Let $x = \overline{abc}$ and $y = \overline{def}$ . The problem can then be restated as finding the solution(s) of

$(x+y)^2 = 1000x+y \qquad 100 \leq x,y < 1000$

Rewriting and solving for $x$ :

$0 = x^2 + 2(y-500)x + y^2-y \\ x = 500-y \pm \sqrt{(y-500)^2 - (y^2-y)} \\ \ \ = 500-y \pm \sqrt{500^2 - 999y}$

A necessary condition is thus that $500^2 - 999y$ is the square of some integer, to be called $k$ . We can write

$999y = 500^2 - k^2 = (500-k)(500+k)$

We are thus looking for a factorization $999y = pq$ , where $p$ and $q$ are positive integers with $p+q=1000$ .

Note that $999 = 3^3 \cdot 37$ . Clearly $p$ and $q$ cannot both be divisible by $3$ , since then $p+q$ would also be divisible by $3$ and hence not $1000$ . Therefore, one of the factors (let us say $p$ ) must be divisible by $27$ , i.e. $p = 27s$ for some integer $s$ .

Furthermore, either $p$ or $q$ (or both) must be divisible by $37$ . However, it cannot be $p$ , since then it's divisible by $999$ and this gives the solution $p=999$ , $q=1$ and thus $y=1$ , which violates $y \geq 100$ . Therefore, $q$ must be divisible by $37$ and we can write $q = 37t$ for some integer $t$ .

We now have the equation

$1000 = p+q = 27s + 37t$

Modulo $9$ , this reads $t \equiv 1$ , which reduces the possible values of $t$ to the set $\{1,10,19\}$ . Trial and error reveals that the only solution is $s=11$ and $t=19$ . This gives $p = 297$ , $q = 703$ , $y = st = 209$ and $k=|500-p|=|500-q|=203$ .

For $x$ we find

$x = 500-y \pm \sqrt{500^2 - 999y} = 500-y \pm k = 291 \pm 203$

The solution $x = 291-203 = 88$ violates $x \geq 100$ , hence $x = 291 + 203 = \boxed{494}$

$(494+209)^2 = 494209$

Let $M = \overline{abc}$ & $N = \overline{def}$ . So, $(M + N)^2 = 1000M + N$

Expansion gives $M^2 + M(2N - 1000) + (N^2 - N) = 0$ which is a quadratic in $M$ .

It's discriminant $\Delta = (2N - 1000)^2 - 4(N^2 - N) = 10^6 - 3996N = 4 (250000 - 999N)$

Clearly, $\sqrt{250000 - 999N} < \sqrt{250000} = 500$

So we may put: $250000 - 999N = (500 - i)^2$ for some positive integer $i$ . Finally, $i(1000 - i) = 999N$

The equation is quite symmetric in $i$ (due to presence of $(1000 - i)$ ),so we may assume: $27 | i \Rightarrow i = 27r$ for some r. Plugging in, we get: $r(1000 - 27r) = 37N$ Now 37 doesn't divide $i$ , for then it would have become 999 itself,not possible. So $1000 - 27r = 37p$ for some p.

Hence, $r = \frac{1000 - 37p}{27} = 37 - p - \frac{10p - 1}{27}$ which is integral for $p = 19$ . Thus, $r = 11, i = 297$ , $N = 209$ & finally $M = 494$ .

let abc be $x$ and def be $y$ (here abc and def are the numbers)

Then $(x+y)^{2}$ = $1000 \times x$ $+$ $y$

$(x+y)^{2}$ = $999x + x + y$

$(x+y) \times (x+y-1)$ = $999 \times x$

$(x+y) \times (x+y-1)$ = $27 \times 37 \times x$

So product of two consecutive positive integers = $27 \times 37 \times x$ .

Now out of two consecutive integers only one is divisible by 3 ,therefore

Numbers $(x+y)$ and $(x+y-1)$ will be of the form 27 $k$ and 37 $m$ where $m \times k$ = $x$ (here m and k should be positive)

Now numbers 27 $k$ and 37 $m$ should be consecutive

If 27 $k$ -37 $m$ =1

Then $k$ = (1+10 $m$ )/27 + $m$

$m$ =8 and $k$ = 11 will be one solution but the product of k and m is 88 but $x$ is 3 digit number

and the next solution is $m$ =35 and $k$ = 48 whose product is 4 digit.

If 37 $m$ -27 $k$ =1

Then $m$ = (1-10 $k$ )/37 + $k$

$k$ =19 , $m$ =26 and so x=494

so abc =494

Let $M= \overline{abc}$ and $N= \overline{def}$ . Then the given equation becomes $(M+N)^2 = 1000M + N$ .

If we expand the left-hand side, subtract $1000M + N$ from both sides, and solve for M using the quadratic equation, we get $M= \frac{1000-2N \pm \sqrt{(1000-2N)^2-4(N^2-N)}}{2}$ ,

which works out to $M = 500-N \pm \sqrt{500^2-999N}$ .

Since M needs to be an integer, $500^2-999N$ must be a perfect square, so we let $500^2-999N=K^2$ for some integer K.

Switching $999N$ with $K^2$ and factoring the resulting difference of squares, we get $999N = (500+K)(500-K)$ . That tells us $999 | (500+K)(500-K)$ .

Now we factor 999 into $3^3*37$ and try to come up with more useful divisibility statements. First, notice that only one of (500+K) and (500-K) can be divisible by 3. That means it must also be divisible by $3^3$ . That, in turn, means the other expression must be divisible by 37. (Otherwise, we'd have 999|(500+K), which leads to the 1-digit solution N=1.)

Let's investigate the case where $500-K \equiv 0\pmod{27}$ and $500+K \equiv 0\pmod{37}$ .

Solving for K in each, we get the set of congruences

$K \equiv 14 \pmod{27}$

$K \equiv 18 \pmod{37}$

Using the second mod relation, we can let $K = 18 + 37L$ for some nonnegative integer L. Plugging into the first relation, we get

$18+37L \equiv 14 \pmod{27}$

$10L \equiv 23 \pmod{27}$

Noticing that $23 \equiv 50 \equiv 10*5 \pmod{27}$ we find $L \equiv 5 \pmod{27}$ . 5 is the only value of L that prevents $500-K$ from going negative, so $L=5$ .

We can now find the values of our other variables:

$K = 18 + 37*5 = 203$

$999N = 703*297$ , so $N=209$

$M = 500-209 \pm 203$ . We use $M=494$ since M must have 3 digits.

We verify that $(494+209)^2=494209$ , and our work is complete! $M =\fbox{494}$ .

(Note: the other case, where $500+K \equiv 0\pmod{27}$ and $500-K \equiv 0\pmod{37}$ , produces no valid solutions).

Let $x=\overline{abc}$ and $y=\overline{def}$ . Then we have $x^2+2xy+y^2=1000x+y$ $y^2+(2x-1)y+x^2-1000x=0$ . We know that $y$ must be integer, so $D=(2x-1)^2-4(x^2-1000x)=3996x+1$ must be square of an integer. Let $3996x+1=m^2$ . Then $3996x=(m-1)(m+1)$ We know that $m-1$ and $m+1$ have same parity, Then since $3996x$ is even, both $m-1$ and $m+1$ are even. $3996=4*27*37$ . From here, we take some cases. $\\$

Case 1. $m-1=2ac$ , and $m+1=2bd$ , with $ab=999$ , and $cd=x$ . If $3\mid m-1$ , then $3\nmid m+1$ . Also if $3\mid m+1$ , then $3\nmid m-1$ So between $a$ or $b$ is divisible by 3.

Case 1a. If $3|a$ , then $3\nmid b$ . So the only possibility for $a$ is $a=27$ , thus $b=37$ , since $27*37=999$ . This will give $27c-1=37d$ . Means that $c=\dfrac{37d+1}{27}=d+\dfrac{10d+1}{27}$ is an integer. We know that for $d=8$ , $c$ is an integer, and also for $d=27e+8$ , for an integer $e$ . For $d=8$ we have $c=11$ , so $x=88$ . It's impossible. For $e>0$ , then we will have $x>1000$ which is also impossibru.

Case 1b. If $3\nmid a$ and $3|b$ , then the only possibility for $b$ is $b=27$ , and $a=37$ . This will give $37c-1=27d$ . And with the same argument from case 1a., we will not have $x$ that we want.

Case 2. If $m-1=4ac$ , and $m+1=bd$ . Same argument, the value between $a$ or $b$ is $27$

Case 2a. If $b=27$ , then $a=37$ . This will gives $148c+2=27d$ . Then $d=\dfrac{148c+2}{27}=5c+\dfrac{13c+2}{27}$ is an integer. We know that for $c=4$ , $d$ is an integer. So we can write $c=27e+4$ , for integer $e$ . For $c=4$ , then $d=22$ , which gives $x=88$ . It's impossible. For $e>0$ , we will have $x>1000$ , which is also impossible.

Case 2b. For $a=27$ and $b=37$ . This will give $108c+2=37d$ . So $d=\dfrac{108c+2}{37}=3c+\dfrac{2-3c}{37}$ is an integer. For $c=13$ , then $d$ is an integer. So we can write $c=37e+13$ , for integer $e$ . If $c=13$ , then $d=38$ , which gives $x=494$ . If $e>0$ , then we will get $x>1000$ , which is impossible.

After all of case, we finally get $x=494$ . For $x=494$ , then $D=1405$ . So $y=\dfrac{-2*494+1+1405}{2}=209$ So $\overline{abc}=x=494$

Let $x= \overline{abc}$ and $y=\overline{def}$ . Then we have $(x+y)^2 = 1000x+y$ .

To simplify things further, let $k=x+y$ . This would give us $y=k-x$ . Note that $k$ should be in the range $0<k<1000$ , since $0<\overline{abcdef}=k^2<1000000$ .

Substituting $k=x+y$ and $y=k-x$ into the original equation, we arrive at $k^2 = 1000x+k-x$ or $k^2-k =999x$ .

And so we see that $k(k-1)$ is divisible by $999=27 \times 37$ .

Since $k$ and $k-1$ are relatively prime, there are four cases:

Case 1: $k$ is divisible by $999$ . The only possible value of $k$ is $999$ since $0<k<1000$ . Thus in this case $\overline{abcxyz}= k^2=998001$ , but $001$ is not a three-digit number, so this is invalid.

$\mbox{}$

Case 2: $k-1$ is divisible by $999$ . The only $k$ in the range $0<k<1000$ that satisfies this is $k=1$ , but this is certainly not valid.

$\mbox{}$

Case 3: $k-1$ is divisible by $37$ and $k$ is divisible by $27$ .

Thus $k \equiv 1 \pmod{37}$ and $k \equiv 0 \pmod{27}$ .

By the Chinese Remainder theorem, $k \equiv 297 \pmod{27\times 37}$ , or $k \equiv 297 \pmod{999}$ . The only value of $k$ in the range $0<k<1000$ is $297$ . Thus $\overline{abcdef} = 297^2 = 88209$ , but $088$ is not a three-digit number, so this is also invalid.

$\mbox{}$

Case 4: $k-1$ is divisible by $27$ and $k$ is divisible by $37$ .

Thus $k \equiv 1 \pmod{27}$ and $k \equiv 0 \pmod{37}$ .

By the Chinese Remainder theorem, $k \equiv 703 \pmod{27\times 37}$ , or $k \equiv 703 \pmod{999}$ . The only value of $k$ in the range $0<k<1000$ is $703$ . Thus $\overline{abcdef} = 703^2 = 494209$ . This satisfies the conditions of the question.

Hence, $\overline{abc} = \fbox{494}$

[This method requires some tedious calculation but it reduces the number of trials to a mere 37]

Since $\overline{abcdef}$ is a 5 digit perfect square, then $10^5 \leq \overline{abcdef} < 10^6$ . Or $316 < \sqrt{\overline{abcdef} } \leq 999$

Let $A = \overline{abc}$ , $B = \overline{def}$

Then essentially, we want to solve $(A + B)^2 = 1000A + B$ with constraints $100 \leq A, B \leq 999$ , $316 \leq A+ B \leq 999$

Take mod of 37 to both sides (knowing that $1000 \bmod{37} = 1$ ), we have $(A + B)^2 \equiv (A + B)$ . Or $A + B \equiv 0, \space 1 \pmod{37}$

For $A + B \equiv 0 \pmod{37}$ , the possible values of $A +B$ are $333,370,409, \ldots, 962, 999$

Similarly, for $A + B \equiv 1 \pmod{37}$ , the possible values of $A +B$ are $334,371,410, \ldots, 963$

With these values, we apply trial and error to $(A + B)^2 = 1000A + B$ gives $A + B = 703$ as the only possible solution

$\Rightarrow 703^2 = 494209 = (494 + 209)^2 \Rightarrow \overline{abc} = A = \boxed{494}$

$999=3^3·37$

• Modulo $3^3=27$ : $(x+y)^2\equiv x+y$ , so $x+y\equiv 0$ or $x+y\equiv 1$ .
• Modulo $37$ : $(x+y)^2\equiv x+y$ , so $x+y\equiv 0$ or $x+y\equiv 1$ .

So there's four possibilities. With Euclid's algorithm, we find $27·11-37·8=1$ , and so, modulo 999, $x+y$ must be:

As $(x+y)^2$ is a 6-digits number, $x+y<1000$ , so we only have to check $x+y$ in 1, 297, 703 and 999.

• $1^2=1$ , not a 6-digits number
• $297^2=88'209$ , not a 6-digits number
• $703^2=494'209$ , so $x=494$ and $y=209$
• $999^2=998'001$ , so $x=998$ and $y=1$ , but $y$ has to be a 3-digits number.

Only possibility left: $\boxed{x=494,\ y=209}$

I used the following C program to find such numbers;

include<stdio.h>

include<conio.h>

void main() { unsigned long i,x,y,rem[6]; clrscr(); for(x=317;x<=999;x++) { y=x x; for(i=0;i<6;i++) { rem[i]=y%10; y=y/10; } if((rem[0]+10 rem[1]+100 rem[2])+(rem[3]+10 rem[4]+100*rem[5])==x) printf("\n%lu",x); } getch(); } The output is 703 and 999. Now 999^2=998001 in which 001 is not of 3 digits. And 703^2=(494+209)^2=494209 Hence 703 is the result.

We let $\overline{abc}=x$ and $\overline{def}=y$ .

The equation becomes $(x+y)^2=1000x+y$ . We expand it to get $x^2+2xy+y^2=1000x+y$ . Treating it into a quadratic in $x$ makes it $x^2+(2y-1000)x+(y^2-y)=0$ . $x=\frac{-(2y-1000) \pm \sqrt{(2y-1000)^2-4(1)(y^2-y)}}{(2)(1)} =\frac{-(2y-1000) \pm \sqrt{4y^2-4000y+1000000-4y^2+4y}}{2}$ , which is equal to $500-y \pm \sqrt{250000-999y}$

Hence $250000-999y$ must be a perfect square. We let $250000-999y=k^2$ . So $500+k)(500-k)=999y$ . We let $500+k=p$ and $500-k=q$ , where k could be negative. $p+q=1000$ and their product is a multiple of $999$ . Hence it must be a multiple of both $27$ and $37$ .

Since $1000$ is not a multiple of $3$ , not both $p$ and $q$ could be multiples of $3$ . Thus the multiple of 37 is not a multiple of 3, and the number that is not the multiple of 37 is a multiple of 27.

WLOG let $p$ be the multiple of 37. Then we let $p=37m$ and $q=27n$ . $37m+27n=1000$ . Reducing $mod 37$ , we know that $27n$ is congruent to $1 mod 37$ .

By Guess and Check, we know that $n=297$ , as $n$ is at most $999$ .

Hence $k=203$ , and $y=\frac{(703)(297)}{999}=209$ , and $x=291 \pm \sqrt{250000-999(209)}=291 \pm 203$ . Since $x$ is a 3 digit number, $x=291+203=494$ .

Since $x=\overline{abc}$ , $\overline{abc}=\boxed{494}$

Since there hasn't been programming questions for a while I decided to solve this one with Python using the code below. I defined $\overline{abc}$ as $x$ and $\overline{def}$ as $y$ . Since we know that they are three digit numbers we can make the range between $100$ and $999$ . In the third line of the code I defined $\overline{abcdef}$ as $(1000x) + y$ as they are equivalent statements. If the equality holds the the next line of code is run which prints the value of $x$ . This is the solution to the problem as $x = \overline{abc}$ .

    for x in range(100, 1000):

        for y in range(100, 1000):

            if (x+y)**2 == (1000*x)+y:

                        print(x)

When the code is executed the output for $x$ is $494$ .

step 1: go to this site Online Python Interpreter

step 2: copy paste this code, then run

for a in range(1,10):

>for b in range(0,10):

>>for c in range(0,10):

>>>for d in range(0,10):

>>>>for e in range(0,10):

>>>>>for f in range(0,10):

>>>>>>_abc=100*a+10*b+c

>>>>>>_def=100*d+10*e+f

>>>>>>_abcdef=100000*a+10000*b+1000*c+100*d+10*e+f

>>>>>>if ( abc+ def)**2 == _abcdef:

>>>>>>>print _abc

>>>>>>>break

Output: 494