The powers of 3

$\begin{aligned} P & = \left(3^{23}-3^{22}\right)\left(3^{24}-3^{23}\right)\left(3^{25}-3^{24}\right) \\ & = 3^{22}(3-1)\times 3^{23}(3-1)\times 3^{24}(3-1) \\ & = 3^{22}(2)\times 3^{23}(2)\times 3^{24}(2) \\ & = 2^3 \times 3^{22+23+24} \\ & = 2^3\times 3^{69} \end{aligned}$

Therefore, $m+n=3+69=\boxed{72}$ .

The given expression is : $\large \color{#20A900} (3^{23} - 3^{22}) \times (3^{24} - 3^{23}) \times (3^{25} - 3^{24})$

$\implies \large 3^{22}(3 - 1) \times 3^{23}(3 - 1) \times 3^{24}(3 - 1)$

$\implies \large {\color{#3D99F6} 2 \times 2 \times 2} \times {\color{#20A900} 3^{22} \times 3^{23} \times 3^{24}}$

$\implies \large 2^3 \times 3^{22 + 23 + 24}$

$\implies \large {\color{#E81990}2}^{\color{#3D99F6}3} \times {\color{#20A900}3}^{\color{#3D99F6}69}$

So, $m = 3, n= 69$

Therefore, $\large \color{#3D99F6}m + n = 72$

It is easy to identify that all the numbers are beginning with 3

$\large \therefore$ all that you have to do is factorize:

$\begin{aligned} \color{#20A900} \large \implies \left(3^{23}-3^{22}\right)\left(3^{24}-3^{23}\right)\left(3^{25}-3^{24}\right) \\ & \color{#69047E} \large = 3^{22}(3-1)\times 3^{23}(3-1)\times 3^{24}(3-1) & \small \color{#3D99F6} \text{Note: 3 - 1 = 2 which co-relates to the question } \\ \\ & \color{#69047E} \large = 3^{22}(2)\times 3^{23}(2)\times 3^{24}(2) \\ & \color{#69047E} \large = 2^3 \times 3^{22+23+24} & \small \color{#3D99F6} \text{Using laws of exponetials} \\ & \color{#D61F06} \large \implies \boxed{ 2^3\times 3^{69}} \end{aligned}$

The question asks us to find the value of ${\color{#3D99F6}m + n}$ when the expression is equal to ${\color{#E81990}2}^{\color{#3D99F6}m} \times {\color{#20A900}3}^{\color{#3D99F6}n}$

Therefore, $\color{#3D99F6} \large m + n = \color{#EC7300} 3 + 69 = \color{#20A900} \boxed{72}$

$\large{\left(3^{23}-3^{22}\right)\left(3^{24}-3^{23}\right)\left(3^{25}-3^{24}\right)}$

$\large{=3^{22}(3-1)\left(3^{23}\right)(3-1)\left(3^{24}\right)(3-1)}$ $\implies$ $\text{Distributive Property}$

$\large{=3^{22}(2)\left(3^{23}\right)(2)\left(3^{24}\right)(2)}$

$\large{=2^{1+1+1}\left(3^{22+23+24}\right)}$ $\implies$ $\text{Laws of Exponents}$

$\large{=2^3\left(3^{69}\right)}$

Therefore, $\large{m+n=3+69=}\huge{\boxed{72}}$

Notice that 3^n - 3^(n-1) = 3^(n-1) * 2. Thus the expression becomes 3^(23+24+25) * 2^3 so m+n = 72.

$2^{m} \times 3^{n}$ =( $3^{23} - 3^{22}$ ) ( $3^{24} -3^{23}$ ) ( $3^{25}-3^{24}$ )

$\Rightarrow$ $3^{22} (3-1)$ $\times$ $3^{23} (3-1)$ $\times$ $3^{24} (3-1)$

$\Rightarrow$ 2 ( $3^{22} \times 3^{23} \times 3^{24}$ )

$2^{m} \times 3^{n}$ = $2^{3} \times 3^{22+23+24}$ = $2^{3} \times 3^{69}$

Therefore, m+n = 3+69 = 72

There's a cheat for this format. Add together the first exponents in each set of brackets. 23+24+25=72

It works with other similar problems as well. I tried it with (4^16 - 4^15)(4^17 - 4^16)(4^18 - 4^17) and it came out to 51, both the proper and improper ways. Feel free to check my method on other possible sets, and comment on this if you come up with an example that it doesn't work for.

$\begin{aligned} P & = \left(3^{23}-3^{22}\right)\left(3^{24}-3^{23}\right)\left(3^{25}-3^{24}\right) \\ & = 3^{22\times 3}(3^0(3-1)\times 3^1(3-1)\times3^2(3-1)) \\ & = 3^{22\times3}(3^{0+1+2}\times (2\times2\times2)) \\ & = 3^{22\times3}(3^{3}\times2^{3}) \\ & = 3^{22\times3+3}\times2^{3} \\ & = 3^{69} \times2^{3} \\ & \end{aligned}$ Therefore, $m+n=3+69=\boxed{72}$ .

(3^23 -3^22) (3^24-3^23) (3^25 -3^24)

we can take 3^22, 3^23, 3^24 as common from 3 terms,then we get

3^22(3 -1) 3^23(3-1) 3^24(2-1)

=3^(22+23+24) *2^3

now m=69 and n=3

so m+n=72

We have: $(3^{23}-3^{22})(3^{24}-3^{23})(3^{25}-3^{24}) = 2^m \times 3^n$

Expanding:

$= (3^{49}-3^{46}-3^{46}+3^{45})(3^{25}-3^{24}) = 3^{72}-3^{71}-3^{71}+3^{70}-3^{71}+3^{70}+3^{70}-3^{69}$

$= 3^{72} - 3 \times 3^{71} + 3 \times 3^{70} - 3^{69}$

$= 3^{72} - 3^{72} +3^{71} -3^{69}$

$= 3^{71} -3^{69}$

Take $= 3^{69}$ as a common factor

$= 3^{69}(9-1)$

$= 3^{69} \times 2^3$

Thus n= 69 and m =3 and m+n=72

$(3^{23}-3^{22})(3^{24}-3^{23})(3^{25}-3^{24})=(3-1)3^22 \times (3-1)3^23 \times (3-10)3^24=(3-1)^3(3^{23} \times 3^{24} \times 3^{25})=2^3 \times 3^{22+23+24}=2^3 \times 3^{69}$ So, $m+n=3+69=\boxed{\large{72}}$

a^x-a^(x-1)=(a-1)^x

So 3^23-3^22=2^23

Further explanation: 3^22=(3^23)/3

3^23-(3^23)/3=(3-3/3)^23=2^23

(I’m not completely sure about this step)

Doing this for each of the parentheses would then provide

2^23x2^24x2^25

2^(23+24+25)=2^72

Edit: nevermind 3^69x2^3>2^72 but I hope I’m on the right path to the 23+24+25 answer phenomenon

Just multiply out the three brackets (a bit messy but it works). You will get eight terms in the form +- $3^x3^y3^z$ (for example the first term is $3^{23}3^{24}3^{25}$ ) Simplifying by realizing that $(a^xa^ya^z) = a^{x + y + z}$ ) gives:

$3^{72} - 3(3^{71}) +3(3^{70}) - 3^{69}$ (1)

$= 3^{69}(3^3 -3(3^2) +3(3) - 1)$

$= 3^{69}(3^3 - 3^3 +3^2 -1)$

$= 3^{69}(8)$

$= 3^{69}(2^3)$

Therefore;

$n = 69$ and $m = 3$

$m + n = 72$

Alternatively you could realise in line (1) that $3(3^{71}) = 3^{72}$ and so the first two terms cancel out. Same method and result but it's that bit cleaner.

I just had a gut feeling that the answer had to be divisible by 9, and only 72 worked.

Please, is just a coincidence that summing 23+24+25 we can get the right result?