Question

Determine the least number which when divided by 24 30 and 54 leaves 5 as remainder in each case.
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Answer 1

Answer:

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Step-by-step explanation:

We also note that 24 & 36 have the common factor 12,

so we are looking for two numbers a & b such that :

a & b are integers and

The smallest pair of non-zero integers & that satisfy this is &

so we can state that :

If you are happy that if you divide your number by either 24 or 36 is zero - then the number you want is 0 (or 1 if you want a non zero remainder).

If you want a number which is divisible by 24 & 36 by at least 1, then 72 is the smallest number which is divisible by both 24 & 36 and have the same remainder (that is 0).

If the remainder is non zero then 73 is the number required.

Answer 2

Answer:

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.

Step-by-step explanation:

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