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If the sum of four consecutive positive integers a three digit multiple of 50, the mean of the these integers must be one of k possible values, where k=

I tried to come up w/ a test case to solve this.. I knew that the ones digits of the 4 consecutive digits had to add up to either 5 or 0 for it to be a multiple of 50..

I went w/ 61, 62, 63, and 64.. all consectuive..and add up to 250 ..

and thus the mean is greater than 10... so anything other than E has to be wrong...

Since a is an integer, you have to eliminate 150, 250,....etc. You have 100, 200,...,900 left. Count them. It is 9. So only 9 possibilities are there. Ans C. _________________

4 consecutive integers cannot add up to 100, try it.

therefore we have 150,250,350 ... 950, so 9

some notes about consecutives:
the average of an even number of consecutives is never a whole number

the average of an odd number of consecutives is a whole number

the sum of an odd number of consecutives is divisible by the number of terms (1,2,3 = 6/3 = 2 or 10,11,12,13,14,15 = 65/5 = 13). this DOES not apply to an even number of terms.

the product of any n consecutives has n as a factor.

I couldn’t help myself but stay impressed. young leader who can now basically speak Chinese and handle things alone (I’m Korean Canadian by the way, so...