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Re: The sum of the odd/even integers [#permalink]
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Let’s fit the range in between any of them. The odd integers from 1 to 199 can be placed between the range of even integers from 2 to 198 or vice versa.

If the range swallows other range, the sum of the numbers of swallowing range must be greater than that of swallowed range. For example, 1, 2, 3,…………..8 or 2,3,,,,,,,,,,,,7.Sum of all numbers in the first range must be greater than that of 2nd range.
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Re: The sum of the odd/even integers [#permalink]
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sandy wrote:
Quantity A
Quantity B
The sum of the odd integers from 1 to 199
The sum of the even integers from 2 to 198




We have:
Quantity A: 1 + 3 + 5 + ... + 197 + 199
Quantity B: 2 + 4 + 6 + ... + 196 + 198

Subtract (2 + 4 + 6 + ... + 196 + 198) from both quantities to get:
Quantity A: (1 + 3 + 5 + ... + 197 + 199) - (2 + 4 + 6 + ... + 196 + 198)
Quantity B: (2 + 4 + 6 + ... + 196 + 198) - (2 + 4 + 6 + ... + 196 + 198)

Simplify:
Quantity A: 199 - 198 + 197 - 196 + . . . . + 5 - 4 + 3 - 2 + 1
Quantity B: 0

Rewrite Quantity A as follows:
Quantity A: (199 - 198) + (197 - 196) + . . . . + (5 - 4) + (3 - 2) + 1
Quantity B: 0

Simplify:
Quantity A: 1 + 1 + 1 + 1 + ... + 1 + 1 + 1 + 1
Quantity B: 0

Answer: A
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Re: The sum of the odd/even integers [#permalink]
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Probably less fast than Carcass answer but we can also see that what we are asked is the sum of an arithmetic progression that is equal to \(S_n=\frac{n}{2}(f+l)\) where n is the number of elements of our progression, f is the first element and l is the last one. Applying this rule to the column A we would get that n = 100 since 199/2=99.5 but the list terminates with an odd number. Thus the formula becomes \(\frac{100}{2}(1+199)=10,000\). Using the same rationale for column B, we get that B = 9,900. Thus, A is larger!
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Re: The sum of the odd/even integers [#permalink]
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sandy wrote:
Quantity A
Quantity B
The sum of the odd integers from 1 to 199
The sum of the even integers from 2 to 198


A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.


Another approach is to compare the sums in parts

We have:
Quantity A: 1 + 3 + 5 + ... + 197 + 199
Quantity B: 2 + 4 + 6 + ... + 196 + 198

Compare the 1st number in each quantity (1 and 2). At this point in the sums, Quantity B is 1 greater than Quantity A.
Compare the 2nd number in each quantity (3 and 4). At this point in the sums, Quantity B is 2 greater than Quantity A.
Compare the 3rd number in each quantity (5 and 6). At this point in the sums, Quantity B is 3 greater than Quantity A.
Compare the 4th number in each quantity (7 and 8). At this point in the sums, Quantity B is 4 greater than Quantity A.
.
.
.
Compare the 48th number in each quantity (195 and 196). At this point in the sums, Quantity B is 98 greater than Quantity A.
Compare the 49th number in each quantity (197 and 198). At this point in the sums, Quantity B is 99 greater than Quantity A.

At this point, Quantity B is 99 greater than Quantity A.
HOWEVER, we have now run out of numbers in Quantity B. Yet we still have the number 199 left to add to Quantity A.
When we add this last value (199) to Quantity A, we help overcome the lead that Quantity B previously had, making Quantity A the bigger quantity.

Answer: A

RELATED VIDEO (I cover the strategy of comparing in parts starting at 2:16)
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Re: The sum of the odd/even integers [#permalink]
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Just make a shorter range of numbers to do that easier, for example from 1 to 11 and from 2 to 10, the idea is the same, or even better. From 1 to 5 and from 2 to 4, even the last number 5 is higher than 4, and you have one more number.
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Re: The sum of the odd/even integers [#permalink]
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GreenlightTestPrep wrote:
sandy wrote:
Quantity A
Quantity B
The sum of the odd integers from 1 to 199
The sum of the even integers from 2 to 198




We have:
Quantity A: 1 + 3 + 5 + ... + 197 + 199
Quantity B: 2 + 4 + 6 + ... + 196 + 198

Subtract (2 + 4 + 6 + ... + 196 + 198) from both quantities to get:
Quantity A: (1 + 3 + 5 + ... + 197 + 199) - (2 + 4 + 6 + ... + 196 + 198)
Quantity B: (2 + 4 + 6 + ... + 196 + 198) - (2 + 4 + 6 + ... + 196 + 198)

Simplify:
Quantity A: 199 - 198 + 197 - 196 + . . . . + 5 - 4 + 3 - 2 + 1
Quantity B: 0

Rewrite Quantity A as follows:
Quantity A: (199 - 198) + (197 - 196) + . . . . + (5 - 4) + (3 - 2) + 1
Quantity B: 0

Simplify:
Quantity A: 1 + 1 + 1 + 1 + ... + 1 + 1 + 1 + 1
Quantity B: 0

Answer:
Show: ::
A




How does the generalized formula n(n+1)/2 adjust to the sequence of the problem? Sum of odd integers from 1 to 199.
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Re: The sum of the odd/even integers [#permalink]
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That's a perfectly valid solution, @Farina
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Re: The sum of the odd/even integers [#permalink]
GreenlightTestPrep wrote:
That's a perfectly valid solution, @Farina


Thank you very much :)
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Re: The sum of the odd/even integers [#permalink]
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Re: The sum of the odd/even integers [#permalink]
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