$A,B,C$ , Easy As $1,2,3$ (Do Re Mi)

Number Theory Level 3

The number $\overline{ABC\dots XYZ}$ consists of 26 distinct variables, consisting of positive digits, all in alphabetical order. When it is multiplied by $4$ , the result is $\overline{ZYX\dots CBA}$ , where the digits of the original number are all reversed.

It is possible for $21999\dots 9978 \cdot 4 = 87999\dots 99912$ . Does there exist another possible expression for $\overline{ABC\dots XYZ}$ ?

Inspiration: (1) , (2) , (3)

Bonus: Prove or disprove there exist more than one distinct expression.

Yes No

2 solutions

David Vreken
Jan 22, 2021

Here are few possible expressions:

21999999999782199999999978
21999999978217821999999978
21999999782199782199999978
21999997821999978219999978
21999997821782178219999978
21999978219999997821999978
21999978219782197821999978
21999782199999999782199978
21999782199782199782199978
21999782178217821782199978
21997821999999999978219978
21997821999782199978219978
21997821978217821978219978
21997821782199782178219978
21978219999999999997821978
21978219999782199997821978
21978219978217821997821978
21978219782199782197821978
21978217821999978217821978
21978217821782178217821978
21782199999999999999782178
21782199999782199999782178
21782199978217821999782178
21782199782199782199782178
21782197821999978219782178
21782197821782178219782178
21782178219999997821782178
21782178219782197821782178

Are there any that don't start with $2$ ?

Chris Lewis - 4 months, 3 weeks ago

No, because if $y$ is the expression, $y$ and $4y$ both have $26$ digits, so $4A < 10$ , which means $A$ is either $1$ or $2$ . However, since $4y$ ends in $A$ and must be even, $A$ must be $2$ . (There was a similar proof for this in the solution section of the third inspiration link.)

David Vreken - 4 months, 3 weeks ago

The problem asks for numbers 'consisting of positive digits.'

Matthew Feig - 4 months, 2 weeks ago

Thanks, I must have missed that! I updated my answer to show only the ones I found with positive digits.

David Vreken - 4 months, 2 weeks ago

Matthew Feig
Jan 25, 2021

Let $S$ be the set of all natural numbers that have the property described in the problem. $S \equiv \{ n \in \N \ | \ 4n \ \text{and} \ n \ \text{have the same decimal digits in reverse order} \}$ The question asks about members of $S$ with $26$ non-zero digits.

Direct calculation shows that $a = 2178, b= 21978, c=219978$ all belong to $S$ . $\begin{aligned} 4 \times 2178 &= 8712 \\ 4 \times 21978 &= 87912 \\ 4 \times 219978 &= 879912 \end{aligned}$
It is also not hard to prove that we can concatenate short members of $S$ to create longer members of $S$ . Let $x \Vert y$ denote the concatenation (or the stringing together end-to-end of the decimal representations) of $x$ and $y$ . $\text{Claim: If } x \in S \text{ and } y \in S, \text{ then } x \Vert x \in S \text{ and } x \Vert y \Vert x \in S .$

Based on the two points above, we can build new members of $S$ out of $a, b, c$ with $26$ digits. So the answer is $\boxed{\text{YES}}$ .

$\begin{aligned} m &= \blue{21978} \ \red{21978} \ \green{219978} \ \red{21978} \ \blue{21978} = b \Vert b \Vert c \Vert b \Vert b \in S &&\text{ and has } 5 + 5 + 6 + 5 + 5 = 26 \text{ digits.} \\ n &= \blue{2178} \ \red{219978} \ \green{219978} \ \red{219978} \ \blue{2178} = a \Vert c \Vert c \Vert c \Vert a \in S &&\text{ and has } 4 + 6 + 6 + 6 + 4 = 26 \text{ digits.} \\ p &= \blue{219978} \ \red{2178} \ \green{219978} \ \red{2178} \ \blue{219978} = c \Vert a \Vert c \Vert a \Vert c \in S &&\text{ and has } 6 + 4 + 6 + 4 + 6 = 26 \text{ digits.} \\ q &= \blue{2178} \ \red{2178} \ \green{21978} \ \green{21978} \ \red{2178} \ \blue{2178} = a \Vert a \Vert b \Vert b \Vert a \Vert a \in S &&\text{ and has } 4 + 4 + 5 + 5 + 4 + 4 = 26 \text{ digits.} \end{aligned}$

A , B , C A,B,C A , B , C , Easy As 1 , 2 , 3 1,2,3 1 , 2 , 3 (Do Re Mi)

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$A,B,C$ , Easy As $1,2,3$ (Do Re Mi)