Find the least number which when divided by 6, 15, 18 leave a reminder of 5 in each case?

Question

Accepted Answer

Ans.

Let's find the L.C.M of 6, 15 and 18.

Therefore, L.C.M of 6, 15 and 18

=

3

�

2

�

3

�

5

=

90

It means that 90 is the smallest number which when divided by 6, 15 and 18 leaves remainder 0 in each case.

Therefore, smallest number which when divided by 6, 15 and 18 leaves remainder 5 = 90 + 5 = 95

Answer

Introduction:
To find the least number that satisfies the given conditions, we need to find the least common multiple (LCM) of the three numbers: 6, 15, and 18. The LCM is the smallest multiple that is divisible by all three numbers.

Step 1: Prime Factorization:
Let's start by finding the prime factorization of each number to determine their common factors.

- Prime factorization of 6: 6 = 2 * 3
- Prime factorization of 15: 15 = 3 * 5
- Prime factorization of 18: 18 = 2 * 3^2

Step 2: Identify Common Factors:
Now, we need to identify the common factors among the three prime factorizations.

- Common factors of 6, 15, and 18: 2 * 3

Step 3: Determine LCM:
To find the LCM, we multiply the common factors identified in the previous step.

- LCM of 6, 15, and 18: LCM(6, 15, 18) = 2 * 3 * 3 * 5 = 90

Step 4: Adding the Remainder:
Since the problem states that the number should leave a remainder of 5 when divided by each of the given numbers, we need to add 5 to the LCM.

- Least number satisfying the given conditions: 90 + 5 = 95

Conclusion:
The least number that satisfies the given conditions, when divided by 6, 15, and 18, leaving a remainder of 5 in each case, is 95.

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