# finding numbers - problem solving !

• Apr 21st 2012, 04:54 PM
ilikepizza
finding numbers - problem solving !
find the largest 7-digit number which satisfy these 2 properties 0-9
- no two digits in the number are the same
- each of the number’s digits divides into the number exactly

a. FIND 3 digits which can't be in this number and EXPLAIN why we cannot use them
b. what is the highest 7 digit number you can make? EXPLAIN why it is the largest 7 digit you can make
c. what is the SMALLEST number you can make which ALSO satisfy the properties

any answers, hints, help discussion etc greatly appreciated!! i'm very stuck on this and don't know how to work it out.
please also try to keep it simple so i can understand thank you.
• Apr 24th 2012, 07:04 AM
Sylvia104
Re: Finding numbers – problem solving!
Let us try this line of attack. In order to make the number as large as possible, we'll use the digits $9,8,7,$ which we will put at the beginning of the number. Then the number must be divisible by $\mathrm{lcm}(9,8,7)=504.$ The largest multiple of $504$ not greater than $9876543$ is $9876384.$ That has a repeated digit and so we reject that. We then keep subtracting $504$ from our number until we get one that fits.

Spoiler:
$9867312$
• Apr 24th 2012, 07:31 AM
Soroban
Re: finding numbers - problem solving !
Hello, ilikepizza!

I don't have this solved, but I have some ideas.

Quote:

Find the largest 7-digit number (0-9) which satisfy these 2 properties:
. . no two digits in the number are the same
. . each of the number’s digits divides into the number exactly

a. Find 3 digits which can't be in this number and explain why we cannot use them.

b. What is the highest 7 digit number you can make?

c. What is the smallest number you can make?

Let $N \;=\;\_\,\_\,\_\,\_\,\_\,\_\,\_$

The zero (0) cannot be in $N$; division by zero is not allowed.

The five (5) cannot be in $N.$
If $N$ is divisible by 5, it must end in 5. .(It can't end in zero.)
. . We have an odd number: . $\_\,\_\,\_\,\_\,\_\,\_\,5$
We select 6 more digits from {1,2,3,4,6,7,8,9} . . . and will include at least one even digit.
And no even digit divides into an odd number.

I'm not sure what the third digit is.

I do know this:
. . If $N$ contains a 9, the sum of the digits must be a multiple of 9.
. . Then we have: {1,2,3,4,6,7,8,9} which has a sum of 40.
. . For the sum to be a multiple of 9, we must discard the 4.
So we have: {1,2,3,6,7,8,9}

To be divisible by 8, the rightmost 3-digit number must be divisible by 8.
. . For example: . $N \;=\; \_\,\_\,\_\,\_\, 2\,1\,6$
Then we must arrange {3,7,8,9} so that $N$ is divisible by 7.

I found one solution: . $N\;=\;8\,9\,7\,3\,2\,1\,6$
. . It could be the largest, but I have no proof.

Edit: I like your solution, Sylvia!
• Apr 24th 2012, 09:30 PM
ilikepizza
Re: finding numbers - problem solving !
thank you both.
I understand that 0 cannot be used as it is not possible to divide by 0, but i'm having trouble understanding the reasoning for 5 (I don't really get it.)
Could you possibly explain it in really simple terms? Also I think the 3rd number is 4 just not sure how to explain it.
any ideas for what the smallest number could be?

Edit: i understand now after reading over thanks!

but still not sure about the smallest number
• Apr 25th 2012, 02:41 AM
Sylvia104
Re: finding numbers - problem solving !
Quote:

Originally Posted by ilikepizza
Also I think the 3rd number is 4, just not sure how to explain it.
Any ideas for what the smallest number could be?

The third number is $4$ but only if you want to use the digit $9.$ We already know that $0$ and $5$ cannot occur in our number; let's show that $4$ and $9$ cannot both occur in our number. Suppose they do. A number is divisible by $9$ if and only if the sum of its digits is divisible by $9$ so let's check for divisibility by $9$ for the candidates for our number:

$9 + 8 + 7 + 6 + 4 + 3 + 2 = 39$

$9 + 8 + 7 + 6 + 4 + 3 + 1 = 38$

$9 + 8 + 7 + 6 + 4 + 2 + 1 = 37$

$9 + 8 + 7 + 4 + 3 + 2 + 1 = 34$

$9 + 8 + 6 + 4 + 3 + 2 + 1 = 33$

$9 + 7 + 6 + 4 + 3 + 2 + 1 = 32$

None of these is divisible by $9.$ Hence if we use the $4$ we can't use the $9$ – but we do want to use the $9$ so as to make our number as large as possible.
• Apr 25th 2012, 02:46 AM
ilikepizza
Re: finding numbers - problem solving !
yes, I get it now, thanks! :)
is the smallest number 1289736? that was what I got
• Apr 25th 2012, 05:24 AM
Wilmer
Re: finding numbers - problem solving !
Quote:

Originally Posted by ilikepizza
is the smallest number 1289736? that was what I got

Yes.
There are 105 total solutions:
1: 1289736
2: 1293768
...
104: 9812376
105: 9867312

Little "looper" that'll get 'em all:
LOOP a,b,c,d,e,f,g from 1 to 9
DOWHILE variables all different and <> 4 or 5
n = a*10^6 + b*10^5 + c*10^4 + d*10^3 + e*100 + f*10 + g
PRINT n if n divisible bt all variables

Have you "seen the light" on the "5" yet?
To be divisible by 5, n must end with 5.
If n ends with 5, then not divisible by even digits.
Capish?