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Word Problems - Consecutive Integers - WHY is this wrong?

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by [email protected] » Wed Sep 18, 2013 6:22 pm
Hi JMSass217,

I think that there might be some typos in your post, so I just want to clarify a few things:

First, Is the question asking you for the numbers 54 to 153, inclusive? (in some of your text, you wrote 53 to 154, which is a different group with a different average and a different number of terms)

Second, assuming that we're dealing with 54 to 153, the rest of your math is correct.

The average is 103.5 and the number of terms is 100

Multiply these values = (103.5)(100) = 10,350

10,350 is NOT evenly divisible by 100 (look at the 350.....)

As far as the technical explanation that accompanies this prompt, the fact is that the above info proves that the sum is NOT evenly divisible by 100.

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by theCodeToGMAT » Wed Sep 18, 2013 7:42 pm
Use Arithmetic Progression:
53......154

S = n(2a + (n-1)d)/2

= 102 ( 2 x 53 + 101 x 1 ) /2
= 51 ( 106 + 101)
= 51 ( 207)
If we multiply.. then the Units Digit is 7; hence not divisible by 100

ANSWER NO
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by theCodeToGMAT » Wed Sep 18, 2013 7:45 pm
The Number of terms between 53 and 154 is NOT 101 but is 102.

154 - 53 = 101 .. This subtraction doesn't consider 53, So you need to add one more to make 53 inclusive
hence, 101 + 1 = 102..

Refer the solution above for explanation that why the Correct Answer is NO only.
JmSass217 wrote:MGMAT book 3 - Word Problems
Chapter 8 - Extra Consecutive Integers
Problem #2

Is the sum of integers from 53 to 154 inclusive divisible by 100?

My answer: Yes.
Average = (first + last)/2 (53+154)/2 = 103.5
# of terms = (last-first)+1 (154-53)+1 = 100
SUM = average x # of terms 103.5 x 100 = 10350
10350 is divisible by 100 (as evidence by previously multipplying by 100 to get there...)

Book Answer: NO, There are 100 integers from 54 to 153 inclusive. For any even number of consecutive integers, the sum of all integers is NEVER a multiple of the number of integers.

I can not for the life of me understand why the solution explanation says this is wrong. I mean I understand the concept of a consecutive set sum of an even number of integers is not divisible by the number of integers, but I even did the math the long way (with handy dandy Excel) to try and see it play out (which btw tells me the sum of consecutive integers method does not always work because the actual sum is 10504 and there are 101 terms "inclusive" - if you include both first and last), and I'm now wondering if MGMAT is wrong... I worked out the problem the way they told me to in the chapter.

I'm willing to accept that I may be blind here, I make a lot of careless mistakes in practice. But could someone please point me in the right direction? Thank You!
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by ganeshrkamath » Wed Sep 18, 2013 8:55 pm
JmSass217 wrote:MGMAT book 3 - Word Problems
Chapter 8 - Extra Consecutive Integers
Problem #2

Is the sum of integers from 53 to 154 inclusive divisible by 100?

My answer: Yes.
Average = (first + last)/2 (53+154)/2 = 103.5
# of terms = (last-first)+1 (154-53)+1 = 100
SUM = average x # of terms 103.5 x 100 = 10350
10350 is divisible by 100 (as evidence by previously multipplying by 100 to get there...)

Book Answer: NO, There are 100 integers from 54 to 153 inclusive. For any even number of consecutive integers, the sum of all integers is NEVER a multiple of the number of integers.
I think the problem here is to find the sum of integers from 54 to 153 inclusive (atleast that's what is mentioned in the original explanation).

Sum of a series = average of that series * number of terms in that series
Sum = (54+153)/2 * 100
= 207/2 * 100
= 103.5*100
= 10350

So answer is NO.

What the book tries to say here is that the sum of consecutive integers is never a multiple of the number of terms when the latter is even.
This is true because, in this case, the first term is odd and the last term is even or vice versa.
So the average is always a fraction.
Therefore the sum will never be an integral multiple of the number of terms.

In mathematical terms,
sum = average * number of terms
sum = (odd + even)/2 * (even)
sum = odd/2 * even
odd/2 is never an integer


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by Brent@GMATPrepNow » Wed Sep 18, 2013 9:07 pm
Is the sum of the integers from 54 to 153, inclusive, divisible by l00?
Here's one more approach.

First determine the number of integers from 54 to 153 inclusive.

We have a rule that says, the number of integers from x to y inclusive equals y - x + 1

So, the number of integers from 54 to 153 inclusive = 153 - 54 + 1 = 100

Now, let's add these 100 numbers in pairs.
To do so, we'll begin adding numbers from the outer edges and work our way in.
So, 54+55+56+57....+150+151+152+153 = (54+153) + (55+152) + (56+151) + . . .
= (207) + (207) + (207) + ...
= (50)(207) [there are 100 numbers altogether, so there must be 50 pairs, each adding to 207]
= 10,350
As you can see, this number is NOT divisible by 100.

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