Question

Show that only one of the numbers n, n+2, and n+3 is divisible by 3​

Answer 1

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Solution:

let n be any positive integer and b=3

n =3q+r

where q is the quotient and r is the remainder

0_ <r<3

so the remainders may be 0,1 and 2

so n may be in the form of 3q, 3q=1,3q+2

CASE-1

IF N=3q

n+4=3q+4

n+2=3q+2

here n is only divisible by 3

CASE 2

if n = 3q+1

n+4=3q+5

n+2=3q=3

here only n+2 is divisible by 3

CASE 3

IF n=3q+2

n+2=3q+4

n+4=3q+2+4

=3q+6

here only n+4 is divisible by 3

HENCE IT IS JUSTIFIED THAT ONE AND ONLY ONE AMONG n,n+2,n+4 IS DIVISIBLE BY 3 IN EACH CASE

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Answer 2

Step-by-step explanation:

Let the number be 5

Then,

n = 5

n + 1 =6

n + 2 = 7

n + 3 = 8

Clearly, only 6 is divisible by 3

This is also called Euclid's division lemma.

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