2
$\begingroup$

It is known that $\overline{abcd}-\overline{dcba}=x$ and $a^3+b^3+c^3+d^3=x$. Find $a+b+c+d$.

From the given we get $999a+90b-90c-999d = a^3+b^3+c^3+d^3$. We could just test out the possibilities, but is there another way to find the sum of the digits?

  • 0
    use that $9$ is a divisor $$a^3+b^3+c^3+d^3$$ – 2017-01-20
  • 0
    we do?${}{}{}{}$ – 2017-01-20
  • 0
    Are we to assume that $a,d $ are nonzero so that the apparent four digit numbers actually are so? Is it required that all four digits are distinct (since they are labelled distinctly)? – 2017-01-20
  • 0
    @hardmath $a$ must be nonzero and so must $d$, since they are leading digits of four-digit numbers. – 2017-01-20
  • 0
    Also, $a\ge d$ and $b\ge c$. Note that as $4\cdot 9^3=2819$, and $90\cdot 9=810$, the difference is only $2106$. Comparing that with $999\cdot 3=2997$ and $999\cdot 2=1998$, we have $0\le a-d\le 2$. – 2017-01-20
  • 1
    @hypergeometric: While $x\gt 0$, how do we know $b\ge c$ (in particular if $a \gt d$)? – 2017-01-20
  • 0
    @hypergeometric Why is $b\ge c$? I don't think that has to be true. – 2017-01-21
  • 0
    @hardmath Not necessary for $b\ge c$ unless $a=d=0$. – 2017-01-21
  • 0
    @IanMiller Not necessary for $b\ge c$ unless $a=d=0$. – 2017-01-21

1 Answers 1

1

Partial Answer

As $x=999a+90b-90c-999d=9(111a+10b-10c-111d)$ then $x\equiv0\pmod{9}$. Looking at cubes modulo 9 we see:

$0^3\equiv0\pmod{9}$, $1^3\equiv1\pmod{9}$, $2^3\equiv8\pmod{9}$, $3^3\equiv0\pmod{9}$, $4^3\equiv1\pmod{9}$, $5^3\equiv8\pmod{9}$, $6^3\equiv0\pmod{9}$, $7^3\equiv1\pmod{9}$, $8^3\equiv8\pmod{9}$, $9^3\equiv0\pmod{9}$.

For the four cubes to be a multiple of $9$ we require:

  • 4 zeros $\big((a,b,c,d)\equiv(0,0,0,0)\pmod{3}\big)$, or
  • 2 zeros, a one and an eight $\big((a,b,c,d)\equiv(0,0,1,2)\pmod{3}\big)$, or
  • 2 ones, 2 eights $\big((a,b,c,d)\equiv(1,1,2,2)\pmod{3}\big)$

We have $999(a-d) = a^3+b^3+c^3+d^3+90(c-b)$. Next consider the maximum value this can take. This occurs when $(a,b,c,d)=(9,0,9,9)$. The right hand side in this case is equal $2997$ (with equality only if $a=d$). Thus $0\le a-d\le2$.

I'll keep adding to this...

Cheating

Via computer the answer is $a=4$, $b=6$, $c=8$, $d=3$. I'll keep thinking about how to get here without cheating. :)

  • 0
    Good start. Also, $a\ge d$ and $b\ge c$. – 2017-01-20
  • 1
    In fact maximizing $b^3+c^3+90(c-b)$ in the permitted range occurs at $c=9,b=0$ leading to $a^3+b^3+c^3+d^3+90(c-b)\le 2997$, but since this requires $a=d=9$ we must have $0\le a-d\le2$, eliminating the $4$ zeroes option. – 2017-01-21
  • 0
    @Joffan I'm not seeing how the 4 zero option is eliminated from that statement. – 2017-01-21
  • 0
    Yeah maybe it needs a few more steps.to eliminate the $a=d$ option. – 2017-01-21