Question

plseeeeeee solve this

Answer 1

Let the no.=x
x when divided by 7 gives remainder 4,hence 7 is a divisor of x-4,
x-4=7k, where k is a natural no.
x=7k+4
Similarly x-3 =9m,where m is a integer.
or x=9m+3
9m+3=7k+4
9m=7k+4-3=7k+1
\[m=\frac{ 7k+1 }{ 9 }\]
smallest value of k when m is a natural no. is =5\[m=\frac{ 7\times5+1 }{ 9 }=\frac{ 36 }{ 9 }=4\]
x=7*5+4=39
it gives 2 as remainder when 39 is divided by 37

Answer 2

we get infinite nos. if we take k=5,14,23,32,.......