Question

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

Hey mate this is ur answer. Hope it helps.

Answer 2

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 :-)