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Difference between revisions of "2015 AMC 10A Problems/Problem 18"

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==Solution==
 
==Solution==
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Notice that 1000 is 3E8 in hexadecimal. The first digit could be 0, 1, 2, or 3 and the second two could be any digit 0-9. 4 • 10 • 10 = 400. However, this includes 3E9, so subtract one. Thus, there are 399 valid <math>n</math>, corresponding to those 399 positive integers less than 1000 with hexadecimal representation less than 1000. (Notice that 399 < 3E8 in hexadecimal.) Our answer is <math>3 + 9 + 9 = 21</math> <math>\textbf{(E) }</math>.
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Notice that 1000 is 3E8 in hexadecimal. The first digit could be 0, 1, 2, or 3 and the second two could be any digit 0-9. 4 • 10 • 10 = 400. However, this includes 000, so subtract one. Thus, there are 399 valid <math>n</math>, corresponding to those 399 positive integers less than 1000 with hexadecimal representation less than 1000. (Notice that 399 < 3E8 in hexadecimal.) Our answer is <math>3 + 9 + 9 = 21</math> <math>\textbf{(E) }</math>.
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==See Also==
 
==See Also==
 
{{AMC10 box|year=2015|ab=A|num-b=17|num-a=19}}
 
{{AMC10 box|year=2015|ab=A|num-b=17|num-a=19}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 21:45, 4 February 2015

Problem 18

Hexadecimal (base-16) numbers are written using numeric digits $0$ through $9$ as well as the letters $A$ through $F$ to represent $10$ through $15$. Among the first $1000$ positive integers, there are $n$ whose hexadecimal representation contains only numeric digits. What is the sum of the digits of $n$?

$\textbf{(A) }17\qquad\textbf{(B) }18\qquad\textbf{(C) }19\qquad\textbf{(D) }20\qquad\textbf{(E) }21$

Solution

Notice that 1000 is 3E8 in hexadecimal. The first digit could be 0, 1, 2, or 3 and the second two could be any digit 0-9. 4 • 10 • 10 = 400. However, this includes 000, so subtract one. Thus, there are 399 valid $n$, corresponding to those 399 positive integers less than 1000 with hexadecimal representation less than 1000. (Notice that 399 < 3E8 in hexadecimal.) Our answer is $3 + 9 + 9 = 21$ $\textbf{(E) }$.

See Also

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

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