answer is 181

let A be xy then B = yx
or A = 10x+y & B = 10y+x
so A+B must be a multiple of 11.
Hence only 181 doesn't fit the bill here.

smallsorrow wrote:If the two-digit integer M and N are positive and have the same digits, but in reverse order, which of the following cannot be the sum of n and m?

a) 181
b) 165
c) 121
d) 99
e) 44

Is there a fast way to solve these kind of problems? It took me ages to find the solution. It is pretty obvious to exclude D and E ... but with the other 3... mmhhh..

No has to be multiple of 11.

smallsorrow wrote:If the two-digit integer M and N are positive and have the same digits, but in reverse order, which of the following cannot be the sum of n and m?

a) 181
b) 165
c) 121
d) 99
e) 44

Is there a fast way to solve these kind of problems? It took me ages to find the solution. It is pretty obvious to exclude D and E ... but with the other 3... mmhhh..

another way at looking at this

its saying that both M and N are positive and their orders are being switched, along with the fact its a 2 digit number

M: 10t + s
N: 10s + t

t and s are just randome variables i picked

NOW, they are saying which one CANNOT be the sum of these two numbers, N & M

10t + s + 10s + t = 11s + 11t

that right there tells us the number MUST be divisible by 11

go through each answer choice and see which answer choice is NOT divisible by 11

a) 181 181/11 = Not divisible
b) 165 165/11 = 15
c) 121 121/11 = 11
d) 99 99/11 = 9
e) 44 44/11 = 4

Hopes this helps, it removes all the guess and check work