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