j_shreyans wrote:If M and N are integers, is 10^M+N/3 an integer?
(1) N=5

(2) MN is even

OAE

Question : Is [(10^M)+N]/3 an integer?

Basically the the question ask whether 10^m+n is divisible by 3. Now, in order 10^m+n to be divisible by 3:
A. It must be an integer, and B. the sum of its digits must be multiple of 3.

(1) N = 5 --> if m<0 (-1, -2, ...) then 10^m+n won't be an integer at all thus won't be divisible by 3, but if m is (0, 1, 2, ...) then 10^m+n will be an integer and also the sum of its digits will be divisible by 3 (for example for m=1 --> 10^m+n=10+5=15 --> 15 is divisible by 3).
Not sufficient.

(2) MN is even --> clearly insufficient, as again m can be -2 and n any integer and the answer to the question will be NO or m can be 0 and n can be 2 and the answer to the question will be YES. Not sufficient.

(1)+(2) From mn=even and n=5 it's still possible for m to be negative even integer (-2, -4, ...), so 10^m+n may or may not be divisible by 3. Not sufficient.

Answer: Option E

Sir ,

If we combined both statement that N=5 and MN is even so M should be (-2,2,-4,4.....)

so if we put the value in 10^M+N/3 so it is divisible .

so the ans should be C .

Pls help if i am taking something wrong.

The problem can be rephrased as follows:

If M and N are integers, is (10^M)+N a multiple of 3?
(1) N=5

(2) MN is even

Both statements are satisfied if N=5 and M=2.
In this case, (10^M) + N = 10² + 5 = 100+5 = 105, which is a multiple of 3.

Both statements are satisfied if N=5 and M=-2.
In this case, (10^M) + N = 10ˉ² + 5 = 0.01 + 5 = 5.01, which is NOT a multiple of 3.

Since (10^M) + N is a multiple of 3 in the first case but not a multiple of 3 in the second case, the two statements combined are INSUFFICIENT.

The correct answer is E.

Hey Guys ,

Thanks for making me correct i was taking in other ways.

Thank u soo much......

A general note here: we know that the sum of the digits of a number tell us the remainder of that number when divided by 3. For instance, 100 has a remainder of 1 when divided by 3, since the sum of its digits (e.g. 1) has a remainder of 1 when divided by 3.

Assuming that m is a non-negative integer, 10� is just a 1 followed by m 0s. This number will ALWAYS have a remainder of 1 when divided by 3. 10� + 5 will thus ALWAYS be divisible by 3. For instance

1 + 5 = 6
10 + 5 = 15
100 + 5 = 105
1000 + 5 = 1005
etc.

However, we've made one crucial assumption here: that m is a nonnegative integer! If m is -1, -2, etc., then 10� + n won't be an integer at all, and hence won't be divisible by 3. So the answer is E.