# Double The Other, Made up the Same

This is inspired by a problem I read today.

Find 2 9-digit numbers (in base-10 notation) such that one is double the other. Both of them are made up of the 9 non-zero digits once each. 123456789 can be one of them, but 122344566 can't be one of them.

I came up with an answer, but I am not sure if it is the only answer. I had a lot of fun finding it.

Tell me in the comments the 2 numbers you came up with, and what process you used. It doesn't have to be a rigorous proof. It could be a simple investigation based on a particular nice insight(that's what I did)

I'll share my answer once one of you comes up with it and says so in the comments.

Have fun!

(P.S.-For anyone who wants to check out the sources of my problems, Adventures In Problem Solving By Shailesh Shirali and Martin Gardner's 'The Colossal Book Of Mathematics' are awesome to start out with) Note by Sachetan Debray
1 year ago

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## Comments

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$abcdefghi = 9$ digit number $(1)$

$2(abcdefghi) = 9$ digit number $(2)$

$a \leq 4$

$1 * 2 = 2, 2 * 2 = 4, 3 * 2 = 6, 4 * 2 = 8, 5 * 2 = 0, 6 * 2 = 2, 7 * 2 = 4, 8 * 2 = 6, 9 * 2 = 8$ - last digits of $n * 2 = 2n$

Using these facts and the restriction on $a$:

$123456789 * 2 = 246913578$ - has $1, 2, 3, 4, 5, 6, 7, 8, 9$ - answer$_1$

$132456789 * 2 = 264913578$ - has $1, 2, 3, 4, 5, 6, 7, 8, 9$ - answer$_2$

$123546789 * 2 = 247093578$ - has $2, 3, 4, 5, 7, 8, 9$ - not an answer

$123457689 * 2 = 246915378$ - has $1, 2, 3, 4, 5, 6, 7 ,8, 9$ - answer$_3$

$123456798 * 2 = 246913596$ - has $1, 2, 3, 4, 5, 6, 7, 8, 9$ - answer$_4$

$213456789 * 2 = 426913578$ - has $1, 2, 3, 4, 5, 6, 7, 8, 9$ - answer$_5$

$214356789 * 2 = 428713578$ - has $1, 2, 3, 4, 5, 6, 7, 8$ - not an answer

$213465789 * 2 = 426931578$ - has $1, 2, 3, 4, 5, 6, 7, 8, 9$ - answer$_6$

$213456879 * 2 = 426913758$ - has $1, 2, 3, 4, 5, 6, 7, 8, 9$ - answer$_7$

$213456798 * 2 = 426913596$ - has $1, 2, 3, 4, 5, 6, 9$ - not an answer

$312456789 * 2 = 624913578$ - has $1, 2, 3, 4, 5, 6, 7, 8, 9$ - answer$_8$

$314256789 * 2 = 628513578$ - has $1, 2, 3, 4, 5, 6, 7, 8$ - not an answer

$312546789 * 2 = 625093578$ - has $2, 3, 5, 6, 7, 8, 9$ - not an answer

$312465789 * 2 = 624931578$ - has $1, 2, 3, 4, 5, 6, 7, 8, 9$ - answer$_9$

$312457689 * 2 = 624915378$ - has $1, 2, 3, 4, 5, 6, 7, 8, 9$ - answer$_{10}$

$312456798 * 2 = 624913596$ - has $1, 2, 3, 4, 5, 6, 9$ - not an answer

$412356789 * 2 = 824713578$ - has $1, 2, 3, 4, 5, 7, 8$ - not an answer

$413256789 * 2 = 826513578$ - has $1,2, 3, 5, 6, 7, 8$ - not an answer

$412536789 * 2 = 825073578$ - has $2, 3, 5, 7, 8$ - not an answer

$412365789 * 2 = 824731578$ - has $1, 2, 3, 4, 5, 7, 8$ - not an answer

$412357689 * 2 = 824715378$ - has $1, 2, 3, 4, 5, 7, 8$ - not an answer

$412356879 * 2 = 824713758$ - has $1, 2, 3, 4, 5, 7, 8$ - not an answer

$412356798 * 2 = 824713596$ - has $1, 2, 3, 4, 5, 6, 7, 8, 9$ - answer$_{11}$

Therefore there are $\fbox{11}$ $9$ - digit numbers $abcdefghi$ that when doubled to make $2(abcdefghi)$, both numbers have all the $9$ non-zero digits once each.

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That's right. I didn't get that, though. This is what I got

123567849*2=247135698

The 'click' I used was that in the string of 9 non-zero numbers, you have exactly 5 odd numbers, and these must be present in the larger number too, so there must be at least 5 carryover 1s. i think you can generate many more numbers using that.

- 1 year ago

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Can anyone write a code for this? It'll be fun to see.

- 1 year ago

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- 1 year ago

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  1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 from itertools import permutations from math import factorial # Record the number of solutions found total_number_of_solutions_found = 0 # The following line reads: Find all permutations of the digits 1 to 9, but the leading digit is at most 4 (otherwise, doubling it will be bigger than a 5-digit number) for this_permutation in ( list(permutations([*range(1,10)], 9))[0: int(4/9 * factorial(9) )] ): # for this specific permutation of digits, we combine these digits into a 9-digit integer s = '' for item in this_permutation: s += str(item) # Don't forget to convert this string into an integer! concatenate_these_digits = int(s) # Now we double it! doubling_this_number = 2 * concatenate_these_digits # Once we double this number, we determine if this number has 9 distinct digits, and all its digits are non-zero if len(set(list(str(doubling_this_number)))) == 9 and (0 not in set(list(str(doubling_this_number)))): # If so, we have determined another solution total_number_of_solutions_found += 1 # Display the output print(total_number_of_solutions_found) # Output: 18432 

- 1 year ago

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Clap Clap!! 👏👏

- 1 year ago

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@Mahdi Raza, you mean 👏👏, right?

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