Riddle #846

logicmathclever

A deliveryman comes to a house to drop off a package. He asks the woman who lives there how many children she has. "Three," she says. "And I bet you can't guess their ages." "Ok, give me a hint," the deliveryman says. "Well, if you multiply their ages together, you get 36," she says. "And if you add their ages together, the sum is equal to our house number." The deliveryman looks at the house number nailed to the front of her house. "I need another hint," he says. The woman thinks for a moment. "My youngest son will have a lot to learn from his older brothers," she says. The deliveryman's eyes light up and he tells her the ages of her three children. What are their ages?
Their ages are 1, 6, and 6. We can figure this out as follows: Given that their ages multiply out to 36, the possible ages for the children are: 1, 1, 36 (sum = 38) 1, 2, 18 (sum = 21) 1, 3, 12 (sum = 16) 1, 4, 9 (sum = 14) 1, 6, 6 (sum = 13) 2, 2, 9 (sum = 13) 2, 3, 6 (sum = 11) 3, 3, 4 (sum = 10) When the woman tells the deliveryman that the children's ages add up to her street number, he still doesn't know their ages. The only way this could happen is that there is more than one possible way for the children's ages to add up to the number on the house (or else he would have known their ages when he looked at the house number). Looking back at the possible values for the children's ages, you can see that there is only one situation in which there are multiple possible values for the children's ages that add up to the same sum, and that is if their ages are either 1, 6, and 6 (sums up to 13), or 2, 2, and 9 (also sums up to 13). So these are now the only possible values for their ages. When the woman then tells him that her youngest son has two older brothers (who we can tell are clearly a number of years older), the only possible situation is that their ages are 1, 6, and 6.
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64 votes

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First create an array A of length 26, representing the counts of each letter of the alphabet, with each value initialized to 0. Iterate through each character in S1 and add 1 to the corresponding entry in A. Once this iteration is complete, A will contain the counts for the letters in S1. Then, iterate through each character in S2, and subtract 1 from each corresponding entry in A. Now, if the each entry in A is 0, then S1 and S2 are anagrams; otherwise, S1 and S2 aren't anagrams. Here is pseudocode for the procedure that was described: def areAnagrams(S1, S2) A = new Array(26) A.initializeValues(0) for each character in S1 arrayIndex = mapCharacterToNumber(character) //maps "a" to 0, "b" to 1, "c" to 2, etc... A[arrayIndex] += 1 end for each character in S2 arrayIndex = mapCharacterToNumber(character) A[arrayIndex] -= 1 end for (i = 0; i < 26; i++) if A[i] != 0 return false end end return true end
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