1.

find the smallest number which when divided by 1088, 640 and 100 leaves 3 as reminder in each case in

Answer» STRONG>To Find :-

  • The smallest number which when divided by 1088, 640 and 100 LEAVES 3 as reminder in each case ?

SOLUTION :-

we know that,

  • The smallest number which when divided by a, b and c leaves r as reminder in each case is = LCM(a,b and c) + r .

so,

Prime factors of 1088 , 640 and 100 are :-

→ 1088 = 2⁶ * 17

→ 640 = 2⁷ * 5

→ 100 = 2² * 5²

LCM = 2⁷ * 5² * 17 = 128 * 25 * 17 = 54400 .

therefore,

→ Required number = 54400 + 3 = 54403 (Ans.)

Learn more :-

वह छोटी से छोटी संख्या बताईये जिसमे 7,9,11 से भाग देने पर 1,2,3 शेष बचे

brainly.in/question/9090122



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