reminder

Is this a GMAT problem?

I would approach the problem in this manner (might not be correct)

Since 1999 ends with 9 the units digit of the squares, cubes, etc will be the same to the squares cubes of 9 alone.
List the squares of 9 and see the remainders

9^1 = 9 ; when divided by 7, r = 2
9^2 = 81 ; when divided by 7, r = 4
9^3 = 729 ; when divided by 7, r = 1
9^4 = 6561 ; when divided by 7, r = 2

Stop here and you see a sequence of r = 2,4,1

so when 9^10 and divided by 7 , r should be = 2

ans is 4

This is different from 4th cycle theorem an can be used with any exponential

1999^10 / 7 reminder ??
(1999x1999x1999......10times) / 7
1999/7 =4
4x4x4x4....10times /7
(4^3 x 4^3 x 4^3 x 4) / 7 OR (4^4 x 4^4 x 4^2) /7
1x1x1x4/7 = 4 OR 4x4x2/7 = 4

Vipul, is this a CAT question or a GMAT question?

I'm not sure that this is solvable using the techniques that the GMAT expects you to know. (It is a consequence of some of them, but so is ... pretty much everything in math. )

The CAT (I think) assumes that you know a bit of modular arithmetic, or at least the property that remainders are multiplicative. For example, the remainder of 10 when divided by 7 is 3, and the remainder of 12 divided by 7 is 5, so the remainder of 10*12 / 7 is the same as the remainder of 3*5 / 7.

Using this, we could say that the remainder of 1999 divided by 7 is 4, so 1999^10 will have the same remainder as 4^10 does when divided by 7. (In modular terms, 1999^10 is congruent to 4^10 mod 7.)

Simplifying further, 4^10 = 16^5. 16 has a remainder of 2 when divided by 7, so 16^5 will have the same remainder as 2^5 when divided by 7. 2^5 = 32, and the remainder of 32 divided by 7 is 4 ... so that's our answer.

Thank Matt for detailed explanation, reminder problems are very popular and this one cant be solved with 4th cycle method so i thought it worth of time, I am not sure whether it is GMAT or CAT type, my associate who is IIM shared this with me.