How many 17s?

$17^2 +17^2+17^2 = 3*17^2 = \sqrt{9*17^2*17^2} =\sqrt{ \color{#D61F06}{9*17^ 2} *17^2}.\\9*17^2 =\boxed{ 2601 }$

We can write the equation as follows:

$\sqrt { { 17 }^{ 2 }+{ 17 }^{ 2 }+{ 17 }^{ 2 }+...+{ 17 }^{ 2 }+{ 17 }^{ 2 }+{ 17 }^{ 2 } } ={ 17 }^{ 2 }+{ 17 }^{ 2 }+{ 17 }^{ 2 }$

$\sqrt { { 17 }^{ 2 }+{ 17 }^{ 2 }+{ 17 }^{ 2 }+...+{ 17 }^{ 2 }+{ 17 }^{ 2 }+{ 17 }^{ 2 } } ={ 17 }\left( 17\cdot 3 \right)$

Because ${17}^{2}$ appears $n$ times under the square root sign, we can write:

$\sqrt { n\cdot { 17 }^{ 2 } } ={ 17 }\left( 17\cdot 3 \right)$

Squaring both sides:

$n\cdot { 17 }^{ 2 }={ { 17 }^{ 2 } }\left( { 17 }^{ 2 }\cdot 9 \right)$

Dividing by ${17}^{2}$ :

$n=\left( { 17 }^{ 2 }\cdot 9 \right)$

$n=\left( 289\cdot 9 \right)$

$\boxed{n=2601}$

We can write the equation as

$\sqrt{n17^{2}} = 17^{2} + 17^{2} + 17^{2}$

therefore

$17\sqrt{n} = 17^{2} + 17^{2} + 17^{2} \sqrt{n} = 17 +17+ 17$

$n= 51^{2}=2601$

$\sqrt{n\times17^{2}} =3\times 17^{2}$

$n \times 17^{2} = 9 \times 17^{4}$

$n = 9\times 17 ^{2} = 2601$

Rewriting the equation as $\sqrt{x}=3(17^2)$

Squaring both sides gives us $x=(3(17^2))^2=17^2(17^2(9))$ So $9(17^2)=\boxed{2601}$

Let $S = \sqrt{17^2+17^2+...+17^2} = \sqrt{k \times 17^2}$ . We know that $S = 17^2+17^2+17^2 = 3 \times 17^2$ . From this, we get $S^2 = k \times 17^2 = (3 \times 17^2)^2 = 9 \times 17^4$ . Let's find k : $k = \frac{9 \times 17^4}{17^2} = \boxed{2601}$ . So 17^2 needs to be 2601 times under the sqare root to let S equals $3 \times 17^2$ .

If we let x equal 17,

$x = 17$

we can write the equation like $x^2 + x^2 +x^2 = 3 \cdot x^2 = \sqrt{x^2 + x^2 + x^2 + \ldots + x^2} = \sqrt{n \cdot x^2}$ Or just

$3 \cdot x^2 = \sqrt{n \cdot x^2}$

We can now square both sides $9 \cdot x^4 = n \cdot x^2$

and isolate n

$n = \frac {9 \cdot x^4}{x^2} = 9 \cdot x^2$

Now we replace our x with 17 $n = 9 \cdot 17^2 = 9 \cdot 289 = 2601$

$(17^2 * 3)^2$ = 751689

$\frac{751689}{289}$ = 2601

This is equal with √(16+16+16+⋯+16+16+16)=16+16+16. There fore to get 48 or 16+16+16 we should have 48*48 in the radical sign. That is 2304. This implies we should have {(2304)/(16)}th 16 in the radical sign. that is 144 16s should have.

$n 17^2 =9 \times 17^4\\ n= 9\times 17^2=2601$

squaring both sides gives $17^{2}$ + $17^{2}$ +...+ $17^{2}$ = $3^{2}$ $\times$ $17^{4}$ = (9 $\times$ $17^{2}$ ) $\times$ $17^{2}$ . The term in the brackets tells us how many $17^{2}$ are required. Let this value be ${n}$ .

$\therefore$ ${n}$ = 9 $\times$ $17^{2}$ = $\boxed{2601}$

Note that: $3 \times 17^2 = 867$

Then $867^2 = 751689$ giving us the result: $751689 / 17^2 = 2601$

|17^2=x| sqrt(x+x+x+x+x....)=sqrt(ax)| a = number of times x is added| sqrt(ax)=3x| ax=9x^2| a=9x| a=(9)17^2| a=(9)289| a=2601| 17^2 should appear 2601 times under the sqrt.|

We need something who’s square root is $17^2+17^2+17^2$ so let’s look at $\sqrt{(3\times 17^{2})^{2}}$ or $\sqrt{3^{2}\times 17^{4}}$ , and $17^{4} = (17^{2})^{2} = 17^{2}\times 17^{2}$ so we need $17^{2}$ lots of $17^{2}$ added together to make a single $17^{4}$ and we need $3^{2}\times 17^{4}$ to be rooted. Or $\sqrt{3^{2}\times 17^{2}\times 17^{2}}$ so there needs to be $3^{2}\times 17^{2} = \boxed{2601}$ $17^{2}$ s added together.

simple solution, we can interpret as:

$\sqrt{x17^{2}} = 867$

$17\sqrt{x} = 867$

$\sqrt{x} = \frac{867}{17}$

$\sqrt{x} = 51$

$x = 2601$

RHS we have 17^{2}+17^{2}+17^{2}=867 Further it is equal to \sqrt{867×867} On factorising we get- \sqrt{17^{2}×3×867} \sqrt{17^{2}×2601}

Hence,17^{2} should have to appear \boxed{2601} times ... Q.E.D.

Let $X=17^{2}$ . The equation in the question can be rewritten as follows:

$\sqrt{nX}=3X$

where $n$ is the number of X's required under the square root. Solving for $n$ :

$n=9X=9\times17^{2}=2601$

as required.

We can rewrite $\sqrt{17^2 + ... + 17^2} = 17^2 + 17^2 + 17^2$ as $\sqrt{a * 17^2} = 3 * 17^2$ . By squaring both sides of the equation, we obtain $a * 17^2 = 9 * 17^4 = a * 17^2 * 17^2 \rightarrow a = 17^2 * 9 = \boxed{2601}$

We can write the equation as √n.17^2=3.17^2 √k.17^2=3.17^2;where n=k.17^2 K=9 n=9×17^2=2601

17^2 * 9 = 2601

Rewrite all the 17s as $\sqrt{a17^{2}}$ and equate this to $3(17^{2})$ Square both sides... $a.17^{2} = (3.17^{2})^{2} = 9.17^{^{2}}$ The number of 17s required is a which is $a = 9.17^{2}= 2601$

we can solve this in this way --- ( 3 * 17^2)^2=(9 * 17^2 * 17^2)=2601 * 17^2 , as we need to keep 3...... squared 17.

Let x be number of times 17^2 will appear
therefore, (x.17^2)^1/2 = 3.17^2
squaring both sides,
x.17^2 = 9.17^2.17^2
x.17^2 = 2601.17^2
x = 2601

We can derive algebraic formula for this. By deriving the formula i.e. 9×a^2 and putting a=17 we get 9×17^2 =9*289=2601

We need LHS to be 9 17^4 and that can happen if there are 9 17^2 = 2601 terms inside the square root.