LCM HCF Divisibility Questions Shortcut Trick

Here you can check shortcut trick to solve questions based on LCM & HCF division. In various competitive exams, questions like “find the least number which when devided by x, leaves remainder y“. The solution to these problems is quite easy if you know the proper basic method. It takes a little time to find the solution to these problems.

Here you can check solution to these type of problems with examples. Read the information given here carefully and completely and then practice yourself.

Shortcut Trick for LCM HCF Divisibility Problems

Example: 1.
Find the least number which should be added to 2497 so that the sum is exactly divisible by 5, 6, 4 and 3:
Solution:
First we have to find the LCM of 5, 6, 4 and 3.
5 = 5
6 = 2, 3
4 = 2, 2
3 = 3
So L.C.M. of 5, 6, 4 and 3 = 60.

When we divide 2497 by 60, we find that the remainder is 37.

So the Number that to be added = 60 – 37 = 23.

Example: 2.
Find the least number which when increased by 5 is divisible by 24, 32, 36 and 54 each.
Solution:
First we have to find the LCM of 24, 32, 36, 54
24 = 2, 2, 2, 3
32 = 2, 2, 2, 2, 2
36 = 2, 2, 3, 3
54 = 2, 3, 3, 3
So LCM = 864
So Required number = (L.C.M. of 24, 32, 36, 54) – 5 = 864 – 5 = 859.

Example: 3.
Find the least multiple of 7, which leaves a remainder of 4, when divided by 6, 9, 15 and 18.
Solution:
First Find LCM.
6 = 2, 3
9 = 3, 3
15 = 3, 5
18 = 2, 3, 3
LCM of 6, 9, 15 and 18 is 90.
Let required number be 90x + 4, which is a multiple of 7.
The Least value of k for which (90x + 4) is divisible by 7 is k = 4.
So the Required number is = (90*4) + 4 = 364.

Example: 4.
Find the least multiple of 23 which when divided by 18, 21 and 24 leaves a remainder 7, 10, 13 respectively.
Solution:
First we have to find the LCM of 18, 21 & 24.
18 = 2 × 3 × 3
21 = 3 × 7
24 = 2 × 2 × 2 × 3
So LCM of 18, 21 & 24 = 504

Now check the given numbers remainder respectively.
18-7=11
21-10=11
24-13=11
We find it that there is 11 difference in each number.

Now the formula is
Required Number = LCM*X-11
Required Number = 504*X-11

by Hit & Trial method:
when putting X=6
We get
Required number = 3013 Ans.