Hi hiteshahire22,

This question can be solved with "bunching."

We're asked for the SUM of all of the EVEN integers from 1 to 99, so we really want the sum of 2, 4, 6, 8.....98.

If you take the smallest and biggest numbers, you have...

2+98 = 100

If you take the next smallest and next biggest numbers, you have...

4+96 = 100

This pattern will continue on, so we're going to have a bunch of 100s, BUT we have to be careful to make sure that we don't miss a number if it's "in the middle" and doesn't get paired up. Since we know that the sum of each pair is 100, we can 'jump ahead' to find the last few pairs...

44+56 = 100

46+54 = 100

48+52 = 100

There IS a middle number: 50; this number does NOT get paired up.

Since 48 is the 24th even integer, we know there are twenty-four 100s + one 50. 2400+50 = 2450

Final Answer:

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Rich

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