PROBLEM E : 6185 from the magazine GAZETA MATEMATICĂ

          From GMB 4/1978 page 164.

In translation :

                         "E : 6185*.  Knowing that the numbers  4830,  448230,  44482230

         are products of two consecutive natural numbers, find these consecutive

                      numbers. Generalization.

[Author : ] I. I. Mihailov, Ivanovo, U.R.S.S."

ANSWER CiP

$4\;830\;=\;69\;\times\;70$

$448\;230\;=\;669\;\times\;670$

$44\;482\;230\;=\;6\;669\;\times\;6\;670$

[CiP addition] $4\;444\;822\;230\;=\;66\;669\;\times\;66\;670$

                                  Generalization :

$\underset{n+1}{\underbrace{4\;\dots\;4}}\;8\;\underset{n}{\underbrace{2\;\dots\;2}}\;30\;=\;\underset{n+1}{\underbrace{6\;\dots\;6}}\;9\times \underset{n}{\underbrace{6\;\dots\;6}}\;70 \tag{G}$

                         Solution CiP

           If a number  $A$  is a product of consecutive numbers, i.e.  $A=k(k+1)$, then we have

$A=k(k+1) \Rightarrow k^2<A<(k+1)^2\Rightarrow k<\sqrt{A}<k+1\Rightarrow k=[\sqrt{A}]$

([X]=the integer part of the number X) We have :

$\sqrt{4830}=69,4...$  and it is verified by calculation that  $4830=69\times 70$

$\sqrt{448230}=669,4...$  and it is verified by calculation that  $448230=669\times 670$

$\sqrt{44482230}=6669,4...$  and it is verified by calculation that  $44482230=6669\times 6670$

The next example in the answer, verified by calculation, is my creation. I also formulated the generalization, which we will verify below.

          To carry out the calculations we need to replace  $1\;\dots\;1$  with the more precise expression

$\underset{n}{\underbrace{1\;\dots\;1}}=\frac{1}{9}\cdot (10^n-1) \tag{1}$

Indeed  $\underset{n}{\underbrace{1\;\dots\;1}}=\frac{1}{9}\cdot \underset{n}{\underbrace{9\;\dots\;9}}=\frac{1}{9}\cdot (1\underset{n}{\underbrace{0\;\dots\;0}}-1)=\frac{1}{9}\cdot(10^n-1).$

          Analyzing the position of the digits of the number  $4\dots482\dots230$, we obtain that the value of the number on the left of the relation  (G) is :

$\underset{n+1}{\underbrace{4\;\dots\;4}}\underset{n+3}{\underbrace{\;8\;\overset{n+2}{\overbrace{\underset{n}{\underbrace{2\;\dots\;2}}\;\underset{2}{\underbrace{30}}}}}}=\underset{n+1}{\underbrace{4\dots4}}\cdot 10^{n+3}+8\cdot 10^{n+2}+\underset{n}{\underbrace{2\dots2}}\cdot 100+30=$

$\overset{(1)}{=}4\cdot \frac{1}{9}\cdot (10^{n+1}-1)\cdot 10^{n+3}+8\cdot 10^{n+2}+2\cdot \frac{1}{9}\cdot(10^n-1)\cdot 100+30=$

$=\frac{4}{9}\cdot 10^{2n+4}-\frac{4}{9}\cdot 10^{n+3}+8\cdot 10^{n+2}+\frac{2}{9}\cdot 10^{n+2}-\frac{2}{9}\cdot 100+30=$

$=\frac{4}{9}\cdot 10^{2n+4}+\frac{34}{9}\cdot 10^{n+2}+\frac{70}{9} \tag{2}$

     On the right side of (G) we have :

$\underset{n+1}{\underbrace{6\dots6}}9\times \underset{n}{\underbrace{6\dots6}}70=\left [\frac{6}{9}\cdot (10^{n+1}-1)\cdot 10+9\right ]\times \left[\frac{6}{9}\cdot (10^n-1)\cdot 100+70\right ]=$

$=\left [\frac{2}{3}\cdot 10^{n+2}+\frac{7}{3}\right]\cdot \left [\frac{2}{3}\cdot 10^{n+2}+\frac{10}{3}\right ]=\frac{4}{9}\cdot 10^{2n+4}+\frac{2}{3}\cdot 10^{n+2}\cdot \left (\frac{7}{3}+\frac{10}{3}\right )+\frac{70}{9}=$

$=\frac{4}{9}\cdot 10^{2n+4}+\frac{34}{9}\cdot 10^{n+2}+\frac{70}{9} \tag{3}$

The results in (1) and (2) prove the equality in (G).

$\blacksquare$