Math Puzzle No. 3: The Ten Divisibilities

This problem, along with a couple of others, was inspired by the work of John H. Conway* and appeared in the October 15, 2020 edition of Quanta Magazine.

Let a, b, c, d, e, f, g, h, i, j be the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 in some order. Each digit appears exactly once. (For those that want a more precise mathematical phrasing, there is a one-to-one correspondence between the set {a, b, c, d, e, f, g, h, i, j} and the set {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}, or better yet, there is a bijection between the two sets.)  We form other integers by concatenating digits, so for example, 'ab' is a two-digit number where 'a' is the tens digit and 'b' is the ones digit.

If abcdefghij is a 10-digit number with the following properties:

a is divisible by 1 (OK, that much is obvious)

ab is divisible by 2 (So, b must be even)

abc is divisible by 3 (Do you know the test for divisibility by 3?)

abcd is divisible by 4

abcde is divisible by 5

abcdef is divisible by 6

abcdefg is divisible by 7

abcdefgh is divisible by 8

abcdefghi is divisible by 9

abcdefghij is divisible by 10

...What number is abcdefghij?

You will need to know some divisibility tests from number theory, and employ a good bit of logic. Good luck.

Update: 1/12/21: read below for the solution.

abcdefghij is divisible by 10, so j = 0

abcde is divisible by 5, and since we already used 0 = j, then e = 5

b, d, f, h must be even, therefore a, c, g, i must be odd. Since abcd is divisible by 4, cd must be divisible by 4, and since c is odd, d must be either 2 or 6. The same holds for h, therefore {d, h} = {2, 6}, {b, f} = {4, 8} and {a, c, g, i} = {1, 3, 7, 9}

Since abc is divisible by 3, then a+b+c is divisible by 3. Since abcdef is divisible by 6, abcdef is also divisible by 3, so a+b+c+d+e+f is divisible by 3. Therefore d+e+f is divisible by 3, and since e = 5, the only possibilities are d = 2, f = 8, h = 6, b = 4 or d = 6, f = 4, h = 2, b = 8

 

Case 1:
b = 4, d = 2, f = 8, h = 6

Since a+4+c is divisible by 3, then {a,c} = {1,7}.  Since 8g6 must be divisible by 8, g = 1 or g = 9, but g can't be 1 since either a or c is 1. Thus g = 9 and i = 3.

Checking the two possibilities: 

• 1472589 is not divisible by 7
• 7412589 is not divisible by 7

This rules out Case 1.

Case 2: 
b = 8, d = 6, f = 4, h = 2, 

Since = 4g2 is divisible by 8, therefore g = 3 or g = 7. Since a+b+c is divisible by 3, we have the following eight possibilities:

• 1896543 is not divisible by 7
• 9816543 is not divisible by 7
• 7896543 is not divisible by 7
• 9876543 is not divisible by 7
• 1836547 is not divisible by 7
• 3816547 is divisible by 7
• 1896547 is not divisible by 7
• 9816547 is not divisible by 7

Leaving 3816547290 as the lone correct solution.

* Here is a bit about the life of John Conway.

John Horton Conway at Princeton University in 2009.    (Photo Credit: Princeton University, Office of Communications, Denise Applewhite)

The previous Math Puzzle and its Solution are available here: https://mathcep.blogspot.com/2020/11/math-puzzle-no-2-monty-hall-problem.html