Question

Find the smallest number which when increased by 11 is exactly divisible by 15, 20, 54

Answer 1

As the smallest number which when increased by 11 is exactly divisible by 15,20 and 54 is the LCM of these numbers - 11
LCM of 15,20 and 54=540
Hence
the smallest number =540-11
=529

Answer 2

Answer:

Smallest number which when increased by 11 is exactly divisible by 15, 20 and 54 =

Smallest number which when increased by 11 is exactly divisible by 15, 20 and 54 = 529

Explanation:

Find the LCM of 15, 20 and 54

Resolve 15,20and 54 as product prime :

15 = 3×5

20= 2×2×5=2²×5

54 = 2×3×3×3=2×3³

Therefore,

LCM(15,20,54) = 2²×3³×5

/* Product of the greatest power of each prime factors of the numbers */

= 540

Now ,

Smallest number which when increased by 11 is exactly divisible by 15, 20 and 54 =

LCM of 15,20 and 54 - 11

= 540 - 11

= 529

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