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Prove the Following Mathematic Form

Albert

Well-known member
Jan 25, 2013
1,225
$A=\overbrace{ 11-------1 }^{m}\underbrace{ 22-------2 }_{m}$

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

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

M R

Active member
Jun 22, 2013
51
[tex]\displaystyle A=\frac{10^m-1}{9} \times 10^m +2 \times \frac{10^m-1}{9}[/tex]

[tex]\displaystyle 9A=(10^m-1) \times 10^m +2 \times (10^m-1)[/tex]

[tex]\displaystyle 9A=10^{2m} -10^m +2 \times 10^m-2[/tex]

[tex]\displaystyle 9A=10^{2m} +10^m - 2[/tex]

[tex]\displaystyle 9A=(10^m-1)(10^m +2) [/tex]

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

I nice start to my day, thank you :)
 

Albert

Well-known member
Jan 25, 2013
1,225
Hi MR

A nice solution (Clapping)
 

Albert

Well-known member
Jan 25, 2013
1,225
$A=\overbrace{ 11-------1 }^{m}\underbrace{ 22-------2 }_{m}$

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

(A can be expressed as the multiplication of two consecutive positive integers)
$A=\overbrace{ 11-------1 }^{m}\times 10^m+\underbrace{ 1-------1 }_{m}\times 2$
$=\overbrace{ 11-------1 }^{m}\times (10^m+2)=\overbrace{ 11-------1 }^{m}\times (\overbrace{ 99-------9 }^{m}+3)$
$=\overbrace{ 11-------1 }^{m}\times (\overbrace{ 33-------3 }^{m}\times 3+3)$
$=\overbrace{ 33-------3 }^{m}\times (\overbrace{ 33-------3 }^{m}+1)=k\times (k+1)$
($k=\overbrace{ 33-------3 }^{m}$)