Find the smallest number which when divided by 6,8,12,15 and 20 leaves the same remainder 5.

To solve this problem, we need to find the smallest number that, when divided by 6, 8, 12, 15, and 20, leaves the same remainder 5.

Step 1: Define the problem
Let the number we are looking for be denoted by x. The condition is that x leaves a remainder of 5 when divided by each of these numbers. This means:

x % 6 = 5
x % 8 = 5
x % 12 = 5
x % 15 = 5
x % 20 = 5

Step 2: Subtract 5 from all the divisors
If we subtract 5 from x in each of the equations, we get the following:

(x - 5) % 6 = 0
(x - 5) % 8 = 0
(x - 5) % 12 = 0
(x - 5) % 15 = 0
(x - 5) % 20 = 0

This tells us that x - 5 must be divisible by 6, 8, 12, 15, and 20. Therefore, x - 5 is a common multiple of these numbers.

Step 3: Find the Least Common Multiple (LCM)
To find the smallest number that is divisible by all of these numbers, we need to calculate the least common multiple (LCM) of 6, 8, 12, 15, and 20.

We start by finding the prime factorization of each number:

6 = 2 × 3
8 = 2³
12 = 2² × 3
15 = 3 × 5
20 = 2² × 5

Now, take the highest powers of each prime factor:

The highest power of 2 is 2³ (from 8)
The highest power of 3 is 3 (from 6, 12, and 15)
The highest power of 5 is 5 (from 15 and 20)
Therefore, the LCM is:

LCM = 2³ × 3 × 5 = 8 × 3 × 5 = 120

Step 4: Solve for x
We know that x - 5 = 120k, where k is some integer. The smallest value of x occurs when k = 1, so:

x - 5 = 120
x = 120 + 5
x = 125

Final Answer:
The smallest number x that, when divided by 6, 8, 12, 15, and 20, leaves a remainder of 5 is 125.