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Let A = {n ∈ Z | |n| ≤ 24}. In how many ways can two distinct numbers be chosen
(simultaneously) from A such that their product is less than their sum?
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We consider the set of integer points {(x,y)} in a coordinate plane,
where -24 <= x <= 24, -24 <= y <= 24, and we want to find the number of points
(x,y) such that
xy < x + y. (1)
This inequality (1) is equivalent to
xy - x - y < 0,
1 + xy - x - y < 1,
(1-x)*(1-y) < 1. (2)
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| This inequality (2) is easy to analyze. |
+-------------------------------------------+
(a) First set of solutions in integer numbers is those pairs (x,y),
where x = 1. It gives all the pairs (1,y) with y = -24, -23, . . . ,23, 24.
In all, there are 24+24+1 = 49 such pairs.
(b) The other set of solutions in integer numbers is those pairs (x,y),
where y = 1. It gives all the pairs (x,1) with x = -24, -23, . . . ,23, 24.
In all, there are 24+24+1 = 49 such pairs.
But the pair (1,1) was just accounted in (a); so, it adds only 49-1 = 48 pairs.
For now, we have 49 + 48 = 97 different pairs.
Thus, we covered cases, when one or both factors in (2) are zero,
so their product is zero.
Now we will consider cases with the negative left side in (2).
Left side of (2) is negative if and only if one factor is positive, while the other factor is negative.
(c) So, third set of solutions is all pairs (x,y), where x > 1, y < 1.
In all, there are (24-1)*(24+1) = 23*25 = 575 such pairs.
They all are different from pairs found in (a) and (b), so we can add 575 to 97.
For now, we have 97 + 575 = 672 different pairs.
(d) Fourth set of solutions is all pairs (x,y), where x < 1, y > 1. ( This set is symmetric to (c) ).
In all, there are (24-1)*(24+1) = 23*25 = 575 such pairs.
They all are different from pairs found in (a), (b) and (c), so we can add 575 to 672.
It gives 672 + 575 = 1247 solutions, in all.
So, inequality (2) has 1247 solutions in the set of ordered pairs {(x,y) | x,y ∈ Z |, |x| ≤ 24, |y| ≤ 24}.
But the problem asks for DISTINCT NUMBERS x and y. In the found set of solutions,
there is only one pair with coinciding x and y: it is the pair (1,1).
By excluding this pair, the ANSWER to the problem 's question is 1247-1 = 1246.
Solved.
This problem is from 22nd Philippine Mathematical Olympiad of 2019.
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Clarification after post by Edwin.
The number of ordered pairs (x,y) with x =/= y is 1246.
So, it considers the pairs (x,y) and (y,x), x=/=y as distinct pairs
(for example, (0,1) and (1,0) are considered as distinct pairs).
Visually, it is the number of integer points/solutions in the square on the coordinate plane.
The number of unordered pairs is 1246/2 = 623.
So, it considers the pairs (x,y) and (y,x), x=/=y as the same pair
(for example, (0,1) and (1,0) are considered as the same pair).
As the incoming post is worded quite cloudy, the problem can be interpreted
in any way (ordered pairs or unordered pairs).
The problem (the context) does not give a preference to any of these two possible interpretations.
They both have equal rights to exist and to be considered.
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| So, which interpretation to accept is not the matter of principle - |
| - it is only the matter of agreement. Any interpretation is valid. |
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| The only important thing for a reader to worry about is that |
| the interpretation of the question and the answer be consistent. |
+------------------------------------------------------------------------+
So, I thankful to Edwin for recognizing my job,
but I do not agree that my solution has an error.
Simply Edwin keeps one interpretation, while I keep another, equally valid,
and there is no any contradiction between them.
