amorphous wrote:

Find the value of \(n\) for each quantity

For A: the numbers are multiples of \(6\); 1st multiple = \(0\) and the last multiple below \(100 = 96\)

value of \(n = \frac{(96-0)}{6} + 1 = 17\) (since both the numbers are divisible by \(6\))

From formula, sum \(= 17 * \frac{18}{2} = 17*9\)

For B: the numbers are multiples of\(8\); 1st multiple \(= 0\) and the last multiple below \(100 = 96\)

value of \(n = \frac{96-0}{8} + 1 = 13\) (since both the numbers are divisible by \(6\))

From formula, sum = \(13 * \frac{14}{2} = 13*7\)

clearly \(A\) is bigger

It is the sum, the number of multiples of 6 is not continuous from

Hi, I found that the you made a mistake with the sum above.

I have a different solution for this question:

- The first multiple of 6 is: 0 = 6*0

- The second multiple of 6 is: 6 =6*1

- The third multiple of 6 is: 12=6*2

............

- The seventeenth multiple of 6 is: 96=6*16

--> the sum of all multiples of 6 is: 6*(0+1+....+16) = 6*(1+...+16)=6*(16*17)/2=8*6*17Similarly,

-->

the sum of all multiples of 8 is: 8*(1+...+12)=8*(12*13)/2=8*6*13So A is bigger