Prove the Following Mathematic Form

[Albert1]

$A=\overbrace{ 11-------1 }^{m}\underbrace{ 22-------2 }_{m}$

prove :$A=k\times (k+1),\,\, where\,\, k\in N$

(A can be expressed as the multiplication of two consecutive positive integers)

 

[M R]

\displaystyle A=\frac{10^m-1}{9} \times 10^m +2 \times \frac{10^m-1}{9}

\displaystyle 9A=(10^m-1) \times 10^m +2 \times (10^m-1)

\displaystyle 9A=10^{2m} -10^m +2 \times 10^m-2

\displaystyle 9A=10^{2m} +10^m - 2

\displaystyle 9A=(10^m-1)(10^m +2)

Each factor on the right is a multiple of 3 and they differ by 3 so on dividing by 9 we get \displaystyle A=k(k+1) where \displaystyle k = \frac{10^m-1}{3} .

I nice start to my day, thank you :)

 

[Albert1]

Hi MR

A nice solution (Clapping)

 

[Albert1]

$A=\overbrace{ 11-------1 }^{m}\underbrace{ 22-------2 }_{m}$

prove :$A=k\times (k+1),\,\, where\,\, k\in N$

(A can be expressed as the multiplication of two consecutive positive integers)

$A=\overbrace{ 11-------1 }^{m}\times 10^m+\underbrace{ 1-------1 }_{m}\times 2$
$=\overbrace{ 11-------1 }^{m}\times (10^m+2)=\overbrace{ 11-------1 }^{m}\times (\overbrace{ 99-------9 }^{m}+3)$
$=\overbrace{ 11-------1 }^{m}\times (\overbrace{ 33-------3 }^{m}\times 3+3)$
$=\overbrace{ 33-------3 }^{m}\times (\overbrace{ 33-------3 }^{m}+1)=k\times (k+1)$
($k=\overbrace{ 33-------3 }^{m}$)