Problem MA78 - Crux Mathematicorum Vol. 46, No. 6 , June, 2020

 

 

 ANSWER CiP : $n=667$

 Verification: $T(667)+3 \cdot 667=6+6+7+3\cdot 667=19+2001=2020$

SOLUTION CiP

Let $n=\overline{abc}$, $a\neq 0$; the given equation is written

$$a+b+c+3\cdot (100a+10b+c)=2020$$

$$\Leftrightarrow a+b+c+30(10a+b)+3c=2020$$

where do we get

(1)$$30\cdot \overline{ab}=2020-a-b-4c.$$

From equation (1) we see that $30\mid2020-a-b-4c$ and because $a, b, c$  are digits in base ten, we have $a+b+4c\geq 9+9+36=54$. Then the right member of (1) is $\geq 1966$ so its values can only be 2010 or 1980. Also from 1 we get

(2)$$\overline{ab}=\frac{2020-a-b-4c}{30} .$$

If $a+b+4c=10$ then $\overline{ab}=\frac{2010}{30}=67$ but $a=6,b=7$ they cannot check the condition $a+b+4c=10$.

If $a+b+4c=40$ then $\overline{ab}=\frac{1980}{30}=66$ so $a=6,b=6$ and $c=\frac{40-6-6}{4}=7$ which is the answer.

$\blacksquare$

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Added February 8, 2021

             Good answer  see V47n01, page 8-9

              They write in their solution that $T(n)=2020-3n$ and since for a three-digit number $n$

$1 \leq T(n) \leq27$

$\Leftrightarrow \; 1 \leq 2020-3n \leq27$

.......$\Leftrightarrow \;\;665 \leq n \leq 673$.

           On the other hand $n \equiv T(n) \;(mod\;3)$ and $T(n)+3n=2020 \; \Rightarrow$

$\Rightarrow \;T(n) \equiv 2020\;(mod\;3)$ $\Rightarrow \;T(n) \equiv 1\;(mod\;3)\;\;\Rightarrow \;n \equiv 1 \; (mod\;3)\;....\;\Rightarrow \;3 \mid (n-1)$.

          From the above two observations, we know that, if such $n$ is possible then it must be either $667,\;670$ or $673$, and we check each possibility ....

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