300 Followers Problem

Number Theory Level 5

The number $300$ is palindromic in bases 7,8 and 9. $300_{10} = 606_{7} = 454_{8} = 363_{9}$ .

What is the total number of bases in which the representation of $300_{10}$ would be palindromic with more than one digit.

9 18 16 3 15 4 13

1 solution

Janardhanan Sivaramakrishnan
Sep 3, 2015

$300$ being divisible by $2,3,4,5,6$ cannot be a palindrome in any of those bases.

It is already given that it is a palindrome in bases $7,8,9$ .

Searching for 3 digit palindromes of type $\overline{xyx}_b$ , with base $b \le \lfloor\sqrt{300} \rfloor = 17$ , We get the equation as $(b^2+1)x+by=300$ . $0 \le x,y < b$ . Solving,

we find valid solutions for base 13. $300_{10}=1a1_{13}$

Now with 2 digit palindromes, i..e $18 \leq b < 300$ , the required equation becomes

$(b+1)x=300$ , which is satisfied for $b=19,24,29,49,59,74,99,149,299$

Thus, there are $\boxed{13}$ such bases.

What about base 13???????

Nabil Moustafa - 5 years, 9 months ago

Yes. It is palindrome in base 13. That has been included in the solution.

Janardhanan Sivaramakrishnan - 5 years, 9 months ago

nb^2 +n = 300. n (b^2 + 1) = 300, then you use factorization. But n has to be smaller than b. for example, if you have base 7, you can't have 909 or something. I got 5 with this limitation.

Bob Yang - 5 years, 9 months ago

@Bob Yang – $300_{10}$ is $606$ in base 7.

Janardhanan Sivaramakrishnan - 5 years, 9 months ago

300 Followers Problem

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