Difference between revisions of "1986 AIME Problems/Problem 7"
Math Kirby (talk | contribs) m |
(→Solutions) |
||
(12 intermediate revisions by 9 users not shown) | |||
Line 2: | Line 2: | ||
The increasing [[sequence]] <math>1,3,4,9,10,12,13\cdots</math> consists of all those positive [[integer]]s which are [[exponent|powers]] of 3 or sums of distinct powers of 3. Find the <math>100^{\mbox{th}}</math> term of this sequence. | The increasing [[sequence]] <math>1,3,4,9,10,12,13\cdots</math> consists of all those positive [[integer]]s which are [[exponent|powers]] of 3 or sums of distinct powers of 3. Find the <math>100^{\mbox{th}}</math> term of this sequence. | ||
− | == | + | == Solutions == |
=== Solution 1 === | === 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. <math>100</math> is equal to <math>64 + 32 + 4</math>, so in binary form we get <math>1100100</math>. However, we must change it back to base | + | 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. <math>100</math> is equal to <math>64 + 32 + 4</math>, so in binary form we get <math>1100100</math>. However, we must change it back to base 10 for the answer, which is <math>3^6 + 3^5 + 3^2 = 729 + 243 + 9 = \boxed {981}</math>. |
=== Solution 2 === | === Solution 2 === | ||
− | Notice that the first term of the sequence is <math>1</math>, the second is <math> | + | Notice that the first term of the sequence is <math>1</math>, the second is <math>3</math>, the fourth is <math>9</math>, and so on. Thus the <math>64th</math> term of the sequence is <math>729</math>. Now out of <math>64</math> terms which are of the form <math>729</math> + <math>'''S'''</math>, <math>32</math> of them include <math>243</math> and <math>32</math> do not. The smallest term that includes <math>243</math>, i.e. <math>972</math>, is greater than the largest term which does not, or <math>854</math>. So the <math>96</math>th term will be <math>972</math>, then <math>973</math>, then <math>975</math>, then <math>976</math>, and finally <math>\boxed{981}</math> |
+ | |||
+ | === Solution 3 === | ||
+ | |||
+ | After the <math>n</math>th power of 3 in the sequence, the number of terms after that power but before the <math>(n+1)</math>th power of 3 is equal to the number of terms before the <math>n</math>th power, because those terms after the <math>n</math>th power are just the <math>n</math>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 <math>3</math> and the terms that come after them, we see that the <math>100</math>th term is after <math>729</math>, which is the <math>64</math>th term. Also, note that the <math>k</math>th term after the <math>n</math>th power of 3 is equal to the power plus the <math>k</math>th term in the entire sequence. Thus, the <math>100</math>th term is <math>729</math> plus the <math>36</math>th term. Using the same logic, the <math>36</math>th term is <math>243</math> plus the <math>4</math>th term, <math>9</math>. We now have <math>729+243+9=\boxed{981}</math> | ||
+ | |||
+ | === Solution 4 === | ||
+ | Writing out a few terms of the sequence until we reach the next power of 3 (27), we see that the <math>2^{nth}</math> term is equal to <math>3^n</math>. From here, we can ballpark the range of the 100th term. The 64th term is <math>3^6</math> = <math>729</math> and the 128th term is <math>3^7</math> = <math>2187</math>. Writing out more terms of the sequence until the next power of 3 again (81) we can see that the (<math>2^n</math>+<math>2^{n+1}</math>)/2 term is equal to <math>3^n</math> + <math>3^{n-1}</math>. From here, we know that the 96th term is <math>3^6</math> + <math>3^5</math> = <math>972</math>. From here, we can construct the 100th term by following the sequence in increasing order. The 97th term is <math>972 + 1 = 973</math>, the 98th term is <math>972 + 3 = 975</math>, the 99th term is <math>972 + 3 + 1 = 976</math>, and finally the 100th term is <math>972 + 9 = \boxed{981}</math> | ||
== See also == | == See also == | ||
Line 18: | Line 25: | ||
[[Category:Intermediate Number Theory Problems]] | [[Category:Intermediate Number Theory Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 09:17, 7 July 2021
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
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.