Question

Determine the smallest no. which when increased by 27 is exactly divisible by 660 and 594

Answer 1

Answer:

5913

Step-by-step explanation:

660=2×2×3×5×11

594=2×3×3×3×11

Therefore, LCM=5940

It increased by 27

so, 5940+27=5913

Therefore, 5913 is the smallest number which when increased by 27 is exactly divisible by 660 and 594.

Answer 2

Concept

The smallest non-zero common number , a multiple of both numbers, is the least common multiple of two numbers.

Given

Given that there is a smallest number which is increased by 27 and   exactly divisible by 660 and 594

Find

We need to find the smallest number

Solution

The numbers given are 660 and 594

Let us calculate the LCM first

The factors are

660=2×2×3×5×11

594=2×3×3×3×11

Therefore, LCM=5940

No according to the statement It  is increased by 27

Therefore, 5940+27=5913

Hence 5913 is the smallest number

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