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