using euclid divisioon lemma show that among the jumbers n , n+ 1 , n + 3 , n + 7 , n + 9 , only one number is divisible by 5 , where n is a natural number

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Solution

Dear Student,

let n be any positive integer and b=5,then by euclid's division lemma,x=5q+r.

positive remainders are 0,1,2and 3,4 since 0 $\leq$ r<5.then x can be 5q,5q+1,5q+2,5q+3,5q+4

1. n=5q,it is divisible by 5

2.n=5q+1, n+1=5q+1+1=5q+2

=5q+1 is not divisible by 5

3. n=5q+1, n+3=5q+1+3=5q+4

=5q+4 is not divisible by 5

4. n=5q+2
n+7=5q+2+7
=5q+9

=5q+9,it is not divisible by 5

5. n=5q+3,
n+9=5q+3+9
=5q+12=5q+12 ,it is not divisible by 5.

therefore one and only one out of n,n+1,n+3,n+7,n+9 is divisible by 5

Regards!

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