Question

वह छोटी से छोटी संख्या ज्ञात करो जिसमें 7 घटाने पर शेष बची संख्या 20, 28, 38 और 105. से पूर्णत:
विभक्त हो।​

Answer 1

Answer:

lcm+7

28=2×2×7

20=2×2×5

38=19×2

105=3×5×7

lcm=2×2×5×7×19×3=7080

7980+7=7980

Answer 2

The required smallest number is 7987.

Step-by-step explanation:

Consider the provided numbers.

First find the LCM of 20, 28, 38 and 105.

20 = 2 × 2 × 5

28 = 2 × 2 × 7

38 = 2 × 19

105 = 3 × 5 × 7

LCM of the numbers are: 2 × 2 × 3 × 5 × 7 × 19 = 7980

Now add 7 in 7980.

7980+7=7987

Hence, the required smallest number is 7987.

#Learn more

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