Question

find the smallest number which when divided by 24, 36, and 54 gives a remainder of 5 each
ume. Hint. Required number (CM of 24, 36, 54) 51
22​

Answer 1

hii friend here is your answer

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.