# Repeatedly dividing $360$ by $2$ preserves that the sum of the digits (including decimals) is $9$

Can someone give me a clue on how am I going to prove that this pattern is true or not as I deal with repeated division by 2? I tried dividing 360 by 2 repeatedly, eventually the digits of the result, when added repeatedly, always result to 9 for example: \begin{align*} 360 ÷ 2 &= 180 \text{, and } 1 + 8 + 0 = 9\\ 180 ÷ 2 &= 90 \text{, and } 9 + 0 = 9\\ 90 ÷ 2 &= 45 \text{, and } 4 + 5 = 9\\ 45 ÷ 2 &= 22.5 \text{, and } 2 + 2 + 5 = 9\\ 22.5 ÷ 2 &= 11.25 \text{, and } 1 + 1 + 2 + 5 = 9\\ 11.25 ÷ 2 &= 5.625 \text{, and } 5 + 6 + 2 + 5 = 18 \text{, and } 1 + 8 = 9\\ 5.625 ÷ 2 &= 2.8125 \text{, and } 2 + 8 + 1 + 2 + 5 = 18 \text{, and } 1 + 8 = 9 \end{align*}

As I continue this pattern I found no error (but not so sure)... please any idea would be of great help!

• One thing to realise is that the sum of the digits of a multiple of 9 is also a multiple of 9: math.stackexchange.com/questions/1221698/… – Matti P. Jan 7 at 11:46
• Rather than dividing by 2, a better way to do it might be to multiply by 5, because both will result in the same decimal digits. Now all you’re doing is (repeatedly) computing that 0*5=0 mod 9. – Shalop Jan 7 at 15:45
• It would work too with division by $5$ (or multiplication by $2$) – Henry Jan 7 at 22:27
• Try dividing by 11 and see what you get :-) – ChatterOne Jan 8 at 7:54

Define the digit sum of the positive integer $$n$$ as the sum $$d(n)$$ of its digits. The reduced digit sum $$d^*(n)$$ is obtained by iterating the computation of the digit sum. For instance $$n=17254,\quad d(n)=1+7+2+5+4=19, \quad d(d(n))=1+9=10, \quad d(d(d(n)))=1+0=1=d^*(n)$$ Note that $$d(n)\le n$$, equality holding if and only if $$1\le n\le 9$$. Also, $$1\le d^*(n)\le 9$$, because the process stops only when the digit sum obtained is a one-digit number.

The main point is that $$n-d(n)$$ is divisible by $$9$$: indeed, if $$n=a_0+a_1\cdot10+a_2\cdot10^2+\dots+a_n\cdot10^n,$$ then $$n-d(n)=a_0(1-1)+a_1(10-1)+a_2(10^2-1)+\dots+a_n(10^n-1)$$ and each factor $$10^k-1$$ is divisible by $$9$$. This extends to the reduced digit sum, because we can write (in the example above) $$n-d^*(n)=\bigl(n-d(n)\bigr)+\bigl(d(n)-d(d(n))\bigr)+\bigl(d(d(n))-d(d(d(n)))\bigr)$$ and each parenthesized term is divisible by $$9$$. This works the same when a different number of steps is necessary.

Since the only one-digit number divisible by $$9$$ is $$9$$ itself, we can conclude that

$$d^*(n)=9$$ if and only if $$n$$ is divisible by $$9$$.

Since $$360$$ is divisible by $$9$$, its reduced digit sum is $$9$$; the same happenso for $$180$$ and so on.

When you divide an even integer $$n$$ by $$2$$, the quotient is divisible by $$9$$ if and only if $$n$$ is divisible by $$9$$.

What if the integer $$n$$ is odd? Well, the digits sum of $$10n$$ is the same as the digit sum of $$n$$. So what you are actually doing when arriving at $$45$$ is actually \begin{align} &45 \xrightarrow{\cdot10} 450 \xrightarrow{/2} 225 && d^*(225)=9 \\ &225 \xrightarrow{\cdot10} 2250 \xrightarrow{/2} 1225 && d^*(1225)=9 \end{align} and so on. Note that the first step can also be stated as $$45 \xrightarrow{\cdot5} 225$$

Under these operations divisibility by $$9$$ is preserved, because you divide by $$2$$ or multiply by $$5$$.

In general if an integer is divisible by 9 then its digital root is $$9$$. Any multiple of an integer divisible by $$9$$ will also be divisible by $$9$$ and have digital root $$9$$.

What's slightly surprising here is that this property is also preserved by division by $$2$$. Note that it doesn't work for, say, division by $$3$$, since $$360/3 = 120$$ has digital root $$3$$.

A relevant property that $$2$$ has, but $$3$$ doesn't, is that it's a factor of $$10$$. Dividing by $$2$$ is equivalent to multiplying by $$5$$, which preserves the property, and then dividing by $$10$$, which also preserves the property because it just removes a final zero, or adds or shifts the decimal point.

More generally, division by an integer $$n$$ will preserve the property of having digital root $$9$$ if all the prime factors of $$n$$ are factors of $$10$$, that is, it equals $$2^a5^b$$ for some $$a,b \geq1$$. This works because division by $$2^a5^b$$ is equivalent to multiplying by $$2^b5^a$$ and then dividing by $$10^{a+b}$$.

• I would say the relevant property is more that 3 is a divisor of 9, and 2 isn't. Dividing/multiplying with 7 has the same effect, though the divisors of $10^k$ will always result in a finite number of non-zero digits. – Paŭlo Ebermann Jan 7 at 18:54
• @PaŭloEbermann Could you please clarify your claim about dividing by 7? Division by 7 of most integers, including 360, results in a number containing an infinite recurring decimal. Surely in such cases the digital root isn't even well-defined? – Adam Bailey Jan 7 at 20:46
• That's what I meant with the second part of my sentence – for divisors of $10^k$ (or equivalently, all prime factors are 2 or 5), you'll always get a finite result (with a defined digital root). For all others, it only holds if the result happens to have a finite number of digits. For example, $945$ has the same digital root $9$ as $945/7 = 135$. – Paŭlo Ebermann Jan 11 at 10:35

A number is divisible by 9 iff its digits sum to a number divisible by 9. That's the underlying phenomenon here.

I think the other answers are unnecessarily complicated.

It's very simple: a number is a multiple of 9 if and only if the sum of its digits equals 9. You started with 360 which is a multiple by 9. When such a number can be divided by 2, the result will still be a multiple of 9 because 2 and 9 are coprime.

And when you finally get to the point where the number is not divisible by 2, as you did with 45, we can simply imagine that you multiplied by 10 before dividing by 2 - so that from 45 you go to 450 and then to 225, which is essentially the same as going to 22.5. And since multiplying by 10 also does not change the property of a number being a multiple of 9 (because 10 and 9 are coprime), your result is still a multiple of 9.

• But 7 is also coprime to 9, and starting from 360 you are immediately at the point where the number is not divisible by 7. And if you multiply by 10, or indeed any power of 10, it's still not divisible by 7. You get a number containing an infinite recurring decimal. – Adam Bailey Jan 8 at 9:31
• @AdamBailey Well, I agree. But what is your point? Why are you considering 7? – Pedro A Jan 8 at 12:31
• @PedroA I'm considering 7 because, like 2, it's coprime to 9, but unlike division by 2, division by 7 does not preserve the property of having a digital root of 9. The conclusion to be drawn is that the reason this property holds for division by 2 is not, or at least not solely, that 2 is coprime to 9. – Adam Bailey Jan 8 at 13:37
• @AdamBailey You are correct, but I think you misread my answer. I think you've read my answer as if I was generalizing for any number that is coprime to 9, but that is not the case. The property of being coprime is necessary for the first part, but the second part of the answer is equally important for my proof to be complete. In other words, division by 7 does preserve the property of being a multiple of 9. But, with 7 the second part of my answer won't work, because after multiplying by 10 you still can't divide by 7. So I never claimed that it should work for 7. – Pedro A Jan 8 at 13:45
• @PedroA Ah, I understand now. – Adam Bailey Jan 8 at 15:24

1) The sum of the digits of a multiple of $$9$$ will be a multiple of $$9$$.

Why? If $$N$$ is a multiple of $$9$$ then $$N - 9k$$ will also be a multiple of $$9$$ for all integers $$k$$.

If $$N = \sum_{k=0}^n a_k 10^k$$ is a multiple of $$9$$ then $$(\sum_{k=0}^n a_k 10^k) - 9(a_1 + 11a_2 + 111a_3 + ..... + 1111.....1111a_n)$$ is a multiple of $$9$$.

And $$(\sum_{k=0}^n a_k 10^k) - 9(a_1 + 11a_2 + 111a_3 + ..... + 1111.....1111a_n)=$$

$$(\sum_{k=0}^n a_k 10^k) - (9a_1 + 99a_2 + 999a_3 + ..... + 9999.....9999a_n)=$$

$$\sum_{k=0}^n [a_k*10^k - \underbrace{999...9}k*a_k]=$$

$$\sum_{k=0}^n a_k(10^k - \underbrace{999...9}k)=\sum_{k=0}^n a_k*1 =$$ the sum of the digits of $$N$$.

2) we can extend that decimals:

If $$N$$ is $$9$$ times some terminating decimal than the sum of the digits are a multiple of $$9$$.

Why? Well, its the same as above except $$N$$ has be divided by a power of $$10$$. But that only affects the position of the decimal point. It doesn't affect the digits.

3) If $$N$$ is a $$9$$ times a terminating decimal than $$\frac N2$$ is $$9$$ times a terminating decimal.

If $$N = 9\times M$$ and $$M$$ is a terminating decimal then $$\frac M2$$ is a terminating decimal. Then $$\frac N2 = 9\times \frac M2$$ is also $$9$$ times a terminating decimal and the digits add up to a multiple of $$9$$.

4) Since $$360 = 9*40$$ we start with a number whose digits add to $$9$$. Successively dividing by $$2$$ will result in $$9*20$$ and $$9*10$$ and so on, always $$9$$ times a terminating decimal with the sum of the digits being a multiple of $$9$$.

Just a quick addendum and question about the division by 7. One can think of that as involving looking not at the decimal equivalent, but rather looking at the decimal digits in one repeating block.

For example, 1/7 = .142857 142857 .... and each individual block has digit sum 27, whose digits sum is then 9. Because this number is a full-period prime, the same pattern exists in the sense that for any fraction k/7 (k=1 to 6) there are blocks of six repeating digits, and in each case, the block is a cyclic permutation of that for 1/7. In each such case, the digit sum is 27 again.

This suggests an extension along this line.

• This serves as a great idea for another pattern to ponder on..hope i can learn more from this... – rosa Jan 9 at 0:58