Square Root Galore

Let $X$ be the sum of all possible $x$ .

The first step is to square both sides:

$x + \sqrt{1000} = a + b + 2\sqrt{ab}$

We rearrange the equation:

$x - a - b = \sqrt{1000} - \sqrt{4ab}$

On the left side of the equation, we have an integer, and on the right side, we have a difference of $2$ square roots. The difference of the $2$ square roots must be an integer. It turns out that the right side is only an integer when the $4ab = 1000$ , but we have to prove this.

Lemma: Given two integers $b$ and $c$ which are not perfect squares, the only solution to the equation $\sqrt{c} - \sqrt{b} = a$ occurs when $b = c$ and $a = 0$ .

Proof of Lemma:

We rearrange the equation and square both sides to obtain:

$a^2 - b + 2a\sqrt{b} = c$

$c - a^2 +b = 2a\sqrt{b}$

On the left side of the equation, we have an integer, and on the right side of the equation, we have a product of an integer and an irrational number. The only time the product of an integer and an irrational number is an integer is when the integer is $0$ . Therefore, $a = 0$ and $b = c$ and our lemma is proven.

Going back to the original problem, we know that $x = a + b$ and $ab = 250$ . The pair $(a, b)$ ranges over all pairs of positive integers where the product is $250$ . Since $x = a + b$ for each distinct, unordered pair $(a, b)$ , we conclude that $X$ is the sum of the factors of $250$ . In order to calculate $X$ , we use the formula for the sum of divisors .

Since $250 = 2*5^3$ , the sum of divisors of $250$ is:

$\frac{2^2 - 1}{2-1}\frac{5^4-1}{5-1} = 3*156 = 468$

The last $3$ digits of $468$ are $\boxed{468}$ .

Suppose $\sqrt{x+\sqrt{1000}}=\sqrt{a}+\sqrt{b}.$ Squaring both sides, $x+\sqrt{1000}=(a+b)+2\sqrt{ab}$ . Naturally, one would like to say that this can only happen if $a+b=x$ and $4ab=1000,$ the following lemma justifies that. Note that $\sqrt{1000}$ is not a rational number, which can be justified in the same way as the famous theorem that $\sqrt{2}$ is not a rational number.

Lemma. Suppose $u+\sqrt{v}=\sqrt{w},$ where $u,v,w$ are rational numbers, and $\sqrt{v}$ is not a rational number. Then $u=0$ and $v=w.$

Proof. Squaring both sides, we get $u^2+v+2u\sqrt{v}=w.$ If $u\neq 0,$ then this implies $\sqrt{v}=\frac{w-v-u^2}{2u},$ which contradicts irrationality of $\sqrt{v}.$ So $u=0,$ which implies $v=w.$

Back to the original problem, if $a+b=x$ and $4ab=1000,$ then $ab=250.$ Note that $a$ and $b$ are positive; we may assume that $a\leq b$ . Then there are four possibilities for $a:$ $1,$ $2,$ $5,$ and $10$ , with $b$ being, respectively, $250,$ $125,$ $50,$ and $25$ . Clearly, all four pairs work, and the corresponding values of $x$ are $251,$ $127,$ $55,$ and $35.$ They add up to $468.$

Squaring gives $x + \sqrt{1000} = a+b + 2\sqrt{ab}$ . Now it's a general result that if $\alpha + \sqrt\beta = \alpha'+ \sqrt{\beta'}$ where $\beta,\beta'$ are not perfect squares, then $\alpha= \alpha'$ and $\beta = \beta'$ . Thus we can equate

$x=a+b, \sqrt{1000} = 2\sqrt{ab} = \sqrt{4ab}$

Thus ab = 250 and it remains to factor 250. This gives 4 solutions:

• (a,b) = (1, 250), which gives x=251;
• (a,b) = (2, 125), which gives x=127;
• (a,b) = (5, 50), which gives x=55;
• (a,b) = (10, 25), which gives x=35;

with a sum of 468.

squaring both sides of equation,we get x + (1000)^1/2 = a+b+2(ab)^1/2 =>x + 2(250)^1/2=a+b+2(ab)^1/2 since x,a,b are positive integers a+b =x=real and ab=250 250=5 5 5*2 therefore possible values of a are 1, 2, 5, 10, 25,50,125,250 and corresponding values of b are 250,125,50,25,10, 5, 2, 1 respectively. and corresponding values of x are-251,127,55,35,35,55,127,251 neglecting repeated values of x,the possible values of x are-251,127,55,35. sum of these values=468,which is the required answer.

By squaring both sides of the equation you end up with x+ sqrt{1000}=a+b+ 2sqrt{a b}. Since a and b must both be integers a+b=x, and 2sqrt{ab}=sqrt{1000}. Simplifying we get ab=250 and a+b=x. Since a and b are 2 factors of 250 that multiply together to 250, x is equal to the sum of all the factors of 250 which sum to 468.

sqrt{x+sqrt{1000}}=sqrt{a}+sqrt{b} sqrt{x+2 sqrt{250}}=sqrt{a}+sqrt{b} x+2 sqrt{250}=(a+b)+2*sqrt{ab} ab=250 a+b=x

• a=1 b=250 x=251
• a=250 b=1 x=251
• a=2 b=125 x=127
• a=125 b=2 x=127
• a=5 b=50 x=55
• a=50 b=5 x=55
• a=10 b=25 x=35
• a=25 b=10 x=35 sum of all x = 251+127+55+35=468

a+b=x ab=250 if a = 1 then b = 250, x=251 if a = 2 then b=125, x=127 if a = 5 then b=50, x=55 if a = 10 then b =25, x=35 if a and b are reversed, then x is the same as it is not reversed (for ex : if a=2 and b=125 are reversed, it become a=125 and b=2 (same x))

Total = 251 + 127 + 55 + 35 =468

Squaring both sides, you get

$x+\sqrt{1000}=a+b+2*\sqrt{ab}$

By similarity of powers,

$a+b=x$ , $\sqrt{1000}=2*\sqrt{ab},$ or $\sqrt{250}=\sqrt{ab}.$

Thus, any integer pair (a,b) that can satisfy ab=250, will give a value of x. This means that a and b are interchangable. Specifically, the pairs are (1,250),(2,125),(10,25), and (50,5), with x values 251, 127, 35, and 55 respectively. Summing the values results in 468.

(√a+√b )^2 => a+2√ab+ b = x+ √1000 => a+b+2√ab = x+ 2√250

x = a+b 250 = ab

250 = 2*5^3

If a = 1 => b = 250 and x = 251 If a = 2 => b = 125 and x = 127 If a = 5 => b = 50 and x = 55 If a = 10 => b = 25 and x = 35

251 + 127 + 55 + 35 = 468

Because $x$ , $a$ , and $b$ are all positive, we can square both sides without missing any solutions: $\sqrt{x+\sqrt{1000}}=\sqrt{a}+\sqrt{b}\Rightarrow x+\sqrt{1000}=a+b+2\sqrt{ab}$ From here we may compare the rational and irration parts of both sides of the equation. Since $x$ is necessarily an integer and $a+b$ is necessarily an integer, and these are the only quantities on both sides that are always integers given the restraints of the problem, we must have $x=a+b$ . Similarly, we get $\sqrt{1000}=2\sqrt{ab}\Rightarrow 250=ab$ . We can now cycle through all the possibilities of $a$ and $b$ concerning the second equation. For $a$ we obtain 1, 2, 5, 10, 25, 50, 125, and 250, all the divisors of 250. The same goes for $b$ , but the $(a,b)$ pairs go in opposite orders of the list presented. If you add $a$ and $b$ for each pair, you will get a possible value of $x$ . Note however, that the sum of the pair $(a,b)$ is the same as the sum of the pair $(b,a)$ , so each value of $x$ was counted twice (since there is no pair with identical values). Thus, the sum of all possible values of $x$ is merely the sum of the divisors of 250, which is 468.

$\sqrt{x+\sqrt{1000}} = \sqrt{a} + \sqrt{b} =\sqrt{( \sqrt{a} + \sqrt{b})^2} = \sqrt{(a+2\sqrt{a}\sqrt{b}+b} = \sqrt{(a+b)+\sqrt{4ab}}$

hence , $4ab=1000$ $ab=250=2\times5^3$ so, there are $(1+1)\times(3+1)=8$ values of x

and $\sum_{}x=\sum_{i=0}^1 2^i\times\sum_{j=0}^3 5^j = \frac{2^{1+1}-1}{2-1}\times\frac{5^{3+1}-1}{5-1}=\boxed{468}$

Rearrange the equation so that, x + (1000)^(1/2) = a + 2(ab)^(1/2) + b. Observe that in order for this equation to be true if a and b are positive integers then 2(ab)^(1/2) must equal (1000)^(1/2), since it is the only method to possibly contribute such a value, therefore x = a + b. then (1000)^(1/2) = 2(250)^(1/2) = 2(ab)^(1/2), so (250)^(1/2) = (ab)^(1/2), from this ab = 250. All possible forms of (a,b) such that this is true are (1,250), (2,125), (5,50), (10,25). The other forms of (a,b) can be ignored as they will produce multiple non-distinct x's, since x = a + b, then add the various possible (a,b), so x = 1 + 250 = 251, x = 2 + 125 = 127, x = 5 + 50 = 55, x = 10 + 25 = 35. Lastly add all distinct possible x's so S = 251 + 127 + 55 + 35 = 468, then the last three digits of 468 are 468.

Calvin, a good problem indeed! $\sqrt{1000} = 10 \sqrt{10}$ so $2 a b = 10 \sqrt {10}$ $a b = 5 \sqrt{10}$ i listed all the possible combinations a , b

$5\sqrt{10} , 1$

$5 , \sqrt{10}$

$\sqrt{2} , 5\sqrt{5}$

$\sqrt{5} , 5\sqrt{2}$

squaring a and b in these combinations and adding them we get 251, 35,127,55. so x can take the values of 251, 35, 127, or 55. so adding these four numbers we get $\boxed{468}$

When we square the equation on b.s we get the eqns x=a+b and ab=250.Then a,b are (1×250), (2×125), (5×50), (10×25).x=251,130,55,35.sum is468.problem solved.

If we square the equation, we get: $x + \sqrt{1000} = a + b + 2\sqrt{ab}$ .

We know that the on the left side, $x$ is rational, whereas $\sqrt{1000}$ is irrational, and as we know, the sum of a rational and irrational number is irrational, so the left side is irrational, so the right side has to be irrational too. Now, $a + b$ is rational, so the thrid term, $2\sqrt{ab}$ has to be irrational, and equal to the irrational part of the left side, i.e:

$2 \sqrt{ab} = \sqrt{1000}$

$ab = 250$

Now, these ordered pairs $a,b$ can be: $(1, 250), (2, 125), (5, 50), (10, 25), (25, 10), (50, 5), (125, 2), (250, 1)$

Plugging these into the first equation, we get $x \in {251, 127, 55, 35}$ . The sum of all of elements of this set yields the solution, $468$ .

By looking at the question, we immediately conclude that [x+2sqrt(250)] is a perfect square.

The factors of 250 being arranged in 2 sets are

[(1)*(250)] ;

{(2)*(125)];

[(5)*(50)];

[(10)*(25)];

Summing these values we gets possible x's as 251, 127, 55, 35 which add to 468

Square root both sides to get $x+\sqrt{1000}=a+b+2\sqrt{ab}$ . Since $\sqrt{1000}$ is irrational on the LHS, we need an irrational part on the RHS as well. This can only be done with the $2\sqrt{ab}$ term. Let $a=25c$ to obtain $x+10\sqrt{10}=25c+b+10\sqrt{bc}$ . Now we just simply consider all possible integer solutions of $bc=10$ , which gives us 4 possible permutations, ie. $(b,c)=\{(1,10),(2,5),(5,2),(10,1)\}$ . Hence,

$\Rightarrow x=25(1)+(10)=35$ $\Rightarrow x=25(2)+(5)=55$ $\Rightarrow x=25(5)+(2)=127$ $\Rightarrow x=25(10)+1=251$

Sum it all up to obtain $35+55+127+251=\boxed{468}$ .

$\sqrt { x+\sqrt { 1000 } } =\sqrt { a } +\sqrt { b } \\ \sqrt { x+2\sqrt { 250 } } =\sqrt { a } +\sqrt { b } \\ The\quad value\quad of\quad x\quad must\quad be\quad a\quad sum\quad of\quad any\quad 2\quad \\ factors\quad of\quad 250\quad so\quad that\quad the\quad equation\quad will\\ be\quad true.\\ The\quad factors\quad of\quad 250\quad are,\\ 1,2,5,10,25,50,125,250\\ 250\times 1=250\quad \longrightarrow \quad 250+1=251\\ 2\times 125=250\quad \longrightarrow \quad 125+2=127\\ 5\times 50=250\quad \quad \longrightarrow \quad 50+5=55\\ 10\times 25=250\quad \longrightarrow \quad 10+25=35\\ 251+127+55+35=468\\ \boxed { Therefore,\\ the \quad last\quad 3\quad digits\quad of\quad the\quad sum\quad of\quad all\quad \\ \quad possible\quad positive\quad integer\quad values\quad of\quad x\quad \\ satisfying\quad the\quad condition\\ is\quad 468. } \\$

My answer, using Turbo Pascal:

var x,i:integer;
begin
i:=0;
for x:=1 to 250 do
if 250 mod x=0 then i:=i+x;
writeln(i);
readln;
end.

A key technique in solving problems like these include knowledge on the comparing coefficients and on Vieta's Formula.

First, we square both sides to get rid of some surds. (\sqrt{x + \sqrt{1000}})