IlCreatore wrote:

We can proceed by steps:

1) number of elements in the two sets: we have to take the first and the last multiple, then compute last - first, divide by the number whose multiples are of interest and sum 1. In our case, in column A, we have \(\frac{100-4}{4}+1 = 25\), while for column B we get \(\frac{100-5}{5}+1 = 20\).

2) Compute the sum. The formula for the sum of an arithmetic progression is \(sum = \frac{n}{2}(first+last)\), where n is the number of elements in the progression and first and last are the first and the last elements. Thus, for column A, \(\frac{25}{2}(4+100) = 1300\), while column B equates \(\frac{20}{2}(5+100) = 1050\).

We conclude that A is greater!

Hi, for step 1, is there a simplified formula that is easier to remember for exam?

Also, it says less that 100, so shouldnt we consider multiples less than 100? Like is 100 still included?

And thanks for the explanation btw