All three-digit numbers that satisfy are of the form $\overline{abc} = 100a + 10b + c = 543 + a^2 + b^2 + c^2,$ with $a,b,c$ being three single-digit (non-negative) numbers. Rearrangement yields $a(100-a) = b(b-10) + c(c-1) + 543.$ We know that $\begin{aligned} -25 &\leq b(b-10) \leq 0, \\ 0 &\leq c(c-1) \leq 72, \end{aligned}$ so $-25 \leq b(b-10) + c(c-1) \leq 72$ and equivalently $518 \leq b(b-10) + c(c-1) + 543 \leq 615$ or $518 \leq a(100-a) \leq 615 \implies a = 6.$ Then $\overline{abc} = 100a + 10b + c = 600 + 10b + c = 543 + 36 + b^2 + c^2,$ implying $10b + c + 21 = b^2 + c^2 \implies c(c-1) = b(10-b) + 21.$ By plugging in different values of $b$ , it is easily verifiable that $(a,b,c) \in \{ (6,1,6), (6,9,6), (6,3,7), (6,7,7) \}.$ The sum of the corresponding numbers is $616 + 696 + 637 + 677 = 2626$ and hence the answer is $\boxed{2626}$ .

Let $\overline{abc}$ denote such three-digit number, where $a,b,c \in \{0,1,\dots,9\}$ and $a\neq 0.$ The required condition is $100a+10b+c=\overline{abc} = 543 + a^2+b^2+c^2.\quad\quad (*)$ Since $1 \le a^2+b^2+c^2 \le 9^2+9^2+9^2 = 243,$ it follows that $5 \le a \le 7$ . Consider $3$ cases:

$\textbf{case 1}$ : If $a= 5,$ then condition $(*)$ becomes $(10b-b^2)+c(1-c) = 68.$ Since $10b-b^2 \le \frac{4(-1)(0)-100}{4(-1)}=25$ and $c(1-c) \le 0$ , it follows that this equation has no solution. Hence $a \neq 5.$

$\textbf{case 2}$ : If $a= 6,$ then condition $(*)$ becomes $b^2-10b+(c^2-c-21) = 0.$ This is a quadratic equation in $b$ with an integer solution, namely $b$ , so the discriminant must be a perfect square. That is $100-4(c^2-c-21) = 4(46-(c-1)c)$ is a perfect square. Since $4$ is a perfect square, we have $46-(c-1)c$ is a perfect square. It can be easily checked that this occurs when $c=6$ or $c=7$ . Consider the following subcases:

$\textbf{case 2.1}$ : If $c=6$ , then by the quadratic formula, we have $b = \frac{10\pm \sqrt{4(46-(c-1)c)}}{2}=\frac{10 \pm 2(4)}{2} = 1,\,9.$ So for this case we have $\overline{abc} = \boxed{616} \quad \text{or} \quad \boxed{696}.$

$\textbf{case 2.2}$ : If $c=7$ , then by the quadratic formula, we have $b = \frac{10\pm \sqrt{4(46-(c-1)c)}}{2}=\frac{10 \pm 2(2)}{2} =3,\, 7.$ So for this case we have $\overline{abc} = \boxed{637} \quad \text{or} \quad \boxed{677}.$

$\textbf{case 3}$ : If $a=7,$ then condition $(*)$ becomes $b^2-10b+(c^2-c-108) = 0.$ This is a quadratic equation in $b$ with an integer solution, namely $b$ , so the discriminant must be a perfect square. That is $100-4(c^2-c-108) = 4(133-(c-1)c)$ is a perfect square. Since $4$ is a perfect square, we have $133-(c-1)c$ is a perfect square. It can be easily checked that this occurs when $c=4.$ By the quadratic formula, we have $b = \frac{10\pm \sqrt{4(133-(c-1)c)}}{2}=\frac{10 \pm 2(11)}{2} =-6,\, 16,$ a contradiction to the assumption that $b$ is a single digit. Hence $a \neq 7.$

It can be easily checked that the numbers $616,\, 696,\, 637$ and $677$ satisfy the condition of the question. Their sum is $616+696+637+677 = 2626,$ having $\boxed{626}$ as the last three digits.

Any 3 digit number may be expressed in the form $100a+10b+c$ where $a$ is the hundreds digit, $b$ is the tens digit and $c$ is the units digit. We can then apply this to give the equation $100a+10b+c-543=a^2+b^2+c^2$ . Rearranging and completing the square gives: $(a-50)^2+(b-5)^2+(c-0.5)^2=1982.25$ .

Claim: $a=6$

Proof:

If $a \leq 5$ , $(a-50)^2 \geq 2025$ $\therefore (b-5)^2+(c-0.5)^2 \leq -42.75$ , but the sum of squares of real numbers can't be negative, we've reached a contradiction, so $a \geq 6$ .

The maximum value of $(b-5)^2$ is $16$ (when $b=1$ or $b=9$ ), the maximum value of $(c-0.5)^2$ is $72.25$ (when $c=9$ ), $\therefore (b-5)^2+(c-0.5)^2 \leq 88.25 \therefore (a-50)^2 \geq 1894$ . But if $a=7, 8$ or $9$ , $(a-50)^2 \leq 1849 < 1894$ , so $a \leq 6$ .

$a \geq 6$ and $a \leq 6 \therefore a=6$ .

$a=6 \Rightarrow (b-5)^2+(c-0.5)^2=46.25$ . Also; $(b-5)^2 \geq 0$ and as discussed earlier, $(b-5)^2 \leq 16$ $\therefore 30.25 \leq (c-0.5)^2 \leq 46.25$ $\therefore c=6$ or $c=7$ .

We can now substitute in $c=6$ and $c=7$ into $(b-5)^2+(c-0.5)^2=46.25$ and solve for $b$ . For $c=6$ , $b=1$ or $9$ . For $c=7$ , $b=3$ or $7$ . Substituting these solutions back into the original equation, we can verify that they all work, hence our solutions are: $616, 696, 637, 677$ , adding these up gives $2626$ , so the last three digits are $626$ .

We let the digits in the hundreds', tens' and ones' place be $a$ , $b$ and $c$ respectively.

Then we will have the following equation: $a^2 + b^2 + c^2 + 543 = 100a + 10b + c$ .

Since $0 \leq a, b, c \leq 9$ , then we have $543 \leq a^2 + b^2 + c^2 + 543 = 100a + 10b + c \leq 543 + 3 \times 9^2 = 786$ . Thus, $a = 5,$ $6$ or $7$ .

Consider $a = 5$ . Then $25 + b^2 + c^2 + 543 = 500 + 10b + c \Rightarrow 68 + b^2 + c^2 = 10b + c$ . Since $68 + b^2 + c^2 \geq 68$ , then $10b + c \geq 68$ and so $b \geq 6$ . With this, we have $10b + c = 68 + b^2 + c^2 \geq 68 + 36 + c^2 = 104 + c^2 \geq 104$ . But $10b + c \leq 10 \times 9 + 9 = 99 < 104$ , so we arrive at a contradiction. Thus $a$ cannot be $5$ .

Next consider $a = 7$ . Then $49 + b^2 + c^2 + 543 = 700 + 10b + c \Rightarrow b^2 + c^2 = 10b + c + 108$ . Since $10b + c + 108 \geq 108$ , then $b^2 + c^2 \geq 108$ and so $b^2 \geq 108 - c^2 \geq 108 - 9^2 = 27$ . This means $b \geq 6$ . With this, we have $b^2 + c^2 = 10b + c + 108 \geq 60 + c + 108 = 168 + c \geq 168$ . But $b^2 + c^2 \leq 9^2 + 9^2 = 162 < 168$ , so we arrive at a contradiction. Thus $a$ cannot be $7$ .

Next consider $a = 6$ . Then $36 + b^2 + c^2 + 543 = 600 + 10b + c \Rightarrow c^2 - c + (b^2 - 10b - 21) = 0$ . So $c = \frac{1 \pm \sqrt{1 - 4(b^2 - 10b - 21)}}{2} = \frac{1 \pm \sqrt{185 - 4(b - 5)^2}}{2}$ . Now we substitute $b = 0, 1, \ldots , 9$ , keeping in mind that $c$ must be a non-negative integer. The only solutions are $(b,c) = (1,6), (3,7), (7,7), (9,6)$ .

Therefore, all the three-digit numbers which satisfy the criteria in the question are $616, 637, 677, 696$ . Their sum is $2626$ , so the answer is $\underline{626}$ .

Let the three digit number be $N=\overline{abc}$ , such that the hundreds, tens, and units digit of $N$ are $a, b,$ and $c$ , respectively. Note that the maximum value of the sum of the squares of the digits of $N$ is $3\cdot9^2=243$ , so $\overline{abc}$ is at most $243+543=786.$ Similarly, the sum or the squares of the digits must be positive, so the minimum value of $N$ is $544$ . Thus, $544 \leq N \leq 786.$ Now, by the condition in the problem, we have $100a+10b+c-(a^2+b^2+c^2)=543$ $a(100-a)+b(10-b)+c(1-c)=543.$ By our earlier bounds on $N$ , $a=5,6,$ or $7.$ Also, note that by completing the square $b(10-b)=-(b+5)^2+25$ . Combining this with $0 \leq b \leq 9$ gives $0 \leq b(10-b) \leq 25.$ Similarly, we get $c(1-c)=-(c-\frac{1}{2})^2+\frac{1}{4}.$ Combining this with $0 \leq c \leq 9$ gives $-72 \leq c(1-c) \leq 0.$ Now, we proceed with casework on the value of $a$ .

Case 1: $a=5$

We substitute $a=5$ into our original equation and rearrange to get $b(10-b)+c(1-c)=68.$ By our bounds, the maximum value of the LHS is $25+0=25.$ Thus, this case yields no solutions.

Case 2: $a=6$

We substitute $a=6$ into our original equation and rearrange to get $b(10-b)+c(1-c)=-21.$ Note that either $c$ or $(1-c)$ is even when $c$ is an integer, so $c(1-c)$ is even. Since the value of the RHS is odd, $b(10-b)$ must be odd. Similarly, we see that $b(10-b)$ is even when $b$ is even and is odd when $b$ is odd. Thus, we only have to test odd values of $b$ , where $0 \leq b \leq 9$ . When $b=1$ , we find that $9+c(1-c)=-21 \Rightarrow$ $c(1-c)=-30$ . By solving the quadratic or guessing and checking values, we find that the only value of $c$ where $0 \leq c \leq 9$ occurs when $c=6$ . This gives the solution $N=616$ . Continuing in this way, we the test $b=3, 5, 7, 9$ to find the solutions $N=637, 677, 696.$

Case 3: $a=7$

We substitute $a=7$ into our original equation and rearrange to get $b(10-b)+c(1-c)=-108.$ However, by our earlier bounds, the minimum value of the LHS is $0+(-72)=-72$ , so this case yields no solutions.

Finally, combining our cases, the only values of $N$ that satisfy the conditions of the problem are $616, 637, 677, 696.$ Thus we have an answer of $616+637+677+696 \equiv \boxed{626} \pmod{1000}.$

All numbers in the set are in the form 100x+10y+z=x^{2}+y^{2}+z^{2}+543, for the digits x, y and z which satisfy this equation.

Rewriting this equation, it becomes x(x-100)+y(y-10)+z(z-1)=543

Defining x' as x-50, y' as y-5, and z' as z-0.5, we can rewrite the left hand side of this equation as (x'-50)(x'+50)+(y'-5)(y'+5)+(z'-0.5)(z'+0.5). Simplified, that equals to (x'^2-50^{2})+(y'^2-5^{2})+(z'^2-0.5^{2}).

Thus, we have that x'^{2} +y'^{2} + z'^{2} = -543+50^{2}+5^{2}+ 0.5^{2} =1982.25

Note that the minimum value of y'^{2} is (5-5)^{2}=0 and the maximum value of y'^{2} is (0-5)^{2}=25, while the minimum value of z'^{2} is (0-0.5)^{2}=0.25 and the maximum value of z'^{2} is (9-0.25)^{2}=72.25.

Thus, y'^{2}+z'^{2} lies between 0+0.25 =0.25 and 25+72.25=97.25 .

This means |x-50|=x' lies between sqrt{1982.25-(y'^2+z'^{2})} and sqrt{1982.25-97.25} =43.4... and sqrt{1982.25-0.25} =44.5...

As this must be an integer, we find that |x-50| =44, which means x must be 50-44=6 .

This simplifies our equation to y'^{2}+z'^{2}=1982.25-44^{2}=46.25.

As before, keeping in mind the minimum and maximum values of y'^{2}, observe that |z-0.5|=|z'|=sqrt{46.25-y'^{2}} lies between sqrt{46.25-25}=4.6.... and sqrt{46.25-0}=6.8.... This means z lies between 0.5+4.6...=5.1 and 0.5+6.8...=7.3..., which tells us z must be either 6 or 7.

Finally, plugging those in, and calculating y=5+sqrt{46.25-(z-0.5)^{2}} and y=5-sqrt{46.25-(z-0.5)^{2}}, we get our four solutions: 616, 696, 637, and 677.

Adding these together and taking the last three digits, our final answer is 626.

Let x,y and z be the digits of the three-digit number. According to the question, 100x+10y+z-(x^2+y^2+z^2)=543 100x+10y+z-x^2-y^2-z^2=543 Thus,x (100-x)+y (10-y)+z (1-z)=0 Substituting x for numbers from 1 to 9(Since x is the first digit and thus cannot be 0),we get x (100-x)=99,196,291,384,475,564,651,736,819=a. Similarly,substituting y and z for numbers from 0 to 9, y (10-y)=0,9,16,21,24,25,24,21,16,9=b and z (1-z)=0,0,-2,-6,-12,-20,-30,-42,-56,-72=c If a=475, 475+b+c will be less than 543 as the highest value for 'b' is 25 and the value of 'c' is either 0 or negative. 475+25=500<543 Thus 'a' should be greater than 475. If a=651, 651+b+c will be more than 543 as the lowest value of 'c' is-72 and the value of 'b' is either 0 or positive. 651+-72=579>543 Thus 'a' should be less than 651. From the above two situations,we get,a=564 If c=-20, 564+b+-20=544+b will be more than 543 as 'b' is either 0 or positive. Thus 'c' should be less than -20. If c=-56, 564+-56+b=508+b will be less than 543 as the greatest value of 'b' is 25 and 508+25=533< 543. Thus 'c' should be more than -56. From the above two situations ,'c' can be -30 or -42. If c=-30, 564+-30+b=543 Then b=9. Then x=6,z=6 and y=1 or 9. Thus the numbers are 616 and 696. If c=-42, 564+-42+b=543 Then b=21. Thus x=6,z=7 and y=3 or 7. Thus the numbers are 637 and 677. Therefore, the required numbers are 616,696,637 and 677. 616+696+637+677=2626. Thus,the last three digits of their sum are 626.

let the 3digit numbers be of the form asd 10>a,s,d & a>0 &a,s,d are whole numbers then 100 a+10 s+d=a^2+s^2+d^2+543
implies a>5 minimum value of s (10-s) is 0 ........1 maximum value of d (d-1) is 72 .........2
maximum value of a^2,d^2,s^2 is 81 ........3 maximum value of s (10-s) is 25 ........4 case-1 (a=6) 600+s (10-s)=d (d-1)+6^2+543 s (10-s)=d (d-1)-21 from 1&4 46>d (d-1)>20 implies 8>d>5 trying d=7,6 when d=6,s=1,9 when d=7,s=3,7 therefore the numbers are 616,696,677,637 case-2 (a=7) 700+s (10-s)=d (d-1)+7^2+543 s (10-s)=d (d-1)-108 from 2&1 therefore 3digit number starting in 7 is not possible as a increases from 1 to 50 100^a-a^2 = a (100-a)also increases(by inverse of am-gm inequality) therefore s (10-s)=d*(d-1)-K where K>108 therefore no other number exist

let the number be of the form ' abc ';

where a,b,c can be any digit from 0-9 ;

or in other words it can be written as '100a+10b+c';

the difference of the number and sum of its digits' square is 543 or

100a-a^2+10b-b^2+c-c^2=543;

=> a(100-a)+b(10-b)+c(1-c)=543;

possible values of c(1-c) are :

c=0 ; c(1-c)=0;

c=1 ; c(1-c)=0;

c=2 ; c(1-c)=(-2);

c=3 ; c(1-c)=(-6);

c=4 ; c(1-c)=(-12);

c=5 ; c(1-c)=(-20);

c=6 ; c(1-c)=(-30);

c=7 ; c(1-c)=(-42);

c=8 ; c(1-c)=(-56);

c=9 ; c(1-c)=(-72);

similarly posibble values of b(10-b) are :

b=0 ;b(10-b)=0;

b=1 ;b(10-b)=9;

b=2 ;b(10-b)=16;

b=3 ;b(10-b)=21;

b=4 ;b(10-b)=24;

b=5 ;b(10-b)=25;

b=6 ;b(10-b)=24;

b=7 ;b(10-b)=21;

b=8 ;b(10-b)=16;

b=9 ;b(10-b)=9;

the number has to be greater than 543; so 'a' can be 5,6,7,8,9; => a(100-a) can be 475,564,651,736,819;

since all possible values of c(1-c) are 0 or negative ; =>b(10-b)+c(1-c) can have minimum value of (-72);

=>a(100-a) cannot be greater than 564;

example: suppose a(100-a)=651;

so the equation becomes;

651+b(10-b)+c(1-c)=543;

=>108+b(10-b)+c(1-c)=0 this not possible because 'b(10-b)+c(1-c)' has minimum value =(-72) and (108-72) not equal to 0;

let a(100-a)=475;

=>475+b(10-b)+c(1-c)=543;

=>b(10-b)+c(1-c)=68;

but this is also not possible as b(10-b)+c(1-c) has maximum value 25;

so only possible value of a(100-a)=564;

now the equation becomes;

564+b(10-b)+c(1-c)=543;

=>21+b(10-b)+c(1-c)=0;

checking for possible values of b(10-b) and c(1-c);

we find that b(10-b) can have value equal to 9 or 21;

and corresponding values of c(1-c) are (-30) and (-42);

=>for b=1 or 9 we have b(10-b)=9 and c(1-c)=(-30) for c=6;

so number can be;

616 or 696;

and for b(10-b)=21 ; b=3 or 7; and c(1-c)=(-42) for c=7;

so number can be ;

637 or 677;

sum of numbers =616+637+677+696= 2626;

=> last three digits are : 626 ;

Let $N= \overline{abc}$ and $a \neq 0, 0 \leq a,b,c \leq 9$

According to the problem we have: $a^2+b^2+c^2+543=100a+10b+c$

Since I can't really find a proper way to factor this, I will complete the square of the individual variables. Thus, we can rewrite the above as:

$(a-50)^2 + (b-5)^2 + \frac{1}{4}(2c-1)^2 =(2500+25+\frac{1}{4}) - 543=1982+\frac{1}{4}$ $(2a-100)^2 + (2b-10)^2 + (2c-1)^2=7929$

Now $(2a-100)^2 <=7929$ and, $(2a-100)^2 <={89}^2$ . The max value of the other two expressions can be $100$ and ${17}^2$ respectively. So $(2a-100)^2 \geq 7929-389 > {86}^2$ . Combining the above gives us that ${86}^2 < (2a-100)^2 \leq {89}^2$ Thus, $a=6$ is the only possible value since $(2a-100)^2$ is an even perfect square and hence can only equal ${88}^2$ .

The new equation transforms to: $(2b-10)^2 + (2c-1)^2=7929-{88}^2= 185$

$185= {13}^2 + {4}^2 ={11}^2+{8}^2$

Keep in mind that $(2b-10)^2$ is an even perfect square.

Case 1: $(2b-10)^2=4^2= \pm 4 \Rightarrow 2b=14,6 \Rightarrow b=7,3$ $(2c-1)^2={13}^2=\pm1 \Rightarrow 2c=14, -12 \Rightarrow c=7$

Possible numbers are : $677, 637$

Case 2: $(2b-10)^2=8^2=\pm 8 \Rightarrow 2b=18,2 \Rightarrow b=9,1$ $(2c-1)^2=11^2=\pm 11 \Rightarrow 2c=12, -10 \Rightarrow c=6,$

Possible numbers are : $696, 616$

Sum of all possible numbers is $=2696$ The next part is trivial.

Because 9^2+9^2+9^2+543=786. The only possible first digit numbers are 5, 6, or 7. We get the equation a^2+b^2+c^2+543=100a+10b+c. When you plug in a=5, you find that there are no solutions. For a=6, you (1,6), (3,7), (7,7), and (9,6). For a=7, you also find no solutions. So the numbers are 616+637+677+696=2626

10 weeks ago August 12 to 18 Level 4 Problem 3 was the same problem!

what i've been through

${ a }^{ 2 }$ + ${ b }^{ 2 }$ + ${ c }^{ 2 }$ + ${ 543 }$ = $100$ ${ a }$ + $10$ ${ b }$ + ${ c }$

The maximum value of sum of three squares can be 243. Thus the maximum value of the number can be 786. Minimum of course is 544. This implies can ${ a }$ can only be 5, 6 or 7.

1. For a = 5

${ b }^{ 2 }$ + ${ c }^{ 2 }$ + ${ 68 }$ = $10$ ${ b }$ + ${ c }$

The RHS is a two digit number greater than 68. It's obvious that no value of b & c will satisfy the equation.

1. For a = 7

${ b }^{ 2 }$ + ${ c }^{ 2 }$ - ${ 108 }$ = $10$ ${ b }$ + ${ c }$

For the LHS to be positive the only combinations of b,c that work are (9,9), (9,8), (8,8) & (9,7). No pair qualifies.

1. For a = 6

${ b }^{ 2 }$ + ${ c }^{ 2 }$ - ${ 21 }$ = $10$ ${ b }$ + ${ c }$

${ (b-5) }^{ 2 }$ + ${ c }^{ 2 }$ = ${ 46 }$ + ${ c }$

${ (b-5) }^{ 2 }$ can only be 0,1,4,9,16. So ${ c }^{ 2 }$ - ${ c }$ can only be 46, 45, 42, 37, 30. Clearly 42 and 30 qualify resulting in ${ c }$ as either 7 or 6. Solve for ${ b }$ and we have 4 numbers.

${ 677 }$ , ${ 637 }$ , ${ 696 }$ , ${ 616 }$ that adds up to ${ 2626 }$

L.H.S.=a^2+b^2+c^2+543 = abc=R.H.S.~~~~abc=100a+10b+c.\\ R.H.S.>543.~~~R.H.S.=543\\ L.H.S.=543+25+16+9=593.\\ But ~if~ L.H.S.>R.H.S.~~uptill~ 100a+10b+9~~\text{there can be no solution.}\\ So ~uptill~549~there~is~no~solution.\\ \text{There is a solution in the rage c=0 to 9 only if YY0-XXX is as given below.}\\ \text{the solution is XXc. \\ ~~~~~Let~~~L.H.S.=XXX~~~~~~~~~~~~R.H.S.=YY0\\ c=0~~~~YY0~-~XXX= \color{#D61F06}{ 0 } ~~~~~~~~~~~R.H.S.=YY0\\ c=1~~~~YY0~-~XXX= \color{#D61F06}{ 0 }~~~~~~~~~~~~R.H.S.=YY1\\ c=2~~~~YY0~-~XXX= \color{#D61F06}{ 2}~~~~~~~~~~~~R.H.S.=YY2\\ c=3~~~~YY0~-~XXX= \color{#D61F06}{ 6|~~~~~~~~~~~~R.H.S.=YY3\\ c=4~~~~YY0~-~XXX= \color{#D61F06}{12}~~~~~~~~~~~~R.H.S.=YY4\\ c=5~~~~YY0~-~XXX= \color{#D61F06}{20}~~~~~~~~~~~~R.H.S.=YY5\\ c=6~~~~YY0~-~XXX= \color{#D61F06}{30}~~~~~~~~~~~~R.H.S.=YY6\\ c=7~~~~YY0~-~XXX= \color{#D61F06}{42}~~~~~~~~~~~~R.H.S.=YY7\\ c=8~~~~YY0~-~XXX= \color{#D61F06}{56}~~~~~~~~~~~~R.H.S.=YY8\\ c=9~~~~YY0~-~XXX=\color{#D61F06}{72}~~~~~~~~~~~~R.H.S.=YY9\\

javascript version

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function sqdigit(n)
{
    var s=0;
    while(n)
    {
        s+=(n%10)*(n%10);
        n=parseInt(n/=10);
    }

    return s;
}

var result=0;
for(var i=100;i<1000;i++)
    if(i-sqdigit(i)==543)
        result+=i;

console.log(result);        

sum=0;

for i=1:999

a=(rem(i,10))^2;

a=a+rem(floor(i/10),10)^2;

a=a+(floor(i/100))^2;

if (i-a==543)

sum=sum+i;

end

end

ans =

    2626

include <stdio.h>

int main() { int a=0, b=0, c=0, sum=0;

for(a=0; a<10; a++)
{
    for(b=0; b<10; b++)
    {
        for(c=0; c<10; c++)
            if( 100*a + 10*b + c == 543 + a*a + b*b + c*c)
            sum = sum + 100*a + 10*b + c;
    }
}
printf("%d", sum);

}

Let x=abc the three digits number. Then x= $a^{2}$ + $b^{2}$ + $c^{2}$ +543 and x=100a+10b+c; we group the terms: $b^{2}$ -10b + $c^{2}$ -c = 100a - $a^{2}$ - 543, or $(b-5)^{2}$ + c(c-1) = 100a - $a^{2}$ - 543 + 25. Number 'a' could take the values 9 to 5. Only for a=7 we obtain solutions: $(b-5)^{2}$ + c(c-1)=83 ; $(b-5)^{2}$ could be 25, 16, 9, 4, 1 or 0 and c(c-1) could be 72, 56, 42, 30, 20, 12, 6, 2, 0. For c=6, b=1 or b=9; for c=7, b=3 or b=7. The thrilling numbers are 616, 696, 637, 677 with their sum 2628. Last three digits ... 626 :)