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My reasoning was a bit long but still within 2 min. range
If you match the numbers in diff. pairs so as to have sums of 100 ( ie 1+99, 2+98, etc... ) You will then have 49 groups of 100. You then add to that 100 and the 50 which is the median number. The sum will be 50x100=5000
5000+50=5050. 5050, of the given numbers, is divisible by 2 only _________________

Beautiful! I'll memorize the sum on n numbers and sum of squares of those n numbers. I think that should be sufficient for the exam. Sum of cubes is a bit outstretched _________________

can somebody provide the list of formulae--basic maths? I have had trouble keeping it at one place. I just don't wanna kill time on something easy/stupid (in the real exam)

can somebody provide the list of formulae--basic maths? I have had trouble keeping it at one place. I just don't wanna kill time on something easy/stupid (in the real exam)