A number theory problem by Shrenik Jobanputra

Let the number be pqr in base 7. It is given it becomes rqp in base 9.

Thus, $49p + 7q + r = 81r + 9q + p$

This gives $8(3p - 5r) = q$

Now as q < 7 (as pqr is a number in base 7), only possibility is $q = 0.$

Thus, $3p = 5r$ . This gives, $p = 5$ and $r = 3$ .

Hence, number is 503 in base 7, 305 in base 9, and 248 in base 10.

We have: $\overline{abc}_7=\overline{cba}_9$

Using the positional notation : $a\times7^2+b\times7+c=c\times9^2+b\times9+a$

Rearraing: $48a=80c+2b$

Since $a$ , $b$ , $c$ form a number in base $7$ , they must be between $0$ and $6$ inclusive.

We will conduct a trial-and-error approach on $48a$ .

• $48\times1=48=80\times0+2\times24\color{#D61F06}✘$
• $48\times2=96=80\times1+2\times8\color{#D61F06}✘$
• $48\times3=144=80\times1+2\times32\color{#D61F06}✘$
• $48\times4=192=80\times2+2\times16\color{#D61F06}✘$
• $48\times5=240=80\times3+2\times0\color{#20A900}✓$
• $48\times6=288=80\times3+2\times24\color{#D61F06}✘$

Therefore, $a=5$ , $b=0$ , and $c=3$ .

Therefore, the number is $503_7=305_9=248_{10}$ .

This is just my method, and not by any means a rigorous proof.

Let the digits be $a, b,$ and $c$ such that $49a+7b+c = a+9b+81c$ . After simplifying, the equation becomes $24a = b+40c$ .

Since $abc$ exists as a base 7 three digit number, we know that $1≤a≤6$ , implying that $b+40c ≤ 144$ . Therefore, $1≤c≤3$ , which further implies that $24a ≤ 126$ , meaning $1≤a≤5$ .

After narrowing it down, I solved by trial and error, finding that the only solution to $24a = b+40c$ was $a=5$ and $c=3$ .

Solving for $b$ , I got $abc = 503_{7}$ . Converting to base 10, the answer is $\boxed{248}$ .