Question

find the least number that when divided by 16,18&20 leaves a remainder 4 in each case, but is completely divisible by 7.

Answer 1

First of all we have to take the L.C.M. of 16, 18 and 20

Prime factorization of 16 = 2*2*2*2

Prime factorization of 18 = 2*3*3

Prime factorization = 20 = 2*2*5

= 2*2*2*2*3*3*5

L.C.M. of 16, 18 and 20 = 720

Now, the L.C.M. of 16, 18 and 20 is 720.

So, the required number will be in the form of (720*x) + 4

Now, we have to apply the hit and trial method to find the least value of x for

which (720*x)+ 4 is divisible by 7 ... by putting x = 1, 2, 3, 4.........n.

First by putting x = 1

⇒ (720*1) + 4

⇒ 720 + 4

⇒ 724

724 is not divisible by 7.

Now, putting x = 2

⇒ (720*2) + 4

⇒ 1440 + 4

⇒ 1444

1444 is also not divisible by 7.

Now, putting x = 3

⇒ (720*3) + 4

⇒ 2160 + 4

⇒ 2164

2164 is also not divisible by 7

Now, putting x = 4

⇒ (720*4) + 4

⇒ 2880 + 4

⇒ 2884

2884 is exactly divisible by 7

So, for the value of x = 4, the required number comes 2884.

2884 is the least number which when divided by 16, 18 and 20 leaves a

remainder 4 in each case but exactly divisible by 7.