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.

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

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!

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)

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.

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:



How does the generalized formula n(n+1)/2 adjust to the sequence of the problem? Sum of odd integers from 1 to 199.

That's a perfectly valid solution, @Farina

Thank you very much

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