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Consecutive integers

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Consecutive integers

by Deepthi Subbu » Thu May 23, 2013 5:02 am
Is the sum of the integers from 54 to 153, inclusive, divisible by l00?

Sum=Average * number of terms
Sum = Average * 100 ( There are 100 numbers between 53 and 154 ) inclusive .
Hence sum is divisible by 100. Is that not true ?

But Manhattan says since there are even number of items, the sum can never be an integer. I believe this is true, but whats the deal with the question above?
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by Atekihcan » Thu May 23, 2013 5:12 am
Deepthi Subbu wrote:Sum = Average * 100 ( There are 100 numbers between 53 and 154 ) inclusive .
Hence sum is divisible by 100. Is that not true ?
You are assuming that the average is an integer which is not true.
In this case, the average of the integers = (first + last)/2 = (54 + 153)/2 = 207/2 = 103.5

So, sum = 103.5*100 = 10350 --> Not multiple of 100
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by Deepthi Subbu » Thu May 23, 2013 5:32 am
Ok, the sum is not an integer . But the 100 in the numerator gets cancelled with the 100 in the denominator. Sum = (Average * 100)/100. Hence sum = Average. If the value can get cancelled out, it means that sum is divisible by 100 rite? It might sound silly, but I am confused.
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by Atekihcan » Thu May 23, 2013 5:34 am
Deepthi Subbu wrote:Ok, the sum is not an integer . But the 100 in the numerator gets cancelled with the 100 in the denominator. Sum = (Average * 100)/100. Hence sum = Average. If the value can get cancelled out, it means that sum is divisible by 100 rite? It might sound silly, but I am confused.
I'm not sure that I understood what you are asking, but Sum = Average*(Number of terms)
So, in this case, sum = average*100

There is no denominator.
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by Deepthi Subbu » Thu May 23, 2013 5:49 am
The sum of consecutive integers can be found by the formula = > Average * number of terms. Here since number of terms is 100 , we get Sum = (Average * 100).

We would like to find if this sum is divisible by 100 , so sum/100 => (Average * 100)/100. The 100 in the numerator and denominator cancel out, thereby giving sum=average. This means sum is divisible by 100. Where am I wrong?
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by Atekihcan » Thu May 23, 2013 6:42 am
Deepthi Subbu wrote:We would like to find if this sum is divisible by 100 , so sum/100 => (Average * 100)/100. The 100 in the numerator and denominator cancel out, thereby giving sum=average. This means sum is divisible by 100. Where am I wrong?
No.
The 100 in the numerator and denominator cancels out only on the right-hand part of the equation, i.e. on the blue part not on the green part.
So, the equation becomes sum/100 = average.

Aside, sum cannot be equal to the average unless the number of terms is equal to 1.

Anyway, sum/100 = average = 103.5 = non-integer
This means sum is not divisible by 100 because if sum was divisible by 100, then sum/100 should have been an integer.
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by aaggar7 » Sun May 26, 2013 2:24 am
Hi Deepthi,

I will continue with your explanation,

Sum = Avg * No. of items

Avg = (first + last)/2 =(153+54)/2= 207/2

No. of items = 153-54 + 1 = 100

So Sum = 100 * (207/2)
= 10350
As last two digits are not 00,it is not a multiple of 100
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by GMATGuruNY » Sun May 26, 2013 4:38 am
Deepthi Subbu wrote: But Manhattan says since there are even number of items, the sum can never be an integer. I believe this is true, but whats the deal with the question above?
The word in red is incorrect.
The SUM of consecutive integers will ALWAYS be an integer.
The AVERAGE will be an integer if there are an ODD number of terms:
(1+2+3)/3 = 2.
The AVERAGE will NOT be an integer if there an EVEN number of terms:
(1+2)/2 = 1.5.
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by Brent@GMATPrepNow » Sun May 26, 2013 5:29 am
Deepthi Subbu wrote: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. In other words,
54+55+56+57 . . . .+150+151+152+153 = (54+153) + (55+152) + (56+151) + . . .
= (207) + (207) + (207) + ...

Since there are 100 numbers altogether, there must be 50 pairs (each adding to 207). So . . .
= (50)(207)
= 10,350

As you can see, this number is NOT divisible by 100.

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