Finding the largest 3-digit number $\; \overline{abc}\;$ s.t $\; \overline{abc}=100a+10b+c \equiv a+b^2+c^3$

2,727

For the first part:

$$100a+10b+c = a+b^2+c^3$$ implies that $$99a + b(10-b)=(c-1)c(c+1)$$

Now, $b(10-b)\leq 25$, with $b(10-b)=0,9,16,21,24,25$. So you want $c$ so that $$(c-1)c(c+1)\equiv 0,9,16,21,24,25\pmod{99}.$$

Now, $(c-1)c(c+1)$ is divisible by $3$, as is $99$, so you really need $$c^3-c\equiv 0,9,21,24\pmod {99}, \,c^3-c\geq 100$$.

This can be done by trial and error.

We know $c>4$. (If $c\leq 4$ then we get $a=0$, which you might want to include.)

$c=5$ gives $c^3-c\equiv 21\pmod {99}$. So $\overline{abc}=135, 175$ are solutions.

$6^3-6\equiv 12\pmod{99}$, so $c\neq 9$.

$7^3-7\equiv 39\pmod{99}$, so $c\neq 7$.

$8^3-8\equiv 9\pmod{99}$ so $(a,b,c)=(5,1,8)$ and $(a,b,c)=(5,9,8)$ are solutions.

$9^3-9\equiv 27\pmod{99}$. So $c\neq 9$.

This yields the four solutions: $135, 175, 518, 598$.

(Allowing $a=0$, we get additional answers $\overline{abc}=000,001,043,063$.)

A start on the bonus question.

For $n$ digits, the largest sum $a_1+a_2^2+\cdots a_2^n$ is $9+9^2+\cdots +9^n = \frac{9}{8}(9^n-1)$. The smallest $n$-digit number is $10^{n-1}$.

So if $10^{n-1}>\frac{9}{8}(9^n-1)$, you can be sure there is no solution. But $$\frac{9}{8}(9^n-1)< \frac{9^{n+1}}{8},$$ so if $10^{n-1}>9^{n+1}/8,$ there is no solution. This is true if $(n-1)\log_9(10)>(n+1)-\log_9 8$ or $$\log_9(10) > 1+\frac{2-\log_9 8}{n-1}\\\log_9(10/9)>\frac {2-\log_9 8}{n-1}$$ or $$n>1+(2-\log_9(8))\log_{10/9}(9)\approx 22.97$$

So there are no solutions with more than 22 digits. That limits it to a finite, if large, problem.

Share:
2,727

Related videos on Youtube

K. Rmth
Author by

K. Rmth

Updated on February 23, 2020

Comments

  • K. Rmth
    K. Rmth over 3 years

    This question comes from a maths contest (infer no calculators or other electronic calculating aids) for 14-16 year olds (infer no use of complicated theorems, but those accessible to high-school students).

    What is the largest three-digit number with the property that the number is equal to the sum of its hundreds digit, the square of its tens digit and the cube of its units digit?

    I cannot find a manual way of doing this, but a little programming got me 4 numbers having such property; namely $135,175,518 \;\text{and}\;598$. Any pointers on how to tackle this problem?


    $\mathbf{Bonus}$:

    Can the above methodology for 3-digit numbers be extended to higher numbers? Using the same program gave the following numbers: $1306,1676,\;\text{and}\;2427$ for 4-digit numbers and there are no equivalent 5-digit and 6-digit numbers. I used the following code:

    store=zeros (1);

    for a=1:9
       for b=0:9
           for c=0:9
               num=a+b^2+c^3;
               dig=(100*a)+(10*b)+(c);
                   if num==dig
                       store=[store dig];
                   end
           end
       end
    

    end store