The Position Matters!

Number Theory Level 2

Is it possible to rearrange the digits of the number 987654321 such that it is divisible by 11?

Yes, it is possible No, it is not possible

1 solution

Zee Ell
Oct 14, 2016

927364581 = 11 × 84305871

Hence, our answer should be:

$\boxed { \text {Yes, it is possible.} }$

We can get many other suitable numbers in the following way:

The divisibility rule for 11:

Add the digits in the odd positions (1st, 3rd, 5th etc. digits) of the integer (let's call this sum O), then add the digits in the even positions of the integer (let's call this sum E). Iff their difference (O - E) is divisible by 11, then our number is also divisible by 11.

Here, the sum of the digits (9 + 8 + ... + 1) is 45.

O = 28 and E = 17 gives us O - E = 11 (it is easy to see that this is the only solution for O and E in this case).

We can get E = 17 by choosing the following digits for the even positions (examples, there are other solutions):

• 2 + 3 + 4 + 8 = 17 (I used this for my solution)

• 1 + 3 + 4 + 9 = 17

• 2 + 3 + 5 + 7 = 17

The rest of the digits of the original number will take the odd positions. We can permutate the odd digits and the even digits, so we can get

5! × 4! = 120 × 24 = 2880 different numbers for each suitable combination of 4 digits (giving us a sum (E) of 17).

Nice solution (+1). I get a total of $9$ ways to obtain $E = 17$ , (i.e., the number of positive integer solutions to $a + b + c + d = 17$ such that $a \lt b \lt c\lt d \le 9$ ).

We can't have $O - E = 44 \Longrightarrow O = 44, E = 1$ as $E \ge 1 + 2 + 3 + 4 = 10$ . The same goes for $O - E = 33$ , as this would require that $O = 39, E = 6$ . Nor can we have that $O - E = 22$ , since along with the fact that $O + E = 45$ this would require that $2*O = 67$ , i.e., that $O$ be non-integral.

On the flip side, $O$ can have a minimum value of $1 + 2 + 3 + 4 + 5 = 15$ and $E$ a maximum value of $6 + 7 + 8 + 9 = 30$ , so we should also look at the case where $O - E = -11$ , i.e., where $O = 17$ and $E = 28$ . There are $2$ scenarios here, namely $1 + 2 + 3 + 4 + 7 = 17$ and $1 + 2 + 3 + 5 + 6 = 17$ .

So in total there are a total of $(9 + 2)*2880 = 31680$ suitable rearrangements.

Brian Charlesworth - 4 years, 7 months ago

The Position Matters!

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