Largest 7-Digit NumberDate: 07/27/2001 at 01:36:24 From: Patrick Rydon Subject: Maths problem involving word probelem and number problem Work out the largest 7-digit number you can applying 2 rules only: 1) every digit in the number must be able to be divided into the number; 2) no digit can be repeated. I think that 0 cannot be one of the numbers because of the infinity result in the division. I think 7 is unlikley to be a digit because 7 seems to divide into so few other numbers. The question was set for us by our advanced maths teacher. Please help. Date: 07/27/2001 at 12:54:24 From: Doctor Peterson Subject: Re: Maths problem involving word probelem and number problem Hi, Patrick. I disagree about 7; in fact, more 7-digit numbers are divisible by 7 than by 9. The difference is merely that it's harder to tell whether a number is divisible by 7. See our FAQ on Divisibility Rules if you don't know how to do this: http://mathforum.org/dr.math/faq/faq.divisibility.html But you're right about 0. We'd like to choose the largest possible set of digits, and then arrange them to form the largest possible number. (That strategy might not work, because the digits we can choose are determined partly by the order in which the digits will end up, since the last digits tell whether it will be divisible by 8, 4, and 2, and 5 also. But this is a good way to start, at least.) So what will it take to get 9 in the number? The sum of the digits has to be a multiple of 9. Since the sum of 1 through 9 is 45, which is divisible by 9, any 9-digit number with all digits different would be a multiple of 9. But we have to drop two digits, so they must add up to 9. To make them as small as possible, I'd like to drop 4 and 5; but it could also be 3 and 6, 2 and 7, or 1 and 8. Let's assume our digits will be 9, 8, 7, 6, 3, 2, 1. Note that 3 and 6 will work automatically, since 3 divides 9 and we have to be even anyway. To be divisible by 8 (and by 4 and 2 as well), the last three digits must be a multiple of 8. We want to use the smallest numbers there; 312 will work. Now we have 9 _ _ _ 3 1 2, and we know that 9, 8, 6, 3, 2, and 1 are all divisors. We just have to decide where to place 8, 7, and 6 to make it a multiple of 7. There are 6 choices, so just try them all. I found two that work, one much larger than the other. I'm not sure yet that this is the very largest answer, but it seems likely. See what you can do. Having this solution, you may want to focus on larger numbers, and just think about how you could choose the last four digits to fit the rules. Let me know if you find a good answer. I'll keep thinking about it. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/ |
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