Question

show that one and only one out of n ,n+ 4 and n+ 8 ,n+ 12 and n + 16 is divisible by 5 Where n is any positive integer....

Answer 1

Answer: Answer: we know that any positive number will be in the form of  5k , 5k+1 , 5k+2 ,5k+3  

5k+4  

now we will make some cases ! to solve our question ,  

 

if n = 5 k  n is perfectly divisible by 5  now sir , n = 5k we will add 4 both sides  

we get 4 +5k = n +4  and if we divide the number it will leave a remiander 4  

when we divide this number by 5 now sir ,  

n = 5k we will add 8 both side we get  

n+8 = 5k +8 which means = >  5k +5 +3 which means = 5(k+1) + 3  

which means it will give a remiander 3 when we divide it by 5  

if n = > 5 k  when we add 12 both sides we get  : - )  

n+12 = 5k + 12 which means = > 5k +10 +2 => 5(k+2) +2 so it means when we divide this quantity  by 5 we get remiander 2  

now again  

5k = n we will ad 16 both sides we get  

5k +16 = n +16 which means 5 (k+3) +1 so it will give a remiander of 1  

now , in case 2  = >  

where n = 5k +1 the number leave a remiander when it is divisible by 5  

now n  =  5k +1 adding both sides 2 we get

 

n+3 = 5k +3 which means it will leave a remiander of 3  

again  

n  = 5k + 1

adding 4 both sides which means  

n+4 = 5(k+1) which means it will leave no remiander  and perfectly divisible by 5  

again ,  

n = 5k +1 adding both sides 8 we get  

n+8 = 5(k+1) +4 which means it will leave 4 as a remiander when we divide it by 5  

again  

n = 5 k +1 adding both sides 12 we get  

n+12 = 5(k+2) + 3 so sir it will leave a remiander 3 when it is divisible by 5  

again

n = 5k +1 adding both sides 16 we get  

n+16 => 5(k+3) +2 which means it will leave a reminder 2 when it is divisible by 5  

so it means only in 1 case 5 is divisible by number

Pls mark me as brainliest :-)

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.

$\fbox \blue{Hence Proved}$