two digits -same digits.HELP!!!!!

answer is 1

any 2 digit number(ab) can be represented as 10a+b (ex 23 as 10*2+3)

so in case let's say M=ab=10a+b
and since N is reverse of m so N=ba=10b+a

so M+N=11(a+b)

so sum has to be amultiple of 11.
all options are mulitple of 11 except 1.

Great explanation, abishekg21!

One other note on this one - if you didn't see a way to think it through algebraically and had to make a quick decision, another way to think this particular problem through is to look at the constraints:

You need two 2-digit numbers to sum to the answer choices:

181
165
121
99
44

Your range of potential 2-digit numbers is between 11 and 99 (10 doesn't work because the inverse would be 01, a one-digit number), so the range of sums is going to be between 22 and 198, with quite a few possibilities to sum 2-digit numbers toward the middle of that range but very few opportunities to hit at the ends (the only way to get 22 is 11 and 11; the only way to get 198 is 99 and 99).

So you may want to start at the ends because there are fewer combinations of 2-digit numbers (independent of the inverse digits) to even try to add to those numbers. 44 should be easy to spot - 22 + 22, but 181 requires:

A max of 99, meaning that the minimum second value is 82.

So you'd have to use the digits 8 and 9 to even have a chance. Your only options are 88+88, 98+89, and 99+99, none of which work, so choice A is a pretty quick elimination.



Now, that may only work on this question, so I'd recommend being comfortable enough to use Abishek's algebraic way, too, but when you're dealing with sums/multiples/etc. it's certainly possible to limit your options by looking at the potential ranges of values, and often you can make quick decisions that way.

thanks guys!!!