Question

Use Euclid division lemma to show that one and only one out of n, n+4,n+8,n+12 and n+16 is divisible by 5, where n is any positive integer.​

Answer 1

Answer:

from the above conditions we can say that only one integer is divisible by 5 in

n, n+4, n+8, n+12, n+16

hope it helps

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Answer 2

Solution:

According to Euclid's division Lemma,

Let the positive integer = n, b=5

n = 5q+r, where q is the quotient and r is the remainder

0 < r < 5 implies remainders may be 0, 1, 2, 3, 4 and 5

Therefore, n may be in the form of 5q, 5q+1, 5q+2, 5q+3, 5q+4

So, this gives us the following cases:

CASE 1:

When, n = 5q

n+4 = 5q+4

n+8 = 5q+8

n+12 = 5q+12

n+16 = 5q+16

Here, n is only divisible by 5

CASE 2:

When, n = 5q+1

n+4 = 5q+5 = 5(q+1)

n+8 = 5q+9

n+12 = 5q+13

n+16 = 5q+17

Here, n + 4 is only divisible by 5

CASE 3:

When, n = 5q+2

n+4 = 5q+6

n+8 = 5q+10 = 5(q+2)

n+12 = 5q+14

n+16 = 5q+18

Here, n + 8 is only divisible by 5

CASE 4:

When, n = 5q+3

n+4 = 5q+7

n+8 = 5q+11

n+12 = 5q+15 = 5(q+3)

n+16 = 5q+19

Here, n + 12 is only divisible by 5

CASE 5:

When, n = 5q+4

n+4 = 5q+8

n+8 = 5q+12

n+12 = 5q+16

n+16 = 5q+20 = 5(q+4)

Here, n + 16 is only divisible by 5

So, we can conclude that one and only one out of n, n + 4, n + 8, n + 12 and n + 16 is divisible by 5.

$\fbox \blue{Hence Proved}$