Question

find the least number that is divisible by 15,18,21,30 leaving 7 remainder in each case​

Answer 1

Answer:

The number ‘x’ should leave remainder 7 when divided by 15,18 and 21. That means  x−7 must be divisible by 15,18 and 21.

(In these type of questions find LCM of the numbers. Add 7 to the LCM and the number you get will be your answer.)

So, LCM of 15,18 and 21 is 630

now,

x−7=630  

x=637  

ANSWER=637

Answer 2

637 is the least number that is divisible by 15,18,21,30 leaving 7 remainder in each case​

Given:

A number is divisible by 15,18,21,30 leaving 7 remainder in each case​

To Find:

Least Number  satisfying

Solution:

LCM  - Least common multiplier of given numbers is the least number which is perfectly divisible by given numbers.

Step 1:

Least number that is divisible by 15,18,21,30 leaving 7 remainder in each case​ will be

LCM ( 15 , 18 , 21 , 30) + 7

Step 2:

Prime Factorize 15, 18 ,21 and 30

15 = 3 x 5

18 = 2 x 3 x 3

21 = 3 x 7

30 = 2 x 3 x 5

Step 3:

Calculate LCM

LCM ( 15 , 18 , 21 , 30) = 2 x 3 x 3 x 5 x 7 = 630

Step 4:

Add 7 in 630 to get desired number

630 + 7 = 637

637 is the least number that is divisible by 15,18,21,30 leaving 7 remainder in each case​