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March 31st, 2019, 05:31 AM   #1
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General Formula Assistance

Hi Guys
Just signed up searching the forums to see if anybody else has come across this but haven't had any luck so was wondering if anybody could assist.

I've checked google searches, facebook, etc., but no luck.

I have been working on a few different assignments / queries that have been posed, but one has really stumped me.


So this is the query:


Take any two digit number (the two digits should be different).
Write down all the different arrangements of the digits. Add them up. Divide this total by the sum of the two digits.
What do you notice? Try again with a different two digit number. Why does this happen?

Extend this by looking at three / four digits/ Is there a way to predict the results?
Can you find a general formula? (What happens if you have repeated digits)?



So the bolded part is what is stumping me. From my working out (please do correct me if I'm wrong), I seem to find that for 2 digits, the total is always 11.

(eg a = 7
b = 5
(5+7 = 12)
75 + 57 = 132 / 12 = 11)

For 3 digits, it seems that the total always equals 222.

From using a written down example, (eg 234), there are 6 different arrangements:
(234, 243, 342, 324, 432, 423)
4 + 2 + 3 = 9
Adding them all makes 1998 / 9 = 222


For 4 digits using same method, it shows 6666 as the total.

However the one part that is causing me an issue is trying to figure out a formula to use to figure out the answer when you choose the number of digits. I'm unable to determine a pattern that will lead to finding this.

I tried for the simplest version (2 digits) but the best I could come up with is:


11(a+b) / (b + a) = 11
However that wouldn't really work in terms of an actual formula as it requires knowing the answer.
Is anybody able to assist please?

Last edited by skipjack; March 31st, 2019 at 10:57 AM.
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March 31st, 2019, 11:12 AM   #2
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11 = 11 × 1!, 222 = 111 × 2!, 6666 = 1111 × 3!, etc.

The formula for the $n$-digit integer in which each digit is 1 is (10$^n$ - 1)/9.
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March 31st, 2019, 03:18 PM   #3
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The general formula was correctly pointed out by skipjack already. For $n$ digit numbers, your procedure will produce: $\frac{(10^n-1)(n-1)!}{9}$. Here is a rough sketch of the proof.

1. Suppose you have a number with the digits $d_{n-1}d_{n-2}\dots d_1 d_0$ which is the base 10 representation of the integer $a = d_0 + d_1\cdot 10 + d_2 \cdot 10^2 + \dots + d_{n-1} \cdot 10^{n-1}$.

2. Evidently, each digit will appear as a coefficient of each power of 10 an equal number of times when you consider every permutation of the digits. For example, if $j$ is fixed, then $d_j$ will be multiplied with $10^k$ the same number of times for each $k$.

3. If we sum subsets of permutations with each $d_j$ appearing exactly once in say the $k^{\rm th}$ position, then the coefficient of $10^k$ in the sum (before carries) is exactly the sum of the digits, $\displaystyle D = \sum_{j = 0}^{n-1} d_j$. This explains why it must divide the sum over all permutations.

4. From (2), we simply need to count how many times these values sum to $D$. This is equivalent to determining how many permutations fix the $j^{\rm th}$ digit and act transitively on the rest. It is not hard to show that the number of such permutations is $(n-1)!$.

5. Now, show that one of our subsets from (3) sums to exactly $\frac{(10^n-1)}{9}$. Combined with (4), we get the formula.

Last edited by skipjack; March 31st, 2019 at 11:22 PM.
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