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When Nancy returned home from the airport, she was unable to open her suitcase. She recalled setting a five-digit lock code using the digits 2, 3, 4, 6, and 8, with no digit repeated. She is certain that the third digit of the code is 3.
Is it necessarily true that the fifth digit of the code is 6?
(1) The first digit of the code is a divisor of the second digit. (2) The fourth digit of the code is a multiple of the second digit.
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Code format: d1 d2 d3 d4 d5 Available digits for positions 1, 2, 4, 5: {2, 4, 6, 8} (each used exactly once) Question: Must d5 = 6?
Statement (1) alone: The first digit is a divisor of the second digit From {2, 4, 6, 8}, find pairs where d1 divides d2: - 2 divides: 4, 6, 8 - 4 divides: 8 - 6 and 8 don't divide any other available digits Possible codes: - d1 = 2, d2 = 4 → remaining {6, 8} for positions 4 and 5 - d1 = 2, d2 = 6 → remaining {4, 8} for positions 4 and 5 - d1 = 2, d2 = 8 → remaining {4, 6} for positions 4 and 5 - d1 = 4, d2 = 8 → remaining {2, 6} for positions 4 and 5 Multiple arrangements are possible, so d5 could be 6, 8, 4, or 2. Statement (1) alone is NOT sufficient.
Statement (2) alone: The fourth digit is a multiple of the second digit From {2, 4, 6, 8}, find pairs where d4 is a multiple of d2: - 2 is a factor of: 2, 4, 6, 8 (but d4 ≠ d2) - 4 is a factor of: 8 - 6 and 8 don't go into any other available digits Similar analysis shows multiple possibilities for d5. Statement (2) alone is NOT sufficient.
Both statements together: Testing all cases where d1 divides d2 AND d4 is a multiple of d2: • d1=2, d2=4: d4 must be 8 (only multiple of 4 left) and d5=6 • d1=2, d2=6: No multiple of 6 available so Impossible • d1=2, d2=8: No multiple of 8 available so Impossible • d1=4, d2=8: No multiple of 8 available so Impossible Only valid code: 2 4 3 8 6
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