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If \(x\) is a two-digit positive odd integer, is \((x - 7)(x - 5)(x - 3)(x - 1)\) divisible by 1920?
(1) \(x + 7\) is a factor of 150
(2) \(x + 17\) is actor of 120
(1) \(x + 7\) is a factor of 150
(2) \(x + 17\) is actor of 120
23 Sep 2021, 04:50
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Official Solution:
If \(x\) is a two-digit positive odd integer, is \((x - 7)(x - 5)(x - 3)(x - 1)\) divisible by 1920?
First of all, factor 1920: \(1920=2^7*3*5\)
Next, notice that since \(x\) is odd, then \((x - 7)(x - 5)(x - 3)(x - 1)\) will be the product of four consecutive even integers: \((2k)(2k+2)(2k+4)(2k+6)=2^4(k)(k+1)(k+2)(k+3)\). Now, \((k)(k+1)(k+2)(k+3)\) is the product of four consecutive integers, thus one of them will be a multiple of 2, one of them will be a multiple of 4 (\(2^2\)) and at least one of them will be a multiple of 3: \((x - 7)(x - 5)(x - 3)(x - 1)=2^4(k)(k+1)(k+2)(k+3)=2^4*2*2^2*3*something=2^7*3*something\). So, basically all we need to find out is whether \((x - 7)(x - 5)(x - 3)(x - 1)\) is divisible by 5 (because we know that it's for sure is divisible by \(2^7*3\)).
(1) \(x + 7\) is a factor of 150
Since \(x\) is a two-digit positive odd integer, then \(x + 7\) is a two-digit positive even integer. Even two-digit factors of 150 are 10, 30 and 50 only. So, \(x\) is either 23 or 43 (\(x\) cannot be 7 because we are told that \(x\) is a two-digit number). In both of these cases \((x - 3)\) will be a multiple of 5, thus\((x - 7)(x - 5)(x - 3)(x - 1)\) will be divisible by 5. Sufficient.
(2) \(x + 17\) is actor of 120
Since \(x\) is a two-digit positive odd integer, then \(x + 17\) is a two-digit positive even integer. Even two-digit factors of 120 are 10, 12, 20, 24, 30, 40, and 60 only. So, \(x\) is either 13, 23 or 43 (\(x\) cannot be -7, -5, 3, or 7 because we are told that \(x\) is a two-digit positive number). In all of these cases \((x - 3)\) will be a multiple of 5, thus\((x - 7)(x - 5)(x - 3)(x - 1)\) will be divisible by 5. Sufficient.
Answer: D
If \(x\) is a two-digit positive odd integer, is \((x - 7)(x - 5)(x - 3)(x - 1)\) divisible by 1920?
First of all, factor 1920: \(1920=2^7*3*5\)
Next, notice that since \(x\) is odd, then \((x - 7)(x - 5)(x - 3)(x - 1)\) will be the product of four consecutive even integers: \((2k)(2k+2)(2k+4)(2k+6)=2^4(k)(k+1)(k+2)(k+3)\). Now, \((k)(k+1)(k+2)(k+3)\) is the product of four consecutive integers, thus one of them will be a multiple of 2, one of them will be a multiple of 4 (\(2^2\)) and at least one of them will be a multiple of 3: \((x - 7)(x - 5)(x - 3)(x - 1)=2^4(k)(k+1)(k+2)(k+3)=2^4*2*2^2*3*something=2^7*3*something\). So, basically all we need to find out is whether \((x - 7)(x - 5)(x - 3)(x - 1)\) is divisible by 5 (because we know that it's for sure is divisible by \(2^7*3\)).
(1) \(x + 7\) is a factor of 150
Since \(x\) is a two-digit positive odd integer, then \(x + 7\) is a two-digit positive even integer. Even two-digit factors of 150 are 10, 30 and 50 only. So, \(x\) is either 23 or 43 (\(x\) cannot be 7 because we are told that \(x\) is a two-digit number). In both of these cases \((x - 3)\) will be a multiple of 5, thus\((x - 7)(x - 5)(x - 3)(x - 1)\) will be divisible by 5. Sufficient.
(2) \(x + 17\) is actor of 120
Since \(x\) is a two-digit positive odd integer, then \(x + 17\) is a two-digit positive even integer. Even two-digit factors of 120 are 10, 12, 20, 24, 30, 40, and 60 only. So, \(x\) is either 13, 23 or 43 (\(x\) cannot be -7, -5, 3, or 7 because we are told that \(x\) is a two-digit positive number). In all of these cases \((x - 3)\) will be a multiple of 5, thus\((x - 7)(x - 5)(x - 3)(x - 1)\) will be divisible by 5. Sufficient.
Answer: D
