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Where is my solution to 2018 AIME I Problem 1 wrong?

a can be any integer belonging to [1,100], so I count the no.of unordered pairs of positive integers whose sum is less than or equal to 100.

i.e., the no.of unordered pairs (0,1),(0,2)...(0,100)(1,1),(1,2)..(50,50) which is equal to {(101•102)/2} -1 [-1 for excluding (0,0)] which is equal to 5150 which far exceeds 2600 which is the total number.

Where is my approach wrong?

Thank you.

Ram Keswani
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  • Why aren't you including $(51,49)$? In all your pairs you have the second term equal or more than the first. Why? – fleablood Aug 02 '19 at 20:54
  • @fleablood so as to not repeat the pairs. For example (2,40) and (40,2) will give same ordered pair of (a,b). So I exclude the repetitive pairs. – Ram Keswani Aug 02 '19 at 21:00
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    $(2,40)$ and $(40,2)$ are not the same thing. – fleablood Aug 02 '19 at 21:01
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    An ordered pair is ordered. $(a,b) \ne (b,a)$. – fleablood Aug 02 '19 at 21:01
  • @fleablood right they are not but both of those ordered pairs will give the same ordered pair of (a,b) = (42,80) – Ram Keswani Aug 02 '19 at 21:02
  • You SAID "so I count the no.of distinct ordered pairs of positive integers whose sum is less than or equal to 100". But you listed the unordered pairs. There are $5150$ ordered pairs but $2600$ unordered pairs. Ask the correct question for what you want. – fleablood Aug 02 '19 at 21:13
  • @fleablood sorry I meant unordered pairs. But how is the no.of unordered pairs 2600? Sorry. – Ram Keswani Aug 02 '19 at 21:18
  • Your approach is wrong because some of those unordered pairs will result in quadratic equations that can't be factored in integers. – Robert Shore Aug 02 '19 at 21:38
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    Well you have $(a,b)$ where $a \le b$ and $a+b \le 100$ so $a=0...50$ and $b = a...100-a$. So we must add $\sum_{a=0}^{50} \sum_{b=a}^{100-a} 1 = \sum_{a=0}^{50} [(100-a)-a + 1]=\sum_{a=0}^{50}(101-2a) = 51101- 2\sum_{a=0}^{50}a = 51101- 2\frac {5051}2 = 51101-5150= 5151 = 2601$. But we must exclude $(0,0)$ so there are $2600$. – fleablood Aug 02 '19 at 21:42
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    Or... Note $(0,0).......(0,100) = 101$ and $(1,1).....(1,99) = 99$ and $(2,2).... (2,98)=97$ ..... $(49,49)...(49,51)=3$ and $(50,50) = 1$. so the sum is $1 + 3 + 5 + ....... + 101=\sum_{k=1}^{51} (2k -1) = 51^2$. – fleablood Aug 02 '19 at 21:46
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    Or... There are $5150$ ordered pairs (excluding $(0,0)$). $50$ of those are of the form $(k,k)$. and $5100$ of them are not. As we want $(a,b) \ne (b,a)$ when $a\ne b$ but we want $(a,b)\equiv (b,a)$ we divide the number of those ordered pairs in half. That is $2550$. We add the $50$ of the form $(k,k)$ so there are $2600$. – fleablood Aug 02 '19 at 21:50
  • .... basically you double counted $(a, b > a)$ and $(b,a)$ separately. Accounting for double counting... there are $2600$; not $5150$. – fleablood Aug 02 '19 at 21:53
  • @fleablood thank you. I realised my mistake. – Ram Keswani Aug 03 '19 at 13:29

1 Answers1

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Note that if we have 0 in the x coordinate then, we will have 101 terms that are possible in the y coordinate (let 0,0 work for now.) Then, for 1 in the x coordinate, have 99 terms that will work for the y coordinate (1-99). Repeat this process and add all the y coordinates possible to get the sum $101+99+...+1$ There are 51 terms in this sequence. Add 1 to every term to get $102+100+...+2$ This sum is equal to $104*51/2$ This is equal to $52*51$. We have to subtract 51 from this expression to get $51^2$ and then subtract 1 from $51^2$ because we cannot have the case (0,0). This means we have $51^2-1$ Which is 2600. Therefore, the answer is 600.

Daniel Shirazi
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