Ask a Teacher

(please note that its something relatsd to 10th class......) Show that only one out of n,n+4,n+8,n+12,n+16 is divisible by n when n is a +ve integer?

Any positive integer will be of the form 5a, 5a + 1, 5a + 2, 5a + 3, or 5a + 4

Case I:

If n = 5a, then n is divisible by 5

Now, n = 5a

    ? n + 4 = 5a + 4

(n + 4), when divided by 5, will leave remainder 4.

also, n = 5a

? n+8 = 5a+8 

? n+8 = 5a+5+3

? n+8 = 5(a+1) + 3

The number (n + 8) will leave remainder 3 when divided by 5.

similarly, n = 5a

? n + 12 = 5a + 12 

? n + 12 = 5a+ 10 + 2

? n + 12 = 5(a+2) + 2

The number (n + 12) will leave remainder 2 when divided by 5.

Again, n = 5a

? n + 16 = 5a + 16 = 5a+15+1 = 5(a + 3) + 1

The number (n + 16) will leave remainder 1 when divided by 5.

Case II:

When n = 5a+1

The integer n will leave remainder 1 when divided by 5.

Now, n = 5a+1

? n+2 = 5a+3

The number (n+2) will leave remainder 3 when divided by 5.

also, n = 5a+1

? n+4 = 5a+5 = 5(a+1)

The number n+4 will be divisible by 5.

Also, n = 5a+1

? n+8 = 5a+9 = 5(a+1)+4

The number n+8 will leave remainder 4 when divided by 5

Again, n = 5a+1

? n+12 = 5a+13 =5a+10+3 =  5(a+2)+3

The number (n+12) will leave remainder 3 when divided by 5.

Again, n = 5a+1

? n+16 = 5a+17 = 5a+15+2 = 5(a+3)+2

The number (n + 16) will leave remainder 2 when divided by 5.

Similarly, we can check the result for 5q + 2, 5q + 3 and 5q + 4.

In each case only one out of n, n + 2, n + 4, n + 8, n + 16 will be divisible by 5.