Find the smallest no. which when increased by 5 becomes divisible by 12, 15 and 20
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Answer:
Factors of 12, 18 and 30
⇒ 12 = 2 × 2 × 3
⇒ 18 = 2 × 3 × 3
⇒ 30 = 2 × 3 × 5
⇒ L.C.M. of 12, 18 and 30 = 180
⇒ Required number = 180 – 5 = 175
∴ Required number is 175.
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