Is it dissolvable! sorry divisible -part 2!

Probability Level 4

The integers a , b a,b are selected from 5 n 5n consecutive integers: 1 , 2 , 3 , 4 , … , 5 n − 1 , 5 n {1,2,3,4, \ldots, 5n-1,5n} . Find the sum of the digits of number of unordered ways can these integers be selected , such that a 2 − b 2 a^2-b^2 is divisible by 5 when n = 100 n=100 .


The answer is 20.

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Gautam Sharma by Gautam Sharma , 23, India

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1 solution

Gautam Sharma
Feb 23, 2015

Every whole number can be expressed in the form of 5 λ + i , i ∈ 0 , 1 , 2 , 3 , 4. 5\lambda+i,i\in0,1,2,3,4.

( a 2 − b 2 ) = ( a + b ) ( a − b ) (a^2-b^2)=(a+b)(a-b)

CASE 1

( a + b ) is divisible by 5 when (a,b) of the form (a+b)\text{is divisible by 5 when (a,b) of the form } ( 5 λ 1 , 5 λ 2 ) ( 5 λ 1 + 2 , 5 λ 2 + 3 ) , (5\lambda_1,5\lambda_2)(5\lambda_1+2,5\lambda_2+3), ( 5 λ 1 + 1 , 5 λ 2 + 4 ) (5\lambda_1+1,5\lambda_2+4)

CASE 2

( a − b ) is divisible by 5 when (a,b) of the form (a-b)\text{is divisible by 5 when (a,b) of the form } ( 5 λ 1 + i , 5 λ 2 + i ) , i ∈ 0 , 1 , 2 , 3 , 4 , 5 (5\lambda_1+i,5\lambda_2+i),i\in0,1,2,3,4,5 5 n C 2 ( C A S E 2 ) + ( n C 1 ) ( n C 1 ) ( C A S E 1 ) 5 ^nC_2 (CASE 2)+(^nC_1)(^nC_1)(CASE 1) I have excluded the case for ( 5 λ 1 , 5 λ 2 ) (5\lambda_1,5\lambda_2) in CASE 1 because it have been included in CASE 2.

5 n C 2 + ( n C 1 ) ( n C 1 ) 5 ^nC_2+(^nC_1)(^nC_1) put n=100 and we get 44750

4 + 4 + 5 + 7 + 0 = 20 4+4+5+7+0=20

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