Largest 7-Digit Number

Date: 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/