41/5=8 with a remainder of 1
41/8=5 with a remainder of 1
so the answer is 41
Hi,
What is the lowest positive integer greater than 1, which when divided by 5 or 8 leaves a remainder of 1
Thanks
4 answers
5 is an integer and if is divided by 5 it will leave a reminder 1
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
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
41/8=5 remainder 1