Difference between revisions of "1961 AHSME Problems/Problem 35"
Rockmanex3 (talk | contribs) (Solution to Problem 35) |
m (→Solution) |
||
Line 14: | Line 14: | ||
Since <math>120 < 695 < 720</math>, divide <math>695</math> by <math>120</math>. The quotient is <math>5</math> and the remainder is <math>95</math>, so rewrite the number as | Since <math>120 < 695 < 720</math>, divide <math>695</math> by <math>120</math>. The quotient is <math>5</math> and the remainder is <math>95</math>, so rewrite the number as | ||
− | <cmath>695 = 5 \cdot | + | <cmath>695 = 5 \cdot 120 + 95</cmath> |
Similarly, dividing <math>95</math> by <math>24</math> results in a quotient of <math>3</math> and a remainder of <math>23</math>, so the number can be rewritten as | Similarly, dividing <math>95</math> by <math>24</math> results in a quotient of <math>3</math> and a remainder of <math>23</math>, so the number can be rewritten as | ||
− | <cmath>695 = 5 \cdot | + | <cmath>695 = 5 \cdot 120 + 3 \cdot 24 + 23</cmath> |
Repeat the steps to get | Repeat the steps to get | ||
− | <cmath>695 = 5 \cdot | + | <cmath>695 = 5 \cdot 120 + 3 \cdot 24 + 3 \cdot 6 + 2 \cdot 2 + 1</cmath> |
The answer is <math>\boxed{\textbf{(D)}}</math>. One can also stop at the second step by noting <math>23 < 24</math>. | The answer is <math>\boxed{\textbf{(D)}}</math>. One can also stop at the second step by noting <math>23 < 24</math>. | ||
Latest revision as of 19:43, 14 September 2023
Problem
The number is to be written with a factorial base of numeration, that is, where are integers such that and means . Find
Solution
This problem can be approached similarly to other base number problems.
Since , divide by . The quotient is and the remainder is , so rewrite the number as Similarly, dividing by results in a quotient of and a remainder of , so the number can be rewritten as Repeat the steps to get The answer is . One can also stop at the second step by noting .
See Also
1961 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 34 |
Followed by Problem 36 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 | ||
All AHSME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.