find the smallest number which when divided by 20 and 12 leaves no reminder
please explain how comes 240
Answers
Answer:
We first need to find the LCM of 12,20,30 and 60.
To find LCM, we first write all numbers as products of their prime numbers.
12=2×2×3
=22×31
20=2×2×5
=22×51
30=2×3×5
=21×31×51
60=2×2×3×5
=22×31×51
We then choose each prime number with the greatest power and multiply them to get the LCM.
LCM=2×2×3×5=4×3×5=60s
Hence, 60 is the smallest number which is exactly divisible by 12,20,30 or 60
So, the smallest number which when divided by 12,20,30 or 60 leaves a remainder 5 each time will be 60+5=65.
Step-by-step explanation:
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Step-by-step explanation:
60 if it can be divided by 12 OR 20
60 / 12 = 5
60 / 20 = 3
get the prime factors of both
20 = 5 * 2 * 2
12 = 2 * 2 * 3
Then merge the factors keeping what’s in common between the 2 and adding the others.
5 * 2 * 2 * 3 = 60 (2 * 2 is in common so we just include that once.)
If however, you say that the number has to be divisible by both 12 AND 20, then you just combine all the factors, or you just simply multiply the 2 numbers together.
So, 5 * 2 * 2 * 2 * 2 * 3 = 240 or 12 * 20 = 240.
240 / 12 = 20
20 / 20 = 1
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