Cracking a Three-Digit Safe Combination
A safe has a three-digit combination. The digits are all different and are chosen from 0 through 9. The first digit cannot be 0. The sum of the three digits is 12. The three-digit number formed by the digits (in order) is divisible by 4. The tens digit is exactly twice the hundreds digit. The owner remembers these clues but has forgotten the exact combination. He needs to find the unique three-digit number that satisfies all the conditions. He starts listing possibilities but gets confused by the interactions among the clues. Help him deduce the combination logically. Also, suppose the sum of the digits were 13 instead of 12. Would there still be a unique three-digit number satisfying all the other clues? If so, what is it? If not, explain why. Provide a clear step-by-step reasoning.
1. Find the three-digit combination. Show your reasoning step by step.
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Let the hundreds digit be h, tens digit be t, and units digit be u. We are told t = 2h. Since t is a digit, h can be 1,2,3,4 (h=0 not allowed as first digit; h=5 gives t=10 invalid). Also all digits different. The sum is h + t + u = h + 2h + u = 3h + u = 12, so u = 12 - 3h. For h=1: u=9, t=2 -> number 129. Check divisibility by 4: last two digits 29, 29 not divisible by 4. So 129 invalid. h=2: u=6, t=4 -> 246. Last two digits 46, not divisible by 4. Invalid. h=3: u=3, t=6 -> 363. Digits not all different (3 appears twice). Invalid. h=4: u=0, t=8 -> 480. Last two digits 80, 80/4=20, divisible by 4. Digits 4,8,0 all different. Valid. So the unique combination is 480.
2. Explain why no other number works for the sum condition 12.
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From t=2h and sum=12, we derived u=12-3h. The only possible h values are 1,2,3,4. We checked each: h=1 gives 129 (not divisible by 4); h=2 gives 246 (not divisible by 4); h=3 gives 363 (digits not all different); h=4 gives 480 (works). Since these are all possibilities, no other number can satisfy all clues. The reasoning is exhaustive, so the solution is unique.
3. If the sum of the digits were 13 instead of 12, would there still be a unique solution? Find it or explain why not.
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With sum=13, u = 13 - 3h. h can be 1,2,3,4. h=1: u=10, impossible (digit >9). h=2: u=7, t=4 -> 247. Last two digits 47, not divisible by 4. h=3: u=4, t=6 -> 364. Last two digits 64, 64/4=16, divisible by 4. Digits 3,6,4 all different. Valid. h=4: u=1, t=8 -> 481. Last two digits 81, not divisible by 4. So only 364 works. Therefore there is still a unique solution: 364.