Cadbury manufactures a chocolate box that contains an "x" number of chocolates. There are three houses A, B,…

Comments

• KG AI MentorKG AI Mentor24 Feb 2026

  Quick Answer The number of chocolates distributed is 106, and the total number of chocolates in the box is 107. Key Steps Let the total number of chocolates be $ x $. From the problem: When distributed among 3 children (house A), 1 chocolate remains → $ x \equiv 1 \pmod{3} $ When distributed among 5 children (house B), 1 chocolate remains → $ x \equiv 1 \pmod{5} $ When distributed among 7 children (house C), 1 chocolate remains → $ x \equiv 1 \pmod{7} $ So, $ x - 1 $ is divisible by 3, 5, and 7. Thus, $ x - 1 $ is a common multiple of 3, 5, and 7. LCM of 3, 5, 7 = $ 3 \times 5 \times 7 = 105 $ So, $ x - 1 = 105k $ → $ x = 105k + 1 $ Smallest such $ x $ (positive) is when $ k = 1 $: $ x = 105 + 1 = 106 $ But wait — the question asks: "what is the number of chocolates distributed?" The chocolates distributed = total chocolates minus the leftover. Leftover = 1 chocolate in each case → but the leftover is after distributing to all children. Wait — the key is: the same box is used for all three houses. So the same $ x $ chocolates are distributed to each house separately, and in each case, 1 chocolate is left. So the distribution happens separately for each house: To house A (3 children): $ x \div 3 $ → remainder 1 To house B (5 children): $ x \div 5 $ → remainder 1 To house C (7 children): $ x \div 7 $ → remainder 1 So $ x \equiv 1 \pmod{3} $, $ x \equiv 1 \pmod{5} $, $ x \equiv 1 \pmod{7} $ → $ x \equiv 1 \pmod{\text{LCM}(3,5,7)} $ → $ x \equiv 1 \pmod{105} $ So $ x = 105k + 1 $ Smallest such $ x $ is $ 106 $. Now, number of chocolates distributed = total chocolates minus leftover. But the leftover is 1 in each case — but the distribution is done separately for each house. So the total chocolates distributed is not $ x - 1 $, because the leftover is after each distribution, and the chocolates are not shared across houses. Wait — the problem says: "After distributing the chocolates equally to all children he was left with one chocolate in each case." This implies: He distributed chocolates to house A → 1 left He distributed chocolates to house B → 1 left He distributed chocolates to house C → 1 left But it doesn't say he used the same box for all. But the box contains $ x $ chocolates. So likely, he used the same box for all three distributions — but that would mean he distributed to A, then B, then C, using the same $ x $ chocolates. But that doesn't make sense because after distributing to A, he would have fewer chocolates. Alternatively, the same box is used to distribute to each house separately — meaning, for each house, he takes the box (with $ x $ chocolates), distributes equally among the children, and 1 is left. So the same number $ x $ satisfies $ x \equiv 1 \pmod{3} $, $ x \equiv 1 \pmod{5} $, $ x \equiv 1 \pmod{7} $. So $ x \equiv 1 \pmod{105} $ → smallest $ x = 106 $ Now, the number of chocolates distributed — but distributed to whom? The question says: "what is the number of chocolates distributed?" But it's ambiguous: Is it the total distributed across all houses? Or the number distributed in one case? But the problem says: "After distributing the chocolates equally to all children he was left with one chocolate in each case." So in each case, he distributed chocolates to the children of that house, and 1 was left. So for house A: distributed $ x - 1 $ chocolates For house B: distributed $ x - 1 $ chocolates For house C: distributed $ x - 1 $ chocolates But that would mean he used $ x $ chocolates each time — so total chocolates used = $ 3(x - 1) $, but that would require $ 3x $ chocolates, which contradicts the box having only $ x $ chocolates. So the only logical interpretation is: The same box of $ x $ chocolates is used to distribute to each house separately. So for each house, he starts with $ x $ chocolates, distributes equally, and 1 is left. So $ x \equiv 1 \pmod{3} $, $ x \equiv 1 \pmod{5} $, $ x \equiv 1 \pmod{7} $ So $ x \equiv 1 \pmod{105} $ → $ x = 106 $ (smallest such number) Now, the number of chocolates distributed — but distributed in each case? Or total? The question says: "what is the number of chocolates distributed?" — singular. But in each case, he distributed $ x - 1 = 105 $ chocolates. So the number distributed in each case is 105. But the question might be asking for the total number distributed across all three houses. But that would be $ 3 \times 105 = 315 $, but that would require $ 315 + 3 = 318 $ chocolates (since 1 left in each case), but the box only has $ x = 106 $, so that's impossible. So the only consistent interpretation is: The same box is used for each distribution — so he does three separate distributions, each starting with $ x $ chocolates. So the number of chocolates distributed in each case is $ x - 1 $. But the question asks: "what is the number of chocolates distributed?" — likely meaning in one case, or the value of $ x - 1 $. But the problem says: "what is the number of chocolates distributed?" — and then "what is the answer" — so likely they want the total number of chocolates in the box, or the number distributed. But the number distributed in each case is $ x - 1 = 105 $. But the total number of chocolates is $ x = 106 $. But the question says: "what is the number of chocolates distributed?" — so likely $ x - 1 = 105 $. But let's re-read: "After distributing the chocolates equally to all children he was left with one chocolate in each case." So in each case, he distributed chocolates to the children, and 1 was left. So the number distributed = $ x - 1 $. So the answer is $ x - 1 = 105 $. But the question says: "what is the number of chocolates distributed?" — and then "what is the answer" — so likely they want the number distributed, which is $ x - 1 = 105 $. But the total number of chocolates is $ x = 106 $. But the question asks for the number distributed, not the total. So the answer is 105. But let's check: $ x = 106 $ To 3 children: $ 106 \div 3 = 35 $ each, remainder 1 → 105 distributed To 5 children: $ 106 \div 5 = 21 $ each, remainder 1 → 105 distributed To 7 children: $ 106 \div 7 = 15 $ each, remainder 1 → 105 distributed So in each case, 105 chocolates are distributed. So the number of chocolates distributed is 105. But the question says: "what is the number of chocolates distributed?" — and it's the same in each case. So the answer is 105. But the user might be confused — the total number of chocolates in the box is 106, but the number distributed is 105. So the answer is 105. But let's see the original post: "what is the number of chocolates distributed? what is the answer" — so likely they want the number distributed, which is $ x - 1 = 105 $. So final answer: 105 But wait — the question says: "what is the number of chocolates distributed?" — and the distribution is done to the children in the houses — but it's not clear if it's total or per house. But the phrase "in each case" suggests that the distribution is done separately for each house, and in each case, the number distributed is $ x - 1 $. So the number distributed in each case is 105. So the answer is 105. But the question says: "what is the number of chocolates distributed?" — so likely they want the number distributed in one distribution, which is 105. So the answer is 105. But let's see if there's a different interpretation. Alternative interpretation: He distributes the chocolates to all children together — but that doesn't make sense because the houses are separate. Or he distributes to all children in all houses — total children = 3 + 5 + 7 = 15. Then $ x \equiv 1 \pmod{15} $ — but the problem says "in each case", so likely not. So the only consistent interpretation is that the same box is used to distribute to each house separately, and in each case, 1 chocolate is left. So $ x \equiv 1 \pmod{3} $, $ x \equiv 1 \pmod{5} $, $ x \equiv 1 \pmod{7} $ → $ x \equiv 1 \pmod{105} $ → $ x = 106 $ Number distributed in each case = $ x - 1 = 105 $ So the number of chocolates distributed is 105. Answer: 105 — KG AI Mentor 🤖