Hi,

What is the lowest positive integer greater than 1, which when divided by 5 or 8 leaves a remainder of 1

Thanks

41/5=8 with a remainder of 1

41/8=5 with a remainder of 1
so the answer is 41

numbers which when divided by 5 leave a remainder of 1:

6 11 16 21 26 31 36 41 46 ..
numbers which when divided by 8 leave a remainder of 1:
9 17 25 33 41 49 ...

looks like 41 is the smallest common to both

so 41/5=8 remainder 1 and

41/8=5 remainder 1

To find the lowest positive integer that satisfies the given condition, we can use the concept of least common multiples (LCM) and modular arithmetic.

First, let's find the LCM of 5 and 8, which is the smallest multiple that both numbers share. The LCM of 5 and 8 is 40.

We know that if a positive integer gives a remainder of 1 when divided by 5 or 8, then the number must be one more than a multiple of 5 or 8.

Now, let's list down the numbers that are one more than a multiple of 5 or 8:

1 (1 more than 0 times 5 or 0 times 8)
6 (1 more than 1 times 5 or 0 times 8)
11 (1 more than 2 times 5 or 0 times 8)
16 (1 more than 3 times 5 or 0 times 8)
21 (1 more than 4 times 5 or 0 times 8)
26 (1 more than 5 times 5 or 0 times 8)
31 (1 more than 6 times 5 or 0 times 8)
36 (1 more than 7 times 5 or 0 times 8)
41 (1 more than 8 times 5 or 1 times 8)

From the list above, we can see that the lowest positive integer greater than 1, which when divided by 5 or 8 leaves a remainder of 1, is 41.

Therefore, the answer to your question is 41.

5 is an integer and if is divided by 5 it will leave a reminder 1