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
3 months, 2 weeks ago

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

2(abcdefghi)=92(abcdefghi) = 9 digit number (2)(2)

a4a \leq 4

12=2,22=4,32=6,42=8,52=0,62=2,72=4,82=6,92=81 * 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 n2=2nn * 2 = 2n

Using these facts and the restriction on aa:

1234567892=246913578123456789 * 2 = 246913578 - has 1,2,3,4,5,6,7,8,91, 2, 3, 4, 5, 6, 7, 8, 9 - answer1_1

1324567892=264913578132456789 * 2 = 264913578 - has 1,2,3,4,5,6,7,8,91, 2, 3, 4, 5, 6, 7, 8, 9 - answer2_2

1235467892=247093578123546789 * 2 = 247093578 - has 2,3,4,5,7,8,92, 3, 4, 5, 7, 8, 9 - not an answer

1234576892=246915378123457689 * 2 = 246915378 - has 1,2,3,4,5,6,7,8,91, 2, 3, 4, 5, 6, 7 ,8, 9 - answer3_3

1234567982=246913596123456798 * 2 = 246913596 - has 1,2,3,4,5,6,7,8,91, 2, 3, 4, 5, 6, 7, 8, 9 - answer4_4

2134567892=426913578213456789 * 2 = 426913578 - has 1,2,3,4,5,6,7,8,91, 2, 3, 4, 5, 6, 7, 8, 9 - answer5_5

2143567892=428713578214356789 * 2 = 428713578 - has 1,2,3,4,5,6,7,81, 2, 3, 4, 5, 6, 7, 8 - not an answer

2134657892=426931578213465789 * 2 = 426931578 - has 1,2,3,4,5,6,7,8,91, 2, 3, 4, 5, 6, 7, 8, 9 - answer6_6

2134568792=426913758213456879 * 2 = 426913758 - has 1,2,3,4,5,6,7,8,91, 2, 3, 4, 5, 6, 7, 8, 9 - answer7_7

2134567982=426913596213456798 * 2 = 426913596 - has 1,2,3,4,5,6,91, 2, 3, 4, 5, 6, 9 - not an answer

3124567892=624913578312456789 * 2 = 624913578 - has 1,2,3,4,5,6,7,8,91, 2, 3, 4, 5, 6, 7, 8, 9 - answer8_8

3142567892=628513578314256789 * 2 = 628513578 - has 1,2,3,4,5,6,7,81, 2, 3, 4, 5, 6, 7, 8 - not an answer

3125467892=625093578312546789 * 2 = 625093578 - has 2,3,5,6,7,8,92, 3, 5, 6, 7, 8, 9 - not an answer

3124657892=624931578312465789 * 2 = 624931578 - has 1,2,3,4,5,6,7,8,91, 2, 3, 4, 5, 6, 7, 8, 9 - answer9_9

3124576892=624915378312457689 * 2 = 624915378 - has 1,2,3,4,5,6,7,8,91, 2, 3, 4, 5, 6, 7, 8, 9 - answer10_{10}

3124567982=624913596312456798 * 2 = 624913596 - has 1,2,3,4,5,6,91, 2, 3, 4, 5, 6, 9 - not an answer

4123567892=824713578412356789 * 2 = 824713578 - has 1,2,3,4,5,7,81, 2, 3, 4, 5, 7, 8 - not an answer

4132567892=826513578413256789 * 2 = 826513578 - has 1,2,3,5,6,7,81,2, 3, 5, 6, 7, 8 - not an answer

4125367892=825073578412536789 * 2 = 825073578 - has 2,3,5,7,82, 3, 5, 7, 8 - not an answer

4123657892=824731578412365789 * 2 = 824731578 - has 1,2,3,4,5,7,81, 2, 3, 4, 5, 7, 8 - not an answer

4123576892=824715378412357689 * 2 = 824715378 - has 1,2,3,4,5,7,81, 2, 3, 4, 5, 7, 8 - not an answer

4123568792=824713758412356879 * 2 = 824713758 - has 1,2,3,4,5,7,81, 2, 3, 4, 5, 7, 8 - not an answer

4123567982=824713596412356798 * 2 = 824713596 - has 1,2,3,4,5,6,7,8,91, 2, 3, 4, 5, 6, 7, 8, 9 - answer11_{11}

Therefore there are 11\fbox{11} 99 - digit numbers abcdefghiabcdefghi that when doubled to make 2(abcdefghi)2(abcdefghi), both numbers have all the 99 non-zero digits once each.

@Sachetan Debray

A Former Brilliant Member - 3 months, 2 weeks ago

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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.

Sachetan Debray - 3 months, 2 weeks ago

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

Mahdi Raza - 3 months, 2 weeks ago

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@Pi Han Goh ?

Mahdi Raza - 3 months, 2 weeks ago

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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

Pi Han Goh - 3 months, 2 weeks ago

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@Pi Han Goh Clap Clap!! 👏👏

Mahdi Raza - 3 months, 2 weeks ago

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@Mahdi Raza @Mahdi Raza, you mean 👏👏, right?

A Former Brilliant Member - 3 months, 2 weeks ago

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