UKMT Special (Problem 22)

The largest of 44 different real numbers is dd.

When the numbers are summed in pairs, the four largest sums are

9,10,12,139, 10, 12, 13

What are the possible values of dd?

[UKMT Hamilton Olympiad 20172017, H44]

Note by Yajat Shamji
8 months, 2 weeks ago

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You have until next Sunday, 3:003:00pm!

Yajat Shamji - 8 months, 2 weeks ago

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Let the other 33 numbers be a,b,ca, b, c in increasing order. Then we know that c+dc+d is the largest sum, and b+db+d is the next largest. Whether a+da+d or b+cb+c is greater cannot be guaranteed. Let us do this in two cases -

Case 11 - a+da+d is larger than b+cb+c

Then, we know that b+cb+c must be fourth largest. Thus, we have the following equations,

c+d=13eq(1)b+d=12eq(2)a+d=10eq(3)b+c=9eq(4)c + d = 13 \ldots \text{eq(1)} \\ b + d = 12 \ldots \text{eq(2)} \\ a+d = 10 \ldots \text{eq(3)} \\ b + c = 9 \ldots \text{eq(4)}

From equations 11 and 44, we can simplify to get b=d4eq(5)b = d - 4 \ldots \text{eq(5)}

From equations 22 and 55 now, we can simplify to get d=8\to d = 8 (and b=4b=4, which can help us solve for aa and cc, but that is not required)

First Case : d=8d = 8


Case 22 - b+cb+c is larger than a+da+d

Then, we know that b+cb+c must be fourth largest. Thus, we have the following equations,

c+d=13eq(1)b+d=12eq(2)b+c=10eq(3)a+d=9eq(4)c + d = 13 \ldots \text{eq(1)} \\ b + d = 12 \ldots \text{eq(2)} \\ b+c = 10 \ldots \text{eq(3)} \\ a+d = 9 \ldots \text{eq(4)}

From equations 22 and 33, we can simplify to get c=d2eq(5)c = d - 2 \ldots \text{eq(5)}

From equations 11 and 55 now, we can simplify to get d=7.5\to d = 7.5 (and c=5.5c=5.5, which can help us solve for aa and bb, but again, that is not required)

Second Case : d=7.5d = 7.5


Thus,

d=8 or 7.5d = 8 \text{ or } 7.5

A Former Brilliant Member - 8 months, 2 weeks ago

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@Yajat Shamji - Where did you find these problems? They are interestingly easy lol

A Former Brilliant Member - 8 months, 2 weeks ago

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Go to this note:

UKMT Specials Note

Yajat Shamji - 8 months, 2 weeks ago

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@Yajat Shamji Nice :)

A Former Brilliant Member - 8 months, 2 weeks ago

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As for your answer, it's correct!

But... slightly different method.

Yajat Shamji - 8 months, 2 weeks ago

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@Yajat Shamji Ok (some text)

A Former Brilliant Member - 8 months, 2 weeks ago

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