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**Solution:**

k+4 is divisible by 7

so k=3+7m=3,10,17,24……………. ( m=0,1,2,3,…………..)

Now let P=k+2n is divisible by 7

P=3+7m+2n

P=7m+3+2n

For to be divisible by 7 the minimum requirement is

3+2n=7t (t=1,2,3…………)

for t=1,n=2

but n>2 so putting t=2

3+2n=14

n=9

Thus the smallest value of n >2 is 9