Find all three digit integers abc = n, such that 2n/3 = a! b! c!
Answer
432
Solution
2n/3 < 1000 < 7!, so a, b, c ≤ 6. Hence n ≤ 666, so 2n/3 ≤ 444, so a ≤ 4. Also 6! > 444, so b, c ≤ 5.
Consider first the case a = 4. Then (2n/3)/a! ≤ 2·455/(3·24) < 13. So b!c! ≤ 12, so b, c ≤ 3. Also 2n is divisible by a!3, so n is divisible by 9. So {b,c} = {2,3}, so n = (3/2)4!3!2! = 432.
Now suppose a = 3. Then n = (3/2) 6 b! c! = 9 b! c!, so we need 300 < 9 b! c! < 355, so 33 < b!c! < 40. Hence b, c ≤ 4. But 4! < 33 and 2·4! > 40, so b, c ≤ 3. Hence b = c = 3. But (3/2)3!3!3! = 324 ≠ 333, so there are no solutions with a = 3.
Suppose a = 2. Then n = 3b!c!, so 200 < 3b!c! < 255. But 3·5! > 255 and 3·3!3! < 200, so at least one of b, c must be 4. If the other is x, then 200/72 ≤ x! ≤ 255/72, so x! = 3. Contradiction, so there are no solutions with a = 2.
Finally, suppose a = 1. Then n = (3/2)b!c!, so (2/3)100 ≤ b!c! ≤ (2/3)155, and so 67 ≤ b!c! ≤ 103. But 5! > 103 and 3!3! < 67, so one of b, c must be 4. If the other is x, then 2.7 < x! < 4.3, so x! = 3 or 4. Contradiction.
Thanks to Suat Namli
© John Scholes
jscholes@kalva.demon.co.uk
13 February 2004
Last corrected/updated 13 Feb 04