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)

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

 

[CellCoree]

To prove this, we can start by breaking down the given expression:

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

$=111...111 \times 22...222$

$=11...11 \times 10^{m} + 1 \times 22...22 \times 10^{m} + 2$

$=11...11 \times 10^{m} + 22...22 \times 10^{m} + 2$

$=11...11 \times (10^{m} + 1) + 22...22 \times 10^{m} + 2$

$=11...11 \times (10^{m} + 1) + 22...22 \times (10^{m} + 1)$

$=(11...11 + 22...22)\times (10^{m} + 1)$

$=33...33 \times (10^{m} + 1)$

Now, we can see that 33...33 is a multiple of 11, since each digit is a 3 and the number of digits is m. Similarly, 10^{m} + 1 is a multiple of 11, since 10^{m} is a multiple of 11 and 1 is also a multiple of 11. Therefore, 33...33 \times (10^{m} + 1) is a multiple of 11.

Additionally, we know that 10^{m} + 1 is also a multiple of 10, since 10^{m} is a multiple of 10 and 1 is also a multiple of 10. Therefore, 33...33 \times (10^{m} + 1) is a multiple of 10.

This shows that A is a multiple of both 11 and 10, and thus, it can be expressed as the multiplication of two consecutive positive integers, k and k+1, where k is the number of digits of 33...33 and k+1 is the number of digits of 10^{m} + 1.

Therefore, we have proved that $A=k\times (k+1),\,\, where\,\, k\in N$.