Using the properties of the sum of digits function, we have that $$S(2^{{24}^{2017}} k) = S( S(2^{{24}^{2017}}) \cdot S(k)).$$ We have that $S(k) \equiv k \pmod{9}$ for all integers $k$. Thus, $S(2^{{24}^{2017}}) \equiv 2^{{24}^{2017}} \pmod{9}$. Then, using the fact that $a^b \equiv a^{b \pmod{\phi(n)}} \pmod{n},$ we have that $2^{{24}^{2017}} \equiv 2^0 \pmod 9$. This comes from the fact that $\phi (9) = 9 \left( 1 - \frac{1}{3} \right)= 6 $ and that $24^{2017} \equiv 0 \pmod{6}$. Thus it suffices to show that $S (S(k)) = S(k)$ has a solution with the given conditions. The first such solution that comes to mind is $k = 1$, where $S(k) = 1$ and $S(S(k)) = 1$ as well.

OmicronGamma wrote: Using the properties of the sum of digits function, we have that $$S(2^{{24}^{2017}} k) = S( S(2^{{24}^{2017}}) \cdot S(k)).$$ Sorry but would you mind explaining how you got this result? I have not seen it before.

Kaskade wrote: OmicronGamma wrote: Using the properties of the sum of digits function, we have that $$S(2^{{24}^{2017}} k) = S( S(2^{{24}^{2017}}) \cdot S(k)).$$ Me neither.

Let $a = c_n \cdot 10^n + c_{n-1} \cdot 10^{n-1} + ... + c_1 \cdot 10^1 + c_0 = \sum ^{n} _{i=0} c_i \cdot 10^i$ and $b = k_m \cdot 10^m + k_{m-1} \cdot 10^{m-1} + ... + k_1 \cdot 10^1 + k_0 = \sum ^{m} _{j=0} k_j \cdot 10^j$. Therefore, $S(a) = \sum ^n _{i=0} c_i$ and $S(b) = \sum ^m _{j=0} k_j$. Therefore, we have that $S(a) \cdot S(b) = \left( \sum ^n _{i=0} c_i \right) \left( \sum ^m _{j=0} k_j \right) = \sum ^n _{i=0} \sum ^m _{j=0} c_i k_j$. And thus $S(S(a) \cdot S(b)) = S \left( \sum ^n _{i=0} \sum ^m _{j=0} c_i k_j \right).$ Now, $S(a \cdot b) = S \left( \left( \sum ^{n} _{i=0} c_i \cdot 10^i \right) \left( \sum ^{m} _{j=0} k_j \cdot 10^j \right) \right)$. Now because we want the digit sum of this product, we can rewrite $S(a\cdot b) = S \left( \left( \sum ^{n} _{i=0} c_i \right) \left( \sum ^{m} _{j=0} k_j \right) \right) = S \left( \sum ^n _{i=0} \sum ^m _{j=0} c_i k_j \right) = S(S(a) \cdot S(b)) .$ Please let me know if my reasoning is faulty.

OmicronGamma wrote: Let $a = c_n \cdot 10^n + c_{n-1} \cdot 10^{n-1} + ... + c_1 \cdot 10^1 + c_0 = \sum ^{n} _{i=0} c_i \cdot 10^i$ and $b = k_m \cdot 10^m + k_{m-1} \cdot 10^{m-1} + ... + k_1 \cdot 10^1 + k_0 = \sum ^{m} _{j=0} k_j \cdot 10^j$. Therefore, $S(a) = \sum ^n _{i=0} c_i$ and $S(b) = \sum ^m _{j=0} k_j$. Therefore, we have that $S(a) \cdot S(b) = \left( \sum ^n _{i=0} c_i \right) \left( \sum ^m _{j=0} k_j \right) = \sum ^n _{i=0} \sum ^m _{j=0} c_i k_j$. And thus $S(S(a) \cdot S(b)) = S \left( \sum ^n _{i=0} \sum ^m _{j=0} c_i k_j \right).$ Now, $S(a \cdot b) = S \left( \left( \sum ^{n} _{i=0} c_i \cdot 10^i \right) \left( \sum ^{m} _{j=0} k_j \cdot 10^j \right) \right)$. Now because we want the digit sum of this product, we can rewrite $S(a\cdot b) = S \left( \left( \sum ^{n} _{i=0} c_i \right) \left( \sum ^{m} _{j=0} k_j \right) \right) = S \left( \sum ^n _{i=0} \sum ^m _{j=0} c_i k_j \right) = S(S(a) \cdot S(b)) .$ Please let me know if my reasoning is faulty. From what you have written, I think I understand the confusion. On one hand, there is the function $S\left(n\right)$ which adds up the digits and gives you the value. For example $S(197) = 1+9+7 = 17$ On the other hand there is a different function, lets call it $D\left(n\right)=S\left(S\left(...\ S\left(n\right)\right)\right)$ which keeps on adding up the digits, then adding up the digits of that etc, until it is constant. For example $D(197) = D(17) = 8$. This is also the same as taking $S(n)$ modulo $9$. Because $S\left(k\right)=k\left(\operatorname{mod}9\right)$, $a\cdot b=S\left(a\right)\cdot S\left(b\right)\left(\operatorname{mod}9\right)$, so $S\left(a\cdot b\right)=S\left(S\left(a\right)\cdot S\left(b\right)\right)\left(\operatorname{mod}9\right)$. So $D\left(a\cdot b\right)=D\left(D\left(a\right)\cdot D\left(b\right)\right)$. I think perhaps you thought that $S(n)$ was $D(n)$ and with that in mind everything you have said is true. However for simply adding up the digits, without modulo $9$, this property is not true.

Exactly, an idea how to show that $k$ exist? continuity?

Kaskade wrote: OmicronGamma wrote: Let $a = c_n \cdot 10^n + c_{n-1} \cdot 10^{n-1} + ... + c_1 \cdot 10^1 + c_0 = \sum ^{n} _{i=0} c_i \cdot 10^i$ and $b = k_m \cdot 10^m + k_{m-1} \cdot 10^{m-1} + ... + k_1 \cdot 10^1 + k_0 = \sum ^{m} _{j=0} k_j \cdot 10^j$. Therefore, $S(a) = \sum ^n _{i=0} c_i$ and $S(b) = \sum ^m _{j=0} k_j$. Therefore, we have that $S(a) \cdot S(b) = \left( \sum ^n _{i=0} c_i \right) \left( \sum ^m _{j=0} k_j \right) = \sum ^n _{i=0} \sum ^m _{j=0} c_i k_j$. And thus $S(S(a) \cdot S(b)) = S \left( \sum ^n _{i=0} \sum ^m _{j=0} c_i k_j \right).$ Now, $S(a \cdot b) = S \left( \left( \sum ^{n} _{i=0} c_i \cdot 10^i \right) \left( \sum ^{m} _{j=0} k_j \cdot 10^j \right) \right)$. Now because we want the digit sum of this product, we can rewrite $S(a\cdot b) = S \left( \left( \sum ^{n} _{i=0} c_i \right) \left( \sum ^{m} _{j=0} k_j \right) \right) = S \left( \sum ^n _{i=0} \sum ^m _{j=0} c_i k_j \right) = S(S(a) \cdot S(b)) .$ Please let me know if my reasoning is faulty. From what you have written, I think I understand the confusion. On one hand, there is the function $S\left(n\right)$ which adds up the digits and gives you the value. For example $S(197) = 1+9+7 = 17$ On the other hand there is a different function, lets call it $D\left(n\right)=S\left(S\left(...\ S\left(n\right)\right)\right)$ which keeps on adding up the digits, then adding up the digits of that etc, until it is constant. For example $D(197) = D(17) = 8$. This is also the same as taking $S(n)$ modulo $9$. Because $S\left(k\right)=k\left(\operatorname{mod}9\right)$, $a\cdot b=S\left(a\right)\cdot S\left(b\right)\left(\operatorname{mod}9\right)$, so $S\left(a\cdot b\right)=S\left(S\left(a\right)\cdot S\left(b\right)\right)\left(\operatorname{mod}9\right)$. So $D\left(a\cdot b\right)=D\left(D\left(a\right)\cdot D\left(b\right)\right)$. I think perhaps you thought that $S(n)$ was $D(n)$ and with that in mind everything you have said is true. However for simply adding up the digits, without modulo $9$, this property is not true. His theorem is always right,you can let $n<m$ and let $c_i =0$ when $n<i<m$,after that the theorem is clear.

OmicronGamma wrote: Using the properties of the sum of digits function, we have that $$S(2^{{24}^{2017}} k) = S( S(2^{{24}^{2017}}) \cdot S(k)).$$ We have that $S(k) \equiv k \pmod{9}$ for all integers $k$. Thus, $S(2^{{24}^{2017}}) \equiv 2^{{24}^{2017}} \pmod{9}$. Then, using the fact that $a^b \equiv a^{b \pmod{\phi(n)}} \pmod{n},$ we have that $2^{{24}^{2017}} \equiv 2^0 \pmod 9$. This comes from the fact that $\phi (9) = 9 \left( 1 - \frac{1}{3} \right)= 6 $ and that $24^{2017} \equiv 0 \pmod{6}$. Thus it suffices to show that $S (S(k)) = S(k)$ has a solution with the given conditions. The first such solution that comes to mind is $k = 1$, where $S(k) = 1$ and $S(S(k)) = 1$ as well. I want to know whether $S_p(X)$ is true or not?

proposed by Bojan Basic, Serbia it was also RMM Shortlist 2017 N1

May I gently b u m p i t

Let $m=24^{2017}$, and let $d(n)=S(2^mn)-S(n)$. First observe that for any $a,b$ there's a large $u$ such that $S(10^ua+b)=S(a)+S(b)$, so for all $a,b$ there's a large $u$ for which $d(10^ua+b)=d(a)+d(b)$. This means that if we can find $p,q$ with $d(p)<0$ and $d(q)>0$ we can compose $d(q)$ copies of $p$ and $-d(p)$ copies of $q$ to get a k for which $d(k)=0$ (also notice that if $p,q$ have no $9$ digits their composition will have none as well). $q=1$ obviously works, so we just need to find $p$. We can inductively construct $l$ with $m$ digits such that $5^m|l$ and all the digits of $l$ are in $\{4,5,6,7,8\}$, and we claim $p=l$ works. Observe that $S(l)\geq 4m$, and $l<10^m$. If we let $l=5^mj$, we have $S(2^ml)=S(10^mj)=S(j)$. Now, since $j=\frac{l}{5^m}<2^m$, we have $S(j)<9\log_{10}(j)<9\log_{10}(2^m)=9m\log_{10}2<4m$, so we are done.