Difference between revisions of "2015 AMC 10A Problems/Problem 18"
Pi is 3.14 (talk | contribs) (→Solution) |
|||
Line 13: | Line 13: | ||
~Clarification by mathboy282 | ~Clarification by mathboy282 | ||
+ | |||
+ | ==Solution 2 (Casework)== | ||
+ | First, we set a bound by writing <math>1000</math> in base-<math>16</math>. <math>1000_{10}=3E8_{16}</math>. Therefore, we are considering numbers with a maximum of <math>3</math> digits, and a maximum of <math>3</math> in the <math>256</math>ths-place (the first place in a <math>3</math>-digit number). | ||
+ | |||
+ | Case <math>1</math>: <math>1</math>-digit numbers: | ||
+ | There are evidently <math>9</math> numbers that fit this category. | ||
+ | |||
+ | Case <math>2</math>: <math>2</math>-digit numbers: | ||
+ | There are <math>9\cdot10=90</math> numbers that fit this category. | ||
+ | |||
+ | Case <math>3</math>: <math>3</math>-digit numbers: | ||
+ | There are <math>3\cdot10\cdot10=300</math> numbers that fit this category | ||
+ | |||
+ | Adding these up, we get <math>9+90+300=399</math> numbers. <math>3 + 9 + 9 = \boxed{\textbf{(E) } 21}</math> ~sosiaops | ||
== Video Solution == | == Video Solution == |
Revision as of 16:59, 24 January 2021
Contents
Problem 18
Hexadecimal (base-16) numbers are written using numeric digits through as well as the letters through to represent through . Among the first positive integers, there are whose hexadecimal representation contains only numeric digits. What is the sum of the digits of ?
Solution
Notice that is when converted to hexadecimal(). We will proceed by constructing numbers that consist of only numeric digits in hexadecimal.
The first digit could be or and the second two could be any digit , giving combinations. However, this includes so this number must be diminished by Therefore, there are valid corresponding to those positive integers less than that consist of only numeric digits. (Notice that is the least hexadecimal number using only decimal digits before .) Therefore, our answer is
Reasoning for the 400 Combinations
This is because we don't want more than for each digit since that would lead to a letter digit(A, B, C, D, E, or F.) Thus we have Then continue as follows.
~Clarification by mathboy282
Solution 2 (Casework)
First, we set a bound by writing in base-. . Therefore, we are considering numbers with a maximum of digits, and a maximum of in the ths-place (the first place in a -digit number).
Case : -digit numbers: There are evidently numbers that fit this category.
Case : -digit numbers: There are numbers that fit this category.
Case : -digit numbers: There are numbers that fit this category
Adding these up, we get numbers. ~sosiaops
Video Solution
https://youtu.be/ZhAZ1oPe5Ds?t=4596
~ pi_is_3.14
See Also
Video Solution: https://www.youtube.com/watch?v=2DVSkWu_H1g
2015 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 17 |
Followed by Problem 19 | |
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 | ||
All AMC 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.