Question

find the least number which when divided by 24 36 45 and 54 leaves a remainder of 3 in each case for class 6

Answer 1

Given:

find the least number which when divided by 24 36 45 and 54 leaves a remainder of 3 in each case

To find:

The required number

Solution:

• To find the least number, first, find the LCM of the given numbers then add the remainder to the LCM.

LCM of  24 36 45 and 54 using prime factorisation:

24 = 2 x 2 x 2 x 3

36 = 2 x 2 x 3 x 3

45 = 3 x 3 x 5

54 = 2 x 3 x 3 x 3

LCM of  24 36 45 and 54 = 2 x 2 x 2 x 3 x 3 x 3 x 5

LCM of  24 36 45 and 54  = 1080

Now, add the remainder:

=> 1080 + 3 = 1083

Hence, the least number which when divided by 24 36 45 and 54 leaves a remainder of 3 in each is 1083

Answer 2

Answer:

221

Step-by-step explanation:

Let's factorize,

24 = 2x2x2x3

36 = 2x2x3x3

54 = 2x3x3x3

The smallest number divisible by all 24, 36, 54 is:

LCM(24, 36, 54) = 216

Thus, to get 5 as remainder, we need to add 5 to 216.

(as 216 is the smallest number which gives remainder 0 when divided by 24 or 36 or 54)

216 + 5 = 221

For 12, 221 gives 18 as quotient and 5 as remainder

For 36, 221 gives 6 as quotient and 5 as remainder

For 54, 221 gives 4 as quotient and 5 as remainder

Hence, 221 is required number.

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