高中數學題請教，真誠感謝！

How many different numbers can be formed by multiplying some or all of the numbers 2, 3, 4, 5, 6, 7, 8?

 

****************************************************************************************************謝謝 tibuko

 

1. The potential products are determined by the prime factors, so from 2-8, we have prime numbers of 2, 3, 5, 7.  The maximum exponent on 2 is (1+2+3+1=7), the maximum exponent on 3 is (1+1=2), and the maximum exponent on 5 and 7 is 1.  So the maximum number of products are: (7+1)*(2+1)*(1+1)*(1+1) = 96.  The smallest is 1 = 2^0 * 3^0 * 5^0 * 7^0, and the largest is 40320 = 2^7 * 3^2 * 5^1 * 7^1 = 2 * 3 * 4 * 5 * 6 * 7 * 8

2. Now, we need to actually remove products that are not possible.  Exponents of 5 and 7 are all possible.  Exponents of 2 and 3 are all possible, because of the factors have exponents of 2 in the form of (1, 1, 2, 3), and exponents of 3 in the form of (1, 1).  However, there is one number 6=2*3, which makes certain combinations of exponents of 2 and 3 not possible.  Namely, if 6 is present, then the exponent of 2 cannot be 0, and the exponent of 3 cannot be 0.  How many of  these impossible combinations are there?  Well, if the exponent of 2 is maximum at 7, then 6 must be present, and the exponent of 3 cannot be 0.  Vice versa, if the exponent of 3 is maximum at 2, then 6 must be present, and the exponent of 2 cannot be 0.  In each case, only the exponents of 5 and 7 are flexible, so each case have 4 numbers, for a total of 8 numbers.

3. We further need to remove the cases that are NOT multiplications. Therefore, we remove the following numbers: 1, 2, 3, 4, 5, 7

4. The total number is therefore: 96 - 8 - 6 = 82

 

[breehm]

回複 'So_Be_It' 的評論 :

謝謝你，這個題看上去簡單，其實還挺複雜的。

[So_Be_It]

對不起，沒考慮重複的積，比如2x3x4=6x4，考慮進去就有點複雜了。

[So_Be_It]

1. number of ways to multiply two numbers: 7!/(2!5!)=21
2. number of ways to multiply three numbers: 7!/(3!4!)=35
3. number of ways to multiply 4 numbers: 7!/(4!3!)=35
4. number of ways to multiply 5 numbers: 7!/(5!2!)=21
5. number of ways to multiply 6 numbers: 7!/6!1!)=7
6. number of ways to multiply all 7 numbers: 7!/(7!0!)=1
Add up: 21+35+35+21+7+1=120
There are a total of 120 ways to multiply some or all the 7 numbers.