Is the sum of the integers from 54 to 153...

[lauren]

In the number properties book, pg. 45, question 10 asks:

Is the sum of the integers from 54 to 153, inclusive divisible by 100?

The answer is no (due to the fact that the special sums rule doesnt apply in the case of even integers). But I answered "yes", given the following logic:

Sum = avg. of set * number of integers

If the number of integers is 100, shouldnt the sum also be divisible by 100?

Thanks!

[Prachi]

The formula that you are using is correct
Sum = averge of set *elements of set.

But in this case average of set is 103.5, so even if you multiply the average with 100, final sum won't be divisible by 100.

[rfernandez]

Nice work, Prachi!

[lauren]

thanks guys!

[StaceyKoprince]

on behalf of everyone, you're welcome! :)

[Guest]

but how do you find this average?

[Guest]

You can use this formula

Sum of n numbers in arithemtic progression = n/2 * 2a1+(n-1)d
where
n-> number of numbers in the progression(this case 100)
a1-> 1st term in the progression(54 in this case)
d-> difference between any 2 successive numbers in the progression(1 in this case)

So S = 100/2 *(2*54+(100-1)1) = 10350 which is not divisible by 100

[esledge]

The previous post gives a correct formula, but it might be more memorable in these forms:

Sum of Set = Average of set * Number of Terms
Sum of Set = Average of Highest and Lowest * Number of Terms
Sum of Set = Sum of Highest and Lowest * Number of Terms/2
Sum of Set = Sum of Highest and Lowest * Number of High-Low Pairs

To solidify this idea, it helps to write out an arithmetic progression (consecutive integers would work), then pair them up starting with the outermost numbers and working in. For example:
Sum of integers 1 to 10 = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10.
We can pair 1 and 10 for a sum of 11.
We can pair 2 and 9 for a sum of 11.
We can pair 3 and 8 for a sum of 11... and so on.
So, 5 pairs can be made that each sum to 11. The grand total sum is 5 (Number of Pairs) * 11 (Sum of each Pair). Likewise, the average of the whole set is simply the average of each pair, as each pair is identical.

By the way, the GMAT typically provides the High and Low terms (i.e. "the integers from 64 to 89, inclusive") rather than the Low Term and the Number of Terms (i.e. "the 17 even integers beginning with 6 and counting up"). The previous formula would be better for the latter.

[Harris]

I find it easy to find the avg the following way:

{(first + last number)/2}* number of terms / numbers of term

just another way of writing avg= S/N.

[JonathanSchneider]

As Emily mentioned earlier, just be careful that you are not dividing unnecessarily at the end. Finding the average of the high and low terms and then multiplying by the number of terms is sufficient.