IIT JEE , AIEEE Forum - Mathematics , Algebra

quart

no of pts having pos vector ai+bj+ck where a,c,b

{1,2,3,4,5} st 2^a +3^b +5^cis divisible by 4

answer is 70 plz explain

Sudheer111

Pls. make the question clear

nadeemoidu

If any of a,b,c differ , then the position vector will change.

So the question is just to find the different values of a, b, c which satisfy the given condition.

Here is the table which shows the remainder when 2,3 and 5 are divided by 4

x= 1 2 3 4 5

2^x) 2 0 0 0 0

3^x) 3 1 3 1 3

5^x) 1 1 1 1 1

For eg. 2^1 gives remainder 2 when divided by 4, 2^2 gives 0 , 3^1 gives 3 , 3^2 gives 1 ,etc

Any combination which gives a sum of 4 will give remainder 0 when divided by 4.

The possible combinations are 0 as remainder for 2^a , 3 as remainder for 3^b and 1 as remainder for 5^c = 4 x 3 x 5 = 60 cases,

and 2 as remainder for 2^a , 1 as remainder for 3^b and 1 as remainder for 5^c = 1 x 2 x 5 = 10

So total 70 cases are possible.

iitkgp_bipin

Well done nadeemoidu.

First of all arrange the tables for 2^x, 3^x, 5^x and pick up the possible combinations to find the total number of possible triplets (a,b,c).

Ask & Discuss Questions with Community & Experts