Lets start with the most important section in Quantitative section,Numbers..

Numbers
================
Numbers has lot of properties to be studied in detail and certain set piece questions are there.from where CAT questions are frequently coming..

Power cycle
Power cycle for digits 2,3,4,5,6,7,8,9

Divisibility checks
2,3,4,5,6,7,8,9,11,13…

Remainder questions
a) Using power cycle
b) Using binomial theorem

Highest power of a number in n!

Problems related to factors of a number
No. of factors.
Sum of factors
No. of co-primes
Product of factors

Express a no. in different forms
Expressing as difference of squares.

Fermats theorem
Eulers no.
Solving remainder questions using Fermat’s Theorem

Miscellaneous properties
1.Properties of a^n- b^n when ‘n’ is odd,even…
2.n^p-p when p is prime.

a. Lets start with Power cycles

****The frequency of the power cycles of various digits is very important.
This helps you to solve remainder questions involving large powers.*****

Let us take some example.

1. Find the remainder of 3^75/5?

Explanation:-

We know the power cycle frequency of 3.Its 4and the power cycle is 3,9,7,1
So convert the large power in the question (here 75) to a much smaller number using this freq. 75=4*18+3
Therefore 3^75 is now 3^{(4*18)+3},this can still be shortened as 3^3 only.
Since the other power frequencies will be repeating only.
Hence our question reduces to finding rem. of 3^3/5
This is quite easy, 27/5,rem. is 2.(or since we know power cycle of 3,3^3 ends with 7,so remainder is 7/5=2).

2.Find the remainder of 2^102/3?

Explanation:-
We know the power cycle frequency of 2.Its 4and the power cycle is 2,4,8,6.
Now we convert the large power in the question (here 102) to a much smaller number using this freq. 102=4*25+2.
Therefore 2^102 is now 2^{(4*25)+2},this can still be shortened as 2^2 only.
Since the other power frequencies will be repeating only.
Hence our question reduces to finding rem. of 2^2/3.Its 4 /3,remainder is 1.

OR

This question can be done in a very easy method.
We knew that rem. of 2/3 is -1.

3.Find the rem. of {(2^203)*(3^506)}/5?

Explanation:-
Here we can use the fact that ,

*****The rem. of a product is the product of the individual rem.****

So the ques. changes to rem(2^203)/5 * rem.(3^506)/5
Rem(2^203)/5 can be found out by using power cycle..
203 = 4*50 +3 (since 4 is power cycle freq of 2).
So rem(2^203)/5 changes to rem2^3/5 ie rem8/5 =3.

Rem(3^506)/5 can also be found out by using power cycle.
506=4*126+2 (since 4 is power cycle freq of 3).
So rem(3^506)/5 changes to rem3^2/5 ie rem9/5 =4

Now indiv. rem. are 3 and 4.
Whole remainder is rem(3*4/5) { since the product is greater than devisor,we need to take rem.of the product again.
So answer is rem(12/5) ie 2.


Now rem. of 2^102/3 is equivalent to (rem. 2/3) ^102 that is (-1)^102 =1.

****For finding out the unit place of a very large power of a no. we can use power cycle****



Points to be remembered.
===============================

The frequency of the power cycles of various digits is very important.
This helps you to solve remainder questions involving large powers.
The rem. of a product is the product of the individual rem.
For finding out the unit place of a very large power of a no. we can use power cycle.