1986 AIME Problems/Problem 7
Contents
Problem
The increasing sequence consists of all those positive integers which are powers of 3 or sums of distinct powers of 3. Find the term of this sequence.
Solutions
Solution 1
Rewrite all of the terms in base 3. Since the numbers are sums of distinct powers of 3, in base 3 each number is a sequence of 1s and 0s (if there is a 2, then it is no longer the sum of distinct powers of 3). Therefore, we can recast this into base 2 (binary) in order to determine the 100th number. is equal to , so in binary form we get . However, we must change it back to base 10 for the answer, which is .
Solution 2
Notice that the first term of the sequence is , the second is , the fourth is , and so on. Thus the term of the sequence is . Now out of terms which are of the form + , of them include and do not. The smallest term that includes , i.e. , is greater than the largest term which does not, or . So the th term will be , then , then , then , and finally
Solution 3
After the th power of 3 in the sequence, the number of terms after that power but before the th power of 3 is equal to the number of terms before the th power, because those terms after the th power are just the th power plus all the distinct combinations of powers of 3 before it, which is just all the terms before it. Adding the powers of and the terms that come after them, we see that the th term is after , which is the th term. Also, note that the th term after the th power of 3 is equal to the power plus the th term in the entire sequence. Thus, the th term is plus the th term. Using the same logic, the th term is plus the th term, . We now have
Solution 4
Writing out a few terms of the sequence until we reach the next power of 3 (27), we see that the term is equal to . From here, we can ballpark the range of the 100th term. The 64th term is = and the 128th term is = . Writing out more terms of the sequence until the next power of 3 again (81) we can see that the (+)/2 term is equal to + . From here, we know that the 96th term is + = . From here, we can construct the 100th term by following the sequence in increasing order. The 97th term is , the 98th term is , the 99th term is , and finally the 100th term is
Solution 5
The number of terms produces includes each power of 3 (), the sums of two power of 3s(ex. ), three power of 3s (ex. ), all the way to the sum of them all. Since there are powers of 3, the one number sum gives us terms, the two number terms, all the way to the sum of all the powers which gives us terms. Summing all these up gives us according to the theorem
Since is the greatest power , then and the sequence would look like {}, where or would be the rd number. The next largest power would be the 64th number. However, its terms contributed extends beyond 100, so we break it to smaller pieces. Noting that plus any combination of lower powers is < , so we can add all those terms() into our sequence:
Our sequence now has terms. The remaining would just be the smallest sums starting with or :
Hence the 100th term would be . ~SoilMilk
Note that there isn't a as choosing 0 numbers will not give you a term.
See also
1986 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 6 |
Followed by Problem 8 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
- AIME Problems and Solutions
- American Invitational Mathematics Examination
- Mathematics competition resources
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.