find the least number which when divided by 10 , 15 , 20 leaves a remainder 4 in each case .
Answers
Your answer is given below *****
Step-by-step explanation:
Given numbers are 12 , 15 , 20 , 54
To Find: least no. which is divided by given nos. and leaves remainder 8
Least no which is divisible by all given no is LCM of all nos.
LCM means least common multiple.
First we find LCM of 12 , 15 , 20 , 54 by prime factorization method
12 = 2 × 2 × 3
15 = 3 × 5
20 = 2 × 2 × 5
54 = 2 × 3 × 3 × 3
LCM( 12, 15 , 20 , 54 ) = 2 × 2 × 3 × 3 × 3 × 5 = 540
To find the required no. we add 8 to LCM
⇒ Required No. = 540 + 8 = 548
Thus, 548 is the least no which when divided by 12 15 20 and 54 leaves in each case a remainder of 8.
Answer: 64 is the least number divided by 10,15 and 20 and leave remainder 4
Step-by-step explanation: LCM of 10 , 15 and 20 = 60
So , 60 is the least number of divisible by 10 ,15 and 20 .
There should be remainder 4 .
So, we will add 4 to 60 which give us 64 . So, 64 is the least number divided by 10,15 and 20 and leave remainder 4 .