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Palindrome Ponder (Posted on 2014-08-04) Difficulty: 3 of 5
The first 20140 base ten positive integers are written in base 3. How many of these ternary representations are palindromes?

Bonus Question: How many ternary palindromes are there when the first 201400 base ten positive integers are written in base 3?

No Solution Yet Submitted by K Sengupta    
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Solution solution Comment 1 of 1
Decimal 20140 is 1000121221 in base 3.  It has 10 digits, so 10-digit base-3 numbers up to this value will be eligible, as well as palindromes with fewer digits.

For palindromes with an odd number of digits, up to 9 digits, any base-3 number up to 22222 may be used as its LHS plus middle digit, which determines its RHS. The number of these is 3^5 - 1, as zero itself is not allowable.

Even length palindromes up to length 8 similarly consist of numbers up to and including 2222 for the LHS, thereby determining the RHS. That's another 3^4 -1 this time.

The limit for 10-digit base-3 palindromes for the LHS is 10001. It looks like we can forget about the preceding paragraph, as we'll incorporate those smaller even-length palindromes here: 3^4+1 - 1.

That makes 3^5 - 1 + 3^4+1 - 1 = 323 such palindromes.


Bonus:

Decimal 201400 is 101020021021 in base 3. It has 12 digits, so 12-digit base-3 numbers up to this value will be eligible, as well as all palindromes with fewer digits.

For palindromes with an odd number of digits, up to 11 digits, any base-3 number up to 222222 may be used as its LHS plus middle digit, which determines its RHS. The number of these is 3^6 - 1, as zero itself is not allowable.

Skipping over the unnecessary paragraph, we go right to:

The limit for 12-digit base-3 palindromes for the LHS is 101012. We get: 3^5+3^3+3+2 - 1.

The total is  3^5+3^3+3+2 - 1 + 3^6 - 1 = 1002.

  Posted by Charlie on 2014-08-04 13:09:37
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