What is the least number which when divided by 6, 15, and 18 leaves a remainder of 5 in each case?

To find the least number that leaves a remainder of 5 when divided by 6, 15, and 18, we can follow these steps:

First, we need to recognize that if a number gives a remainder of 5 when divided by another number, it can be rewritten in the form:

Number = k * divisor + remainder

For our case, let’s denote the least number we’re looking for as N. So, we can express it as:

N = 6a + 5

N = 15b + 5

N = 18c + 5

where a, b, and c are integers.

Subtracting 5 from each equation gives us:

N – 5 = 6a

N – 5 = 15b

N – 5 = 18c

Thus, the number N – 5 must be a common multiple of 6, 15, and 18. Now, we need to find the least common multiple (LCM) of these numbers.

The prime factorization of each number is as follows:

• 6 = 2 × 3
• 15 = 3 × 5
• 18 = 2 × 3²

To find the LCM, we take the highest power of each prime:

• From 6: 2¹
• From 15: 3¹ and 5¹
• From 18: 3² (which is higher than 3¹)

So, the LCM is:

LCM = 2¹ × 3² × 5¹ = 2 × 9 × 5 = 90

Therefore, N – 5 = 90k for some integer k. To find the least number N, we set k = 1:

N – 5 = 90 × 1

N – 5 = 90

N = 90 + 5 = 95

Finally, the least number which when divided by 6, 15, and 18 leaves a remainder of 5 is 95.