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are given a set of 100 distinct positive integers with the property that for any three of them, say $ a, b, c $, the number $ a + b - c $ is also among the 100 numbers. We are to **prove that the set must be an arithmetic progression**. --- Let the set be $ S = \{a_1, a_2, \dots, a_{100}\} $, with $ a_1 < a_2 < \dots < a_{100} $, all positive integers. ### Step 1: Understand the condition For any three elements $ a, b, c \in S $, the number $ a + b - c \in S $. Note: even if $ a, b, c $ are not distinct? The problem says "any three of them" — probably meaning any three elements, not necessarily distinct? But since the set has **distinct** integers, and we're choosing three elements from the set, they could be repeated only if we're allowed to pick the same element multiple times — but that would be a multiset. But the problem says "a set of 100 distinct positive integers", and "any three of them" — in combinatorics problems like this, "any three" usually means **any three distinct elements**. So we assume $ a, b, c $ are **distinct**. So the condition: for any distinct $ a, b, c \in S $, $ a + b - c \in S $. We must show $ S $ is an arithmetic progression. --- ### Step 2: Consider the minimal and maximal elements Let $ m = a_1 $ = smallest element, $ M = a_{100} $ = largest element. Pick $ a = M $, $ b = M $, $ c = m $. Wait — but $ a, b, c $ must be **distinct**? Then we cannot pick two copies of $ M $. So choose $ a = a_{100} = M $, $ b = a_{99} $, $ c = a_1 = m $. All distinct. Then $ a + b - c = M + a_{99} - m \in S $ Since $ M $ is the maximum, and $ a_{99} \leq M $, we have: $ M + a_{99} - m \leq M + M - m = 2M - m $ But this result must be in $ S $, so: $ M + a_{99} - m \leq M \Rightarrow a_{99} - m \leq 0 \Rightarrow a_{99} \leq m $ But $ a_{99} \geq a_1 = m $, so $ a_{99} = m $. Contradiction unless $ n = 2 $, which it isn't. Wait — that leads to contradiction. So our assumption that $ M + a_{99} - m > M $ must be invalid? But why? Wait, let me compute the value: We know $ M = a_{100} $, $ a_{99} < M $, $ m = a_1 $ Then: $ a + b - c = M + a_{99} - m $ This must be in $ S $, so it must be ≤ $ M $, because $ M $ is maximum. So: $$ M + a_{99} - m \leq M \Rightarrow a_{99} - m \leq 0 \Rightarrow a_{99} \leq m $$ But $ a_{99} \geq a_1 = m $, so $ a_{99} = m $ But that's impossible since there are 100 distinct integers; $ a_1 = m $, $ a_{99} > a_1 $, so $ a_{99} \geq m + 98 $ Therefore, **$ M + a_{99} - m > M $** (since $ a_{99} - m \geq 98 $, so $ M + a_{99} - m \geq M + 98 > M $) But this contradicts the maximality of $ M $: the result is larger than maximum element, yet must be in $ S $. Contradiction? Wait — **this suggests our condition implies that $ M + a_{99} - m \leq M $**, which forces $ a_{99} \leq m $, which is impossible. So we must have made a wrong assumption. Wait — maybe $ a + b - c $ doesn’t have to be ≤ M? But if $ M $ is the maximum element in the set, and $ S $ contains only these 100 elements, then **any value $ a + b - c $ must be ≤ M**, otherwise it's not in $ S $. But we have shown: $ a + b - c = M + a_{99} - m \geq M + (m + 98) - m = M + 98 > M $ So this number is **greater than M**, so it **cannot** be in S. But the condition says it must be in S. Contradiction. This is a problem — unless our assumption that S is any arbitrary set is wrong? Or we chose bad values? Wait — we assumed the set is not an AP and reached contradiction. But we haven't used the AP property yet. Actually, maybe the only way this contradiction doesn’t happen is if $ a_{99} - m $ is very small — but how? Wait, **maybe the set is an arithmetic progression**? Let’s test it. ### Step 3: Try an arithmetic progression. Suppose $ S = \{ a, a + d, a + 2d, \dots, a + 99d \} $ Take any three elements: $ x = a + id $, $ y = a + jd $, $ z = a + kd $ Then $ x + y - z = (a + id) + (a + jd) - (a + kd) = a + (i + j - k)d $ Which is again in the set, since $ i + j - k $ is an integer between $ 0 + 0 - 99 = -99 $ and $ 99 + 99 - 0 = 198 $, but we need the result to be in $ S $, which only has indices from 0 to 99. So we need $ 0 \leq i + j - k \leq 99 $ — which is **not always true**! For example, take $ i = 99 $, $ j = 99 $, $ k = 0 $: then $ i + j - k = 198 $, which is way beyond 99. So $ x + y - z = a + 198d $, which is **not in the set**, even

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