The least number, which when divided by 48, 6 | vedclass

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(B) First, observe the difference between each divisor and its corresponding remainder:$48 - 38 = 10$$60 - 50 = 10$$72 - 62 = 10$$108 - 98 = 10$$140 -

Vedclass · Accepted Answer

$15110$

Vedclass · Answer

(B) First, observe the difference between each divisor and its corresponding remainder:$48 - 38 = 10$$60 - 50 = 10$$72 - 62 = 10$$108 - 98 = 10$$140 - 130 = 10$Since the difference is constant $(10)$, the required number is $(LCM 	ext{ of } 48, 60, 72, 108, 140) - 10$.Prime factorization:$48 = 2^4 	imes 3$$60 = 2^2 	imes 3 	imes 5$$72 = 2^3 	imes 3^2$$108 = 2^2 	imes 3^3$$140 = 2^2 	imes 5 	imes 7$$LCM = 2^4 	imes 3^3 	imes 5 	imes 7 = 16 	imes 27 	imes 35 = 15120$.Therefore, the required number $= 15120 - 10 = 15110$.

The least number, which when divided by $48, 60, 72, 108$ and $140$ leaves $38, 50, 62, 98$ and $130$ as remainders respectively is

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