What is the reminder when 11^122 is divided by 100?

Hello sorry for late reply
Leaving it as (20+1)^61/100 will give you the same answer

Now what I was doing is to treat the remainders as dividends
.: 20^60 is divisible by 100 right and we have 20^61 So remainder 20 and 1^61 remainder 1 ,

We have (20+1)/100 = 21/100
Hope it helps

Thanks Staphyk for your reply.

I'm little unable to understand why 20^60 is divisible by 100 and 20^61 leaves remainder 20. I think both are divisible by 100 without reminder.

What do I miss?

Thanks

Mo2men You can very well apply the concept of cyclicity here. 11^1 gives 11 and 11^2 gives 21 as the last two digits and similarly, the cyclicity of this can be easily found to be 10.

So, 122 can be written in the form = 12*10 + 2 OR 10*N + 2

Now, it's very easy to see that 11^122 will have 21 as the last two digits and on division by 100, will give 21 as the remainder.

Simple solution. Will get you to the answer anywhere between ~45 secs and 1 min.

PLEASE HIT KUDOS IF YOU LIKE MY SOLUTION

I already did it and found the cyclicity of 10 as you mentioned but I'm interested more in first post. I find it easier and time efficient.
Thanks

Also tried the pattern method and the one I posted but don’t know how to further explain to you

Posted from my mobile device

I do wonder if people writing these questions have ever seen an actual GMAT. This question is so far outside the scope of the test that there isn't any reason to study it.

But if you did want to solve: we just want to know the last two digits (ten and units digits) of 11^122, since those will give us the remainder when we divide 11^122 by 100. If you've ever solved a question that asks something like "what is the units digit of 8^31", you will have seen that as we work out the units digits of 8^1, 8^2, 8^3 and so on, we start to see a repeating pattern. The same thing will happen here or in any other remainders question:

11^1 ends in 11
11^2 ends in 21
11^3 ends in 31
11^4 ends in 41

and so on, and you can probably predict now how the pattern develops. When we reach 11^10, we'll have a number ending in "01", and then the pattern will begin to repeat; 11^11 ends in 11, 11^12 ends in 21, and so on. Since the pattern repeats in blocks of ten terms, 11^120 is just like 11^10 (so will end in "01"), and 11^122 will be exactly like 11^2, and its last two digits are "21", so that's the answer.

It's not remotely close to being a realistic GMAT question though.

Hi IanStewart

First thanks for you quick prompt. I solved as you did but I wonder if there is a short cut like first poster did but I lost him in certain step and do not understand how he proceeded. I wish if you can help with it

Second, I suspect that it is out of GMAT scope so I trust your view and your directions, although I try to build some stamina.

Thanks Staphyk a lot my friend for your help and support at least you tried

I don't think that solution is correct.

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