Question

What is the smallest number when divided by 24.36 and 54 gives a remainder of 5 each time?

Answer 1

Answer:

The smallest such number has to be 5! But if we’re not considering that, then here’s the solution:

The numbers which leave remainder 5 when divided by 24 are: 29, 53,.. [(a multiple of 24)+5].

Similarly, the numbers which leave remainder 5 when divided by 36 are: 41, 77,... [(a multiple of 36)+5]

And finally, the numbers which leave remainder 5 when divided by 54 are: 59, 113,.. [(a multiple of 54)+5].

Here we want one number that does all three! So it has to be a number that is =[(multiple of 24 and 36 and 54)+5] = (A common multiple)+5.

But because we want the least such number, we take least common multiple of 24, 36 and 54 and then add 5.

Let’s now work on finding the lcm of these three numbers.

24=2*2*2*3

36=2*2*3*3

54=2*3*3*3

Now the lcm is the number obtained by taking the highest available power of each prime factor.

So the lcm is = (2^3) * (3^3) =216.[The Highest power available is 3 for 2(in 24), 3 for (in 54)].

Answer = 216+5=221.

Step-by-step explanation:

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.