find the smallest number when divided by 24 , 36 and 54 gives a remainder of 5 each time
first and most explained answer will be marked brainliest
Answers
Answer: 221
Step-by-step explanation:
Prime factorization of
24 = 2 × 2 × 2 × 3
36 = 2 × 2 × 3 × 3
54 = 2 × 3 × 3 × 3
Required LCM = 2 × 2 × 2 × 3 × 3 × 3 = 216
The smallest number which is exactly divisible by 24, 36 and 54 is 216
In order to get remainder as 5
Required smallest number = 216 + 5 = 221
THANKS!
I hope it will help you.
Good LucK!
Answer:
221
Step-by-step explanation:
Prime factorization of
24 = 2 × 2 × 2 × 3
36 = 2 × 2 × 3 × 3
54 = 2 × 3 × 3 × 3
So the required LCM = 2 × 2 × 2 × 3 × 3 × 3 = 216
The smallest number which is exactly divisible by 24, 36 and 54 is 216
In order to get remainder as 5
Required smallest number = 216 + 5 = 221
Therefore, the smallest number which when divided by 24, 26 and 54 gives a remainder of 5 each time is 221.
PLS MARK IT AS THE BRAINLIEST
:)
:)
:)