Question

the least number which when divided by 42,72 and 84 leaves the remainder 25,55 and 67 respectively is equals to

Answer 1

42-25 = 17

72-55 = 17

84-67 = 17

L.C.M of 42,72 and 84 = 504

so,

Required no. = 504 - 17 =487

ok

thanks

Answer 2

Answer:

let A be such number that satisfies above condition

then ,

A = 42k + 25-----------(1)

A = 72Y + 55-----------(2)

A = 84Z + 67------------(3)

From 1 and 2

42k + 25 = 72Y + 55

42k = 72Y +30

7K = 12Y + 5

K = (12Y+ 5)/7

we need to find Y such that we get K as positive integer.(apply Hit and trial method)

possible value of Y is 6

then A = 72(6) + 55

= 487

When we divide 487 by 84 we get 67 as Remainder.

So, 487 is Required No.

or

42-25 = 17

72-55 = 17

84-67 = 17

L.C.M of 42,72 and 84 = 504

so,

Required no. = 504 - 17 =487