awkward base problem

Question

Call a positive integer $N$ a "7-10 double" if the digits of the base-7 representation of $N$ form a base-10 number that is twice $N$. For example, 51 is a 7-10 double because its base-7 representation is 102. What is the largest 7-10 double? Express your answer in decimal.

I am really confused about how to apporach this... My current thoughts are if N contains digit 4/5/6, then the quantity of 2N transformed to base 7 will not be the same but it seems like it doesnt help much. Can someone give me some clue about what to do next? Thank you!

Take 51 for example, multiply by two and it will be 102. And 102(base7)=51(base 10), so it is a 7-10 double.

Answer 1

Say your number is $x$; then we have $$ x=a_0+a_1\cdot 7+\dots+a_k\cdot 7^k $$ and the requirement is that $$ a_0+a_1\cdot 10+\dots+a_k\cdot 10^k=2x $$ We so have $$ a_0+a_1(2\cdot 7-10)+a_2(2\cdot 7^2-10^2)+\dots+a_k(2\cdot 7^k-10^k)=0 $$ or $$ a_0+4a_1= a_2(10^2-2\cdot 7^2)+\dots+a_k(10^k-2\cdot 7^k) $$ Note that the left-hand side is at most $6+4\cdot 6=30$ and $$ 10^t-2\cdot7^t>30 $$ as soon as $t>2$ (prove it). Thus the number can be written $$ x=a_0+a_1\cdot 7+a_2\cdot 7^2 $$ with $0\le a_i\le 6$.

Can you finish?

Answer 2

Here is the hint.

Let's first consider a number which has a three-digit base-7 representation. Then the 7-double number will have to satisfy $$2(49 a_2 + 7 a_1 + a_0) = 100 a_2 + 10 a_1 + a_0,$$ which simplifies to $$2a_2=4a_1 + a_0.$$ Clearly, $a_0$ needs to be even, and $0 \le a_i \le 6$. With these constraints, you can find some 7-double numbers.

Now apply these ideas to those numbers which have four-digit base-7 representations. You will immediately find what is the biggest 7-double number.

Answer 3

It's $315=630_7$

Indeed $$2 \left(7^2 a+7 b+c\right)=10^2 a+10 b+c$$ implies $$98 a + 14 b + 2 c == 100 a + 10 b + c$$ that is $$b=\frac{1}{4} (2 a-c);\;\text{ with } a=6\to b=\frac{12-c}{4},\;c=0,1,2,3,4,5,6$$

The maximum value for $b$ is achieved when $c=0$ so $b=3$

Hope this helps

PS

No larger numbers exists satisfying the rule

If we try with $4$ digit numbers we get that the first digit should be $\dfrac{1}{314} (-2 a+4 b+c)$ which makes no sense