AIME 2001 I · 第 8 题

Solution

We let $N_7 = \overline{a_na_{n-1}\cdots a_0}_7$; we are given that

$$2(a_na_{n-1}\cdots a_0)_7 = (a_na_{n-1}\cdots a_0)_{10}$$ (This is because the digits in $N$ ' s base 7 representation make a number with the same digits in base 10 when multiplied by 2)

Expanding, we find that

$$2 \cdot 7^n a_n + 2 \cdot 7^{n-1} a_{n-1} + \cdots + 2a_0 = 10^na_n + 10^{n-1}a_{n-1} + \cdots + a_0$$ or re-arranging,

$$a_0 + 4a_1 = 2a_2 + 314a_3 + \cdots + (10^n - 2 \cdot 7^n)a_n$$ Since the $a_i$s are base-$7$ digits, it follows that $a_i < 7$, and the LHS is less than or equal to $30$. Hence our number can have at most $3$ digits in base-$7$. Letting $a_2 = 6$, we find that $630_7 = \boxed{315}_{10}$ is our largest 7-10 double.

Solution 2 (Guess and Check)

Let $A$ be the base $10$ representation of our number, and let $B$ be its base $7$ representation.

Given this is an AIME problem, $A<1000$. If we look at $B$ in base $10$, it must be equal to $2A$, so $B<2000$ when $B$ is looked at in base $10.$

If $B$ in base $10$ is less than $2000$, then $B$ as a number in base $7$ must be less than $2*7^3=686$.

$686$ is non-existent in base $7$, so we're gonna have to bump that down to $666_7$.

This suggests that $A$ is less than $\frac{666}{2}=333$.

Guess and check shows that $A<320$, and checking values in that range produces $\boxed{315}$.

Solution 3

Since this is an AIME problem, the maximum number of digits the 7-10 double can have is 3. Let the number be

$$abc$$ in base 7. Then the number in expanded form is

$$49a+7b+c$$ in base 7 and

$$100a+10b+c$$ in base 10. Since the number in base 7 is half the number in base 10, we get the following equation.

$$98a+14b+2c=100a+10b+c$$ which simplifies to

$$2a=4b+c.$$ The largest possible value of a is 6 because the number is in base 7. Then to maximize the number, $b$ is $3$ and $c$ is $0$. Therefore, the largest 7-10 double is 630 in base 7, or $\boxed{315}$ in base 10.