 as reminder when divided by 2 means all odd multiple of 3.
So n= 3, 9, 15 and so on.. reminder when n divided by 6 is 3,3,3,... respectively.. so [Sufficient]

Ans: C
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Bunuel
n is a positive integer. What is the remainder when n is divided by 6?

(1) n is a multiple of 3.
(2) When n is divided by 2, the remainder is 1.


n = 6x + R , where R<6. We need to find R.

1) n = 3T, does not give a definite value for R. Insufficient

2) n = 2y + 1, on multiplying b/s by 3, we get

3n = 6y + 3, can be modified to 3n - 2n = 6y + 3 - 2n which gives n = 6y + 3 - 2n. on its own statement 2 is also insufficient to solve for R

Combining 1) & 2) n = 6y + 3 - 2(3T) or n = 6y + 3 - 6(T) or n = 6(y - T) + 3, which is in the form of n = 6x + R so, R = 3, Sufficient. Answer is C.


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n is a positive integer. What is the remainder when n is divided by 6?

(1) n is a multiple of 3.
(2) When n is divided by 2, the remainder is 1.[/quote]

From (1) , n=3a i.e 3,6,9,12,15,18....etc.

This divided by 6 will give either 0 or 3 as remainder. Hence NOT SUFFICIENT

From (2) , n=2b+1. Consider putting values of b as 1,2,3....etc, n=3,5,7,9

Values of n are multiples and remainders too will be different for each one of them.

Combining (1) + (2) , n = 6K + 3, which when divided by 6 will give remainder as -

6K/6 + 3/6 => 0 + 3 => 3 (remiander). Correct Answer is C.

Pls note n =6K +3 can be found as follows -

3a = 2b+1
=> a = (2b+1)/3
Putting values of b as 1,2,3...etc.
b=1 ,a =1
b=2,a = 5/3 (not an integer)
b=3,a= 7/3 (not an integer)
b=4,a=3

We see least value of n can be found when a =1 or b =1. Incorporating this n=3
N = least number obtained + (LCM of divisors)*K , where K is a non negative integer.
= 3 + 3a*2b = 3+6ab =(3+6K)
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Given: n is a positive integer. Asked: What is the remainder when n is divided by 6?

(1) n is a multiple of 3.
remainder when n is divided by 3 = 3 or 6
Not sufficient

(2) When n is divided by 2, the remainder is 1.
n= 2k+1
Remainder when n is divided by 6 = {1,3,5}
Not sufficient

(1)+(2)
Remainder when n is divided by 6 = 3
Sufficient

IMO C

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