by RonPurewal Wed Jul 23, 2014 4:35 am
That problem is definitely not happening with algebra, so we'll have to test cases.
As always when testing cases, try for "not sufficient". In other words, try to find...
... a case in which ≥14 items were sold on one day,
... a case in which, every day, ≤13 items were sold.
The "border" cases (for the greatest # of items sold in a day) are 13 and 14, so those are the cases of greatest interest to us here. (No sense trying to get a number bigger than 14 or less than 13 if we can get those.)
Statement 1: The greatest and least numbers of items are 3 apart.
Ok. If the greatest number is 14, the least number must be 11. Can we do that?
Well, there are 25 items left. So... sure. 14, 14, 11, 11 works. So does 14, 13, 12, 11.
If the greatest number is only 13, then the smallest number must be 10.
Can we do that?
50 -Â 13 -Â 10 = 27 items left.
Won't work"”"”you can't split 27 into two numbers that are both 13 or less.
Clearly won't work with any number smaller than 13, either"”"”you'll just have the same problem, only worse. So, sufficient.
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