Switching Bases

First of all we find all the 3-digits numbers such that they have only the digits $0$ , $1$ or $2$ . They are: $120$ , $200$ and $112$ . So we notice that $120=15 \cdot 8 = f_3(15)$ , $112=14 \cdot 8 = f_3(14)$ but $200=25\cdot 8 =f_3(18) \neq f_3(25)$ . Therefore we found that $15$ and $14$ are solutions. There are no bigger solutions because we notice that the number of digits in the number $f_3(n)$ is higher than the number of digits of $8 \cdot n$ as $n$ gets big. [This seems obvious. How would you justify this statement? - Calvin] Therefore the answer is $\fbox{29}$ .

[Note: By "Order of magnitude", Ferdinand means power of 10 / number of digits in base 10 notation. Otherwise, the term by itself is extremely vague, since it doesn't necessarily specify a ratio. For example, 1, 2, 4, 8, 16, etc are on a scale of magnitude.- Calvin]

We know that: $3^1 = 3 = 10_3$ $3^2 = 9 = 100_3$ $3^3 = 27 = 1000_4$ $3^4 = 81 = 10000_4$ $3^5 = 243 = 100000_4$ The function given was $f_3 (n) = 8 * n$ If we multiply any number by 8, we can only increase its "order of magnitude" by 1 at most. Therefore, since 27 is a number with order of magnitude of 2, we can only have at most its product with 8 as a number with order of magnitude 3 but the correct value is $1000_3$ which has an order of magnitude of 4. Therefore, we limit ourselves to numbers less than 27. [His implicit assumption is that all numbers above 27 will fail for this reason, not just numbers with an order of magnitude of 4. - Calvin]

If we list the base 3 format of the numbers and find the numbers that is divisible by 8 in base 10 and the quotient is the base 10 equivalent, we will get the numbers $120_3 = 15$ and $112_3 = 14$ . If we add these we will get 29.

First, we must know that $f_3(n) = p$ has the $p$ which digits are not more than 2, example : $12012 , 11022 , 10001$ , but the value of $14323 , 13013$ is not permitted. $p$ also must be a multiple of 8 too. We split it into some cases:

a. If the number $p$ has only one digit. There is no $p$ such that the digit is 0,1, or 2

b. If the number $p$ has two digits. With some checking we may know that no $p$ such that all the digits are 0,1, or 2.

c. If the number $p$ has three digits. With some checking, we know that all $p$ that fulfills all the criteria which we have mentioned above are $112, 120$ , which occurs when $n = 14 , 15$ . Now we check is $n = 14 , 15$ fullfill the equation $f_3(n) = 8n$ For $n=14$ : $f_3(14) =112 = 1 . 3^2 + 1 . 3 + 2 = 14$ is true. For $n=15$ $f_3(15) = 120 = 1 . 3^2 + 2 . 3 + 0 = 15$ is true.

d) If the digit $p$ has 4 or more digit. We know that some $p$ that fullfills the criteria are $1000, 1112,1120 , 2000 , 2112 , 2120$ which occurs when $n=125,139,140,250,264,265$ But let's see, If $p=1000$ , then $n = 1 . 3^3 + 0 . 3^2 + 0 . 3 + 0 = 27$ if we use $f_3(n) = p$ , and $27 . 8 = 216$ , if we use $f_3(n) = 8n$ , which is not true. For more clearly, the first digit on $p$ in a four digit number can only make the maximum of number n = 54 which is obtained by $n = 2 . 3^3$ .

e) for $p$ 5 or more digit, it is clearly impossible to make the equation $f_3 (n) = 8n$ is true. Then the value of $n$ that satisfy the equation are $n= 14 , 15$ The sum is = $14+15=29$

To start off, I got all the numbers, which, when multiplied by 8, would result in the last digit being either 0 or 2. These numbers ended in 0, 4, 5 and 9. Then I created an array of these numbers starting off with 4. Slowly the base 3 numbers became way larger than the number multiplied by 8. Then I stopped and found the numbers were 14 and 15. Here's the start of the array.

$4, 11$

$5, 12$

$10, 101$

$14, 112$

$15, 120$

$20, 202$

$...$

Sorry if this seems impractical but that seems the only way to do it.

We see that 15 and 14 both work. I'd be interested to see how you get this answer.