Question

Find the smallest no. which when we divided by 24 ,36 and 54 gives a remainder 7 each

Answer 1

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.

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Answer 2

Answer:

223

Step-by-step explanation:

Complete step-by-step solution:

Here, we have to find the least number which when divided by 24,36 and 54 leaves 5 as the remainder in each case.

Before solving this question, we must know what the division theorem is. Division theorem states that “if ‘n’ is an integer and ‘d’ is a positive integer, there exist unique integers ‘q’ and ‘r’ such that.

n=dq+r where 0⩽r<d .

Here, ‘n’ is the number or dividend , ‘d’ is the divisor. ’q’ is the quotient and ‘r’ is the remainder.

We will find the least common multiple of the given numbers 24,36 and 54 .

2∣∣24,36,54−−−−−−−−

3∣∣12,18,27−−−−−−−−

3∣∣4,6,9−−−−−

2∣∣4,2,3−−−−−

2∣∣2,1,3−−−−−

3∣∣1,1,3−−−−−

1,1,1

We will find the product of the above multiplies to find the least common multiple.

⇒L.C.M.=2×2×2×3×3×3

On multiplying the terms, we get

⇒L.C.M.=216

Since we get 7 as the remainder from all the numbers, we will now find the required number.

Adding the number 7 from the obtained least common multiple, we get

=216+7

=223