Advanced Maths Doubt?

Question

Hi, i m in college. i have few question types dealing with large nos. that i believe are Quickly solvable:- I need the general methods to solve these quickly.

1)What is the last 2 digits when a numebr is divided by another? Eg. When (23^222) is divided by 4.

2)What is the last digit when a numebr is divided by another?

-Thhanks.

Can u please also link me to a Web Resource for similar maths questions(w ans). I want examples

Or better still, please solve this Example itself:-
"find last digit & last-2-digits when 23^123 is divided by 4"

-Thanks

sorry, just take a similar example - (23^x)/4 where 23^x is completely div. by 4

Answer 1

You just have to figure a pattern...
23^1 = 23
--> 23 / 4 = 5.75
Last digit = 5

23^2 = 529
--> 529 / 4 = 132.25
Last digit = 2

23^3 = 12167
--> 12167 / 4 = 3041.75
Last digit = 1

23^4 = 279841
--> 279841 / 4 = 69960.25
Last digit = 0

23^5 = 6436343
--> 6436343 / 4 = 1609085.75
Last digit = 5

23^6 = 148035889
--> 148035889 / 4 = 37008972.25
Last digit = 2

23^7 = 3404825447
--> 3404825447 / 4 = 851206361.75
Last digit = 1

23^8 = 78310985281
--> 78310985281 / 4 = 19577746320.25
Last digit = 0

So the pattern of the last digit is:
5, 2, 1, 0, 5, 2, 1, 0, etc.

Take the exponent and divide by 4, then take the remainder:
1/4 --> r1 --> 5
2/4 --> r2 --> 2
3/4 --> r3 --> 1
4/4 --> r0 --> 0
5/4 --> r1 --> 5
6/4 --> r2 --> 2
7/4 --> r3 --> 1
8/4 --> r0 --> 0
etc.

See the pattern?

So if we take the exponent (222) and divide by 4, you get 55 r2. Anytime you have r2 --> 2

So 23^222 will have 2 as the last digit.

Answer 2

can u clarify ur question as 23^222 is not divisible by 4.