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Remainder Q - Any Other Faster Approach?

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Remainder Q - Any Other Faster Approach?

by madhukumar_v » Tue Jul 13, 2010 7:32 pm
1. What is the remainder when n is divided by 10?

1). The ten'sdigit of 11^n is 4

2). The hundred'sdigit of 5^n is 6

I found the below approach, but its time consuming to find the next power of eleven. Does any one know a better approach?

[spoiler]11^4 = 14641

similarly 11 ^ 14 = ....841

we get n = 4 , 14,24,34,44..... every time dividing by 10 leaves 4..
so stem(1) alone sufficient.

stem(2) after n>2

5 ^ 3 = 125
5^4 = 625
5^5 = 3125
5^6 = 15625 so for every n >2 and n is even we get 6 in hundreds place

so n can have 4, 6, 8,10,12... when divided by 10 leaves diffirent values so not sufficient..

clearly answer is (A) ie stem(1) alone sufficient and stem(2) not sufficient.

[/spoiler]
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by Brian@VeritasPrep » Wed Jul 14, 2010 4:32 pm
Hey madhukumar_v,

Interesting question - one thing that I always revert to with really big numbers (especially when exponents are involved) is that the GMAT loves to test your ability to find patterns and think about how numbers react. Two things stand out to me here:

1) I bet I can find a pattern...I just know that the GMAT wants me to.
2) 11 is a unique number when you multiply it because the tens and units digits are the same, so it's probably easier to multiply than meets the eye. 11 is 10+1, so if you multiply it you take the initial number (multiply by one) and add it to the same number but with a 0 on the end of it (multiply by 10). So:

11^1 = 11
11^2 = 11 + 110 = 121
11^3 = 121 + 1210 = 1331
11^4 = 1331 + 13310 = 14641
11^5 = 14641 + 146410 = 161051

Note: I didn't have this memorized, so I wouldn't ask anyone to memorize it, but if you train yourself to think about finding patterns and looking how numbers might lend themselves to patterns, I think you can replicate it.

Now, a pattern starts to show up - the tens digit is the same as the exponent. You can watch it tick over each time in the progression from 11 to 121 to 1331 to 14641...it goes 1, 2, 3, 4. At this point, I'd say that:

1) It's pretty likely that this pattern will continue as it's held for 5 straight exponents, so it seems pretty hard-and-fast. My only concern would be that something would change with exponents greater than one digit, but that's a lot of math to keep this up 5 more times, so I'd either take my chances that it's held this far and I don't see why it wouldn't, or:

2) Ask myself WHY the pattern holds. Doing this in practice you can get pretty quick at recognizing those little nuances in numbers that make these things work. Why does it hold? Because essentially what we're doing when we go to the next power of 11 is that we take the number we just had and add it to itself-times-ten. Well, we always have a units digit of 1, so "itself-times-ten" will always have a units digit of 0 and a tens digit of 1, so we're always just adding 1 to the previous tens digit. Knowing that, we know that the tens digit will continue to increase by one each time, meaning that when the tens digit is 4, that our exponent is either 4, 14, 24, 34, etc. The remainder will always be 4, and so statement 1 is sufficient.


I guess to summarize, when you know you're dealing with huge numbers, you're likely to be dealing with number properties. In those cases, look for a combination of:

1) Patterns in numbers
2) Unique ways to perform calculations with those numbers to help expose patterns
3) Reasons that the patterns hold so that you can confidently extrapolate them to bigger numbers
Brian Galvin
GMAT Instructor
Chief Academic Officer
Veritas Prep

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by madhukumar_v » Wed Jul 14, 2010 7:20 pm
Thanks Brian.
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