mgmat practice question

[varunkh70]

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

Approach 1(explanation in mgmat guide) -

No. of terms =>153 - 54+1=100. No. of terms are even so special sums rule does not hold therefore the answer is "NO"

Special sums rule - the sum of odd no. of integers is always divisible by the no. of terms in the set but this does NOT work for even no. of terms(which is our case)

Approach 2(trying out the other way coz of which I get a different answer)
1) No. of terms is 100
2) Average of the 1st and last term is (54+153)/2= 103.5
3) 100 * 103.5 gives a multiple of 100
therefore the answer is "YES"

Can anyone explain the difference in the results?

[ssmiles08]

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

Approach 1(explanation in mgmat guide) -

No. of terms =>153 - 54+1=100. No. of terms are even so special sums rule does not hold therefore the answer is "NO"

Special sums rule - the sum of odd no. of integers is always divisible by the no. of terms in the set but this does NOT work for even no. of terms(which is our case)

Approach 2(trying out the other way coz of which I get a different answer)
1) No. of terms is 100
2) Average of the 1st and last term is (54+153)/2= 103.5
3) 100 * 103.5 gives a multiple of 100
therefore the answer is "YES"

Can anyone explain the difference in the results?

I think your first approach is right.

For your second approach: the "SUM" is 10350

10350 is NOT divisible by 100. for a number to be divisible by 100, it must have at least 2 zeros at the end and the result should be an integer.

[varunkh70]

oh..my bad i got it now..10350 is just not divisible by 100..

[sreak1089]

I do it this way :

54 + 55 + .. + 153 ==

1 + 2 + 3 + .. *+ 153 - (1 + 2 + 3 + .. .+ 53)

= (153 * 154) / 2 - (53 * 54) / 2
= 1/2 [ (100 + 53)(100 + 54) - 53 * 54) ]
= 1/2 (100*100 + 100 * 54 + 100 * 53)

when divided by 100:
= 1/2(100 + 54 + 53)

hence not divisibly by 100

Is there a shorter method?

[Ian Stewart]

There are many ways to add evenly spaced sums; if you have a method that works for you and you're comfortable with it, there's no reason to learn a new one. They can all be done without any formulas as follows:

54 + 55 + 56 + ... + 151 + 152 + 153
= (53 + 1) + (53 + 2) + (53 + 3) + ... + (53 + 98) + (53 + 99) + (53 + 100)
= 53n + (1 + 2 + 3 + ... + 98 + 99 + 100)

where n is the number of terms. From the sum in brackets, we can see that n = 100, so our sum is

5300 + 101*50 = 5300 + 5050 = 10350.

(I added 1 + 2 + 3 + ... + 98 + 99 + 100 by regrouping as follows: (1 + 100) + (2 + 99) + (3 + 98) + ... = 101 + 101 + 101 + ... + 101 = 50*101, because we have 50 pairs of numbers which each add to 101).

Of course there are formulas you can learn for these problems that also work well.

[sreak1089]

Hi Ian,

That was cool....I really appreciate your logic.