Question

which is smallest number that can be subtracted from 1936v so that on being divided by 9, 10, 15, the remainder is 7 every time

Answer 1

you need to solve below :-
1936-x = 9p + 7
1936-x = 10q + 7
1936-x = 15r + 7

Answer 2

combining first two equations, gives u :-

1936-7-x = 90m

1929 - x = 90m

that gives u : x = 39

Answer 3

since x = 39 is satisfied by 3rd equation also, we're done.
39 is the required smallest number

Answer 4

let me knw if smthng doesnt make sense

Answer 5

thanks

Answer 6

yw