Question

1) Find the smallest number which when divided by 24 and 36 leaves a remainder of 5 in each case. (it is correct question.)

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Answer 1

Answer:

the smallest number which when divided by 24, 36 and 54 gives a remainder of 5 each time is 221.

Step-by-step explanation:

Prime factorization of

Prime factorization of24 = 2 × 2 × 2 × 3

Prime factorization of24 = 2 × 2 × 2 × 336 = 2 × 2 × 3 × 3

Prime factorization of24 = 2 × 2 × 2 × 336 = 2 × 2 × 3 × 354 = 2 × 3 × 3 × 3

Prime factorization of24 = 2 × 2 × 2 × 336 = 2 × 2 × 3 × 354 = 2 × 3 × 3 × 3So the required LCM = 2 × 2 × 2 × 3 × 3 × 3 = 216

Prime factorization of24 = 2 × 2 × 2 × 336 = 2 × 2 × 3 × 354 = 2 × 3 × 3 × 3So the required LCM = 2 × 2 × 2 × 3 × 3 × 3 = 216The smallest number which is exactly divisible by 24, 36 and 54 is 216

Prime factorization of24 = 2 × 2 × 2 × 336 = 2 × 2 × 3 × 354 = 2 × 3 × 3 × 3So the required LCM = 2 × 2 × 2 × 3 × 3 × 3 = 216The smallest number which is exactly divisible by 24, 36 and 54 is 216In order to get remainder as 5

Prime factorization of24 = 2 × 2 × 2 × 336 = 2 × 2 × 3 × 354 = 2 × 3 × 3 × 3So the required LCM = 2 × 2 × 2 × 3 × 3 × 3 = 216The smallest number which is exactly divisible by 24, 36 and 54 is 216In order to get remainder as 5Required smallest number = 216 + 5 = 221

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Answer 2

Answer:

77

Step-by-step explanation:

LCM of 24 & 36 = 72

so the smallest number which when divided by 24 and 36 leaves a remainder 5 = 72 + 5 = 77