Math Is Fun Forum

tony123

Member

Registered: 2007-08-03

Posts: 229

ten numbers

Given ten numbers 2, 3, 5, 6, 7, 8, 10, 11, 12, 13,
one must cross
out several of them so that the total of any of the remaining numbers would
not be an exact square
(that is, the sum of any two, three, four, : : :, and of
all the remaining numbers would not be an exact square).

At most how many numbers can remain?

JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Re: ten numbers

Well, we have these sets

The sum of the numbers in each set is a perfect square; therefore we must strike out at least one number from each set. Hence we must remove at least 4 numbers from the lot. Do we have to remove any more?

Well, yes.

Note that 6 + 10 = 16, so one of 6 and 10 has to go. Similarly, as 3 + 5 + 8 = 16, one of 3, 5 and 8 has to go. If it’s the 8 or the 10 that goes, then that's the 5th number we’ve removed. Suppose both the 10 and 8 stay. Then the 6 has to go, and so must one of 3 and 5. If the 3 goes then we’ve removed both numbers in set B, and so we’ve removed 5 numbers already. We shall see that if we insist on keeping the 8 and the 10, then one of the sets above will have to have both its numbers removed.

Now look and some more square sums. We have 7 + 8 + 10 = 25 and so the 7 must go. If the 2 also goes, then we’ve removed two numbers from set A. So suppose the 2 stays put. Now, what do we have? 2 + 11 + 12 = 25. Hence, one of 11 and 12 must go.

(i) If the 11 goes and the 5 remains, then the 3 must go. (One of 3 and 5 must go, remember.) The 6 has already gone. Hence set B has been emptied.

(ii) If the 12 goes and the 13 remains, then now we have 3 + 13 = 16, so again the 3 must go. Same result: set B is emptied.

QED. We have shown that at least 5 numbers must be removed.

Unfortunately, I don’t think that’s the end of the story. I think we still have to remove one more number. I’m tired now, so I’ll continue with this tomorrow, okay?

Note that 6 will definitely be the most we need to remove. It is perfectly possible to be left with 4 numbers with square-free sums:

JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Re: ten numbers

Wait! Wait! Look what I found!

So forget what I said at the end of my last post. The minimum number of numbers that must be removed is 5. The maximum number of numbers that can remain is 5. Problem solved!