Can Someone Solve This?

Question

given 7 distinct positive integers that add up to 100, prove that some three of them add up to at least 50. (hint: arrange them in order: a

Answer 1

If the three largest numbers add up to less than 50, then the THIRD largest number has to be at most 15. It can't be 16, because then the three largest numbers would add up to at least 16 + 17 + 18 = 51.

However, if the third largest number is 15, then the four smallest numbers add up to at most 14 + 13 + 12 + 11 = 50.

So, you have at most 50 for the four smallest numbers, less than 50 for the three largest numbers - the sum has to be less than 100.

Answer 2

Assume a, b, c, d, e, f, g, are positive integers that add up to 100, with the order a < b < c < d < e < f < g. It then follows that

a + b + c + d + e + f + g = 100

{We want to prove that there exists three of those 7 numbers which add up to at least 50.}

Since each number is distinct and positive, it follows that:
a >= 1, b >= 2, c >= 3, d >= 4, e >= 5, f >= 6, g >= 7. Therefore, if we add some of these inequalities together,

a + b + c + d >= 10

From the equation a + b + c + d + e + f + g = 100, since
100 = 10 + 90, it follows that

a + b + c + d + e + f + g = 10 + 90, so
a + b + c + d + e + f + g - 90 = 10

Therefore, given

a + b + c + d >= 10, and using what we just solved for 10, we have

a + b + c + d >= a + b + c + d + e + f + g - 90

Subtracting a + b + c + d both sides,

0 >= e + f + g - 90, or
90 >= e + f + g
e + f + g <= 90

Hmm.. can't figure this out. Will work on it more later.
Edit: Looks like someone else figured it out; bailing out.

Answer 3

If we arrange the solution sets in order we have:

The first set of 7 positive integers adding to 100 are (1, 2, 3, 4, 5, 6, 79).
The last set of three positive integers adding to 100 are (11, 12, 13, 14, 15, 16, 19).

The first set has the lowest 4 digits we can begin with (1,2,3,4) which mean that the 3 remaining numbers must add to 90.

The last set has the highest 4 highest digits we can start with (11, 12, 13, 14) which mean the 3 remaining numbers must add to 50.

Hence, the last three digits in an ordered set must add to at least 50.

Answer 4

you have g+f+5e greater than 110 because e-a greater than 4 etc.

If e less than 15 then g+f+e greater than 50. If e>15, then f is at least 17 and g 18 so e+f+g>50.

Answer 5

2, 3, 10, 12, 13, 15, 45

2 + 3 + 45 = 50

Answer 6

1,2,3,6,19,59,10


59+1+2=62
62>50
proved