Infinite Sets of Solutions and Almost Solutions of the Equation N⋅M = reversal(N⋅M) II

Show more

1. Introduction

In this paper, motivated by their intrinsic interest and by applications to the study of numeric palindromes and other sequences of integers, we discover a method for producing infinite sets of solutions and almost solutions of the equation:

$N\cdot M=reversal\left(N\cdot M\right).$ (1)

where if N is an integer written in base b, which is understood from the context then reversal(N) is the base b integer obtained from N writing its digits in reverse order.

An almost solution of (1) is a pair of integers $\left(M,N\right)$ for which the equality (1) holds up to a few of digits for which we understand their position. Our results are valid in a general numeration base $b>2$ and complement the results in [1]. Recently one of us showed in Nitica [2] that, in any numeration base b, for any integer N not divisible by b, the Equation (1) has an infinite set of solutions $\left(N,M\right)$. Nevertheless, as one can see from [3], finding explicit values for M can be difficult from a computational point of view, even for small values of N, e.g. $N=81$. We show in [1] for many numeration bases explicit infinite families of solutions of (1). This families of solutions here complement and are independent of those shown in [1].

Another application of our results may appear in the study of the classes of b-multiplicative and b-additive Ramanujan-Hardy numbers, recently introduced in Nitica [4]. The first class consists of all integers N for which there exists an integer M such that ${S}_{b}\left(N\right)$, the sum of base b-digits of N, times M, multiplied by the reversal of the product, is equal to N. The second class consists of all integers N for which there exists an integer M such that ${S}_{b}\left(N\right)$, times M, added to the reversal of the product, is equal to N. As showed in Nitica [2] [4], the solutions of Equation (1) for which we can compute the sum of digits of ${S}_{b}\left(N\right)\cdot M+reversal\left({S}_{b}\left(N\right)\cdot M\right)$ or of ${S}_{b}\left(N\right)\cdot M\cdot reversal\left({S}_{b}\left(N\right)\cdot M\right)$, can be used to find infinite sets of above numbers.

2. Statements of the Main Results

The heuristics behind our results is that the product of a palindrome by a small integer still preserves some of the symmetric structure of the palindrome; if, in addition, the palindrome has many digits of 9, many times the results observed in base 10 can be carried over to an arbitrary numeration base b replacing 9 by b − 1.

Let $b\ge 2$ be a numeration base. If x is a string of digits, let ${\left(x\right)}^{^k}$ denote the base b integer obtained by repeating x k-times. Let ${\left[x\right]}_{b}$ denote the value of the string x in base b.

Next theorem is one of our main results.

Theorem 1. Let $b\ge 2$ be a numeration base. Let $0<A,B,c,d\le b$ integers such that $A\cdot B={\left[cd\right]}_{b}$ and $c+d=A$. Then,

$A{^}^{k}\cdot B={\left[cA{^}^{k-1}d\right]}_{b}.$

Proof of Theorem 1 is covered in Section 3. Similar proof to that of Theorem 1 gives also the somewhat stronger statement Theorem 3.

The above table illustrates the result from Theorem 1 if $b=10$ and $\left(A,B\right)=\left(9,9\right)$, ${\left[cd\right]}_{b}={\left[81\right]}_{10}$, and $k\in \left\{2,3,4,5,6,7,8\right\}$. Note that $9\times 9=81$ and $8+1=9$.

Theorem 2. Let $b>2$ numeration base and $k,l>1$ integers then one has:

$\begin{array}{l}{\left(b-1\right)}^{^k}\cdot {\left[{a}_{1}{a}_{2}{a}_{3}\cdots {a}_{l}\right]}_{b}\\ ={\left[{a}_{1}{a}_{2}{a}_{3}\cdots {a}_{l}\right]}_{b}{\left[{a}_{1}{a}_{2}{a}_{3}\cdots {a}_{l}-1\right]}_{b}\left(b-1\right)^\left(k-l\right)-{\left[{b}^{l}-{a}_{1}{a}_{2}{a}_{3}\cdots {a}_{l}\right]}_{b}\end{array}$ (2)

in particular if b is odd and ${\left[{a}_{1}{a}_{2}{a}_{3}\cdots {a}_{l}\right]}_{b}=\left({b}^{l}-1\right)/2$.

Then (2) gives a solution of (1).

The proof of Theorem 2 is done in Section 4.

The following examples illustrate the statement of Theorem 2.

Example:

${9}^{^130}\cdot {\left[123\right]}_{10}={\left[122\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{9}^{^1327}83\right]}_{10}$

${7}^{^130}\cdot {\left[123\right]}_{8}={\left[{1227}^{^127}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}489\right]}_{8}$

${9}^{^130}\cdot {\left[123\right]}_{10}={\left[122\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{9}^{^127}389\right]}_{8}$

Theorem 3. let $b>2$ umeration base. Let $0<A,B,c,d,\alpha \le b$ integers such that $A\cdot B={\left[cd\right]}_{b}$ and $c+d=\alpha $. Then,

${A}^{^k}B={\left[c{\alpha}^{^k-1}d\right]}_{b}=A{B}^{^k}$

Next theorem shows for all numeration bases examples of pairs $\left(A,B\right)$ that satisfy the hypothesis of Theorem 1.

Theorem 4. Let $b\ge 2$ be a numeration base. Then the pairs $\left(AB\right)={\left[\left(b-1\right)\left(b-k\right)\right]}_{b},1\le k\le b$ satisfy the hypothesis of Theorem 1.

Proof:

${\left[\left(b-1\right)\left(b-k\right)\right]}_{b}$

${b}^{2}-bk-b+k=b\left(b-k-1\right)+k={\left[\left[b-k-1\right],k\right]}_{b}$

$\Rightarrow b-k-1+k=b-1.$

Corollary. Let $b\ge 2$ be numeration base. Then $\left[\left(b-1\right)\left(b-2\right)\right]b$.

Consequently, satisfies the hypothesis of Theorem 1, consequently

$\left(b-1\right){^}^{k}\left(b-2\right)={\left[\left(b-3\right){\left(b-1\right)}^{^\left(k-1\right)}2\right]}_{b}.$

Proof: apply Theorem 4 to the pair $\left(AB\right)=\left(b-1\right)\left(b-2\right)$.

The above table illustrates the result from Theorem 1 & Theorem 3 if $b=7$, $b-1=6$, $b-2=5$, ${\left[cd\right]}_{b}={\left[42\right]}_{7}$, thus $A=6,B=5$ and $k\in \left\{2,3,4,5,6,7,8\right\}$. Note that ${\left[6\cdot 5\right]}_{7}={\left[42\right]}_{7}$ and ${\left[4+2\right]}_{7}=6$.

$\left|\begin{array}{cc}b& \left(A,B\right)\\ 2& \\ 3& \left(2,2\right)\\ 4& \left(2,3\right),\left(3,2\right),\left(3,3\right)\\ 5& \left(2,3\right),\left(2,4\right),\left(3,2\right),\left(3,4\right),\left(4,2\right),\left(4,3\right),\left(4,4\right)\\ 6& \left(2,5\right),\left(3,5\right),\left(4,5\right),\left(5,2\right),\left(5,3\right),\left(5,4\right),\left(5,5\right)\\ 7& \left(2,4\right),\left(2,6\right),\left(3,3\right),\left(3,5\right),\left(3,6\right),\left(4,2\right),\left(4,4\right),\left(4,6\right),\\ & \left(5,3\right),\left(5,6\right),\left(6,2\right),\left(6,3\right),\left(6,4\right),\left(6,5\right),\left(6,6\right)\\ 8& \left(3,7\right),\left(4,7\right),\left(5,7\right),\left(6,7\right),\left(7,2\right),\left(7,3\right),\left(7,4\right),\left(7,5\right),\left(7,6\right),\left(7,7\right)\\ 9& \left(2,5\right),\left(2,8\right),\left(3,4\right),\left(3,8\right),\left(4,3\right),\left(4,5\right),\left(4,6\right),\left(4,7\right),\left(4,8\right),\\ & \left(5,2\right),\left(5,4\right),\left(5,6\right),\left(5,8\right),\left(6,4\right),\left(6,5\right),\left(6,8\right),\left(7,4\right),\left(7,8\right),\\ & \left(8,2\right),\left(8,3\right),\left(8,4\right),\left(8,5\right),\left(8,6\right),\left(8,7\right),\left(8,8\right)\\ 10& \left(2,9\right),\left(3,4\right),\left(3,7\right),\left(3,9\right),\left(4,6\right),\left(4,9\right),\left(5,9\right),\left(6,4\right),\left(6,7\right),\\ & \left(6,9\right),\left(7,3\right),\left(7,6\right),\left(7,9\right),\left(8,9\right),\left(9,2\right),\left(9,3\right),\left(9,4\right),\left(9,5\right),\\ & \left(9,6\right),\left(9,7\right),\left(9,8\right),\left(9,9\right)\end{array}\right|$

The above table shows all pairs $\left(A,B\right)$ that satisfy the hypothesis of Theorem 1 for small numeration bases. We observe that for $b=2$ there are no pairs $\left(A,B\right)$ that satisfy the hypothesis of Theorem 1.

3. Proof of Theorem 1

$\begin{array}{c}\underset{l=1}{\overset{k}{{\displaystyle \sum}}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}A{b}^{l}\cdot B=\underset{l=1}{\overset{k}{{\displaystyle \sum}}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}A\cdot B{b}^{l}=\underset{l=1}{\overset{k}{{\displaystyle \sum}}}\left(cb+d\right){b}^{l}=\underset{l=1}{\overset{k}{{\displaystyle \sum}}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}c\cdot {b}^{l+1}+d\cdot \underset{l=1}{\overset{k}{{\displaystyle \sum}}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}{b}^{l}\\ =c\cdot {b}^{k+1}+\underset{l=1}{\overset{k-1}{{\displaystyle \sum}}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}c\cdot b+\underset{l=1}{\overset{k-1}{{\displaystyle \sum}}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}d\cdot b+d\cdot {b}^{k}\\ =c\cdot {b}^{k+1}+\underset{l=1}{\overset{k-1}{{\displaystyle \sum}}}\left(c+d\right)\cdot {b}^{l}+d\cdot {b}^{k}\\ =c\cdot {b}^{k+1}+\underset{l=1}{\overset{k-1}{{\displaystyle \sum}}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}A\cdot b+d\cdot {b}^{k}={\left[c\left(A\right){^}^{k-1}d\right]}_{b}\end{array}$

4. Proof of Theorem 2

Using that ${\left(b-1\right)}^{k}={b}^{k}-1$ and that ${\left(b-1\right)}^{k-l}={b}^{k-l}-1$.

One has that:

$\begin{array}{l}{\left(b-1\right)}^{k}\cdot {\left[{a}_{1}{a}_{2}{a}_{3}\cdots {a}_{l}\right]}_{b}=\left({b}^{k}-1\right)\cdot {\left[{a}_{1}{a}_{2}{a}_{3}\cdots {a}_{l}\right]}_{b}\\ ={\left[+{b}^{k}{a}_{1}{a}_{2}{a}_{3}\cdots {a}_{l}\right]}_{b}-{b}^{l}{\left[{a}_{1}{a}_{2}{a}_{3}\cdots {a}_{l}\right]}_{b}\\ =+{\left[+{b}^{k}{a}_{1}{a}_{2}{a}_{3}\cdots {a}_{l}\right]}_{b}-1+{b}^{k}+{b}^{l}-{b}^{l}\\ =+{\left[+{b}^{k}{a}_{1}{a}_{2}{a}_{3}\cdots {a}_{l}\right]}_{b}-1+{b}^{l}\left({b}^{k-l}-1\right)+{\left[{b}^{l}-{a}_{1}{a}_{2}{a}_{3}\cdots {a}_{l}\right]}_{b}\\ =-1\left(b-1\right)^\left(k-l\right)-{\left[{b}^{l}-{a}_{1}{a}_{2}{a}_{3}\cdots {a}_{l}\right]}_{b}\end{array}$

5. Conclusion

Motivated by possible applications to the study of palindromes and other sequences of integers we discover a method for producing infinite families of integer solutions and almost solutions of the equation $N\cdot M=reversal\left(N\cdot M\right)$. Our results complement the results in [1] and are valid in all numeration bases $b>2$.

Acknowledgements

While working on this project C. E. was an undergraduate student at West Chester University of Pennsylvania.

References

[1] Nitica, V. and Junius, P. (2019) Infinite Sets of Solutions and Almost Solutions of the Equation *N*⋅*M* = *reversal* (*N*⋅*M*). Open Journal of Discrete Math, 9, 63-67.

https://doi.org/10.4236/ojdm.2019.93007

[2] Nitica, V. (2019) Infinite Sets of *b*-Additive and *b*-Multiplicative Ramanujan-Hardy Numbers. The Journal of Integer Sequences, 22, Article number: 9.4.3.

[3] World of Numbers.

http://www.worldofnumbers.com/em36.htm

[4] Nitica, V. (2018) About Some Relatives of the Taxicab Number. The Journal of Integer Sequences, 21, Article number: 18.9.4.