https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Ctw0611&feedformat=atom AoPS Wiki - User contributions [en] 2021-04-22T11:47:29Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=2003_AMC_10B_Problems/Problem_25&diff=113972 2003 AMC 10B Problems/Problem 25 2020-01-01T20:50:02Z <p>Ctw0611: /* Solution 5 (Even Faster than extremely fast) */</p> <hr /> <div>==Problem 25==<br /> <br /> How many distinct four-digit numbers are divisible by &lt;math&gt;3&lt;/math&gt; and have &lt;math&gt;23&lt;/math&gt; as their last two digits?<br /> <br /> &lt;math&gt;\textbf{(A) } 27 \qquad\textbf{(B) } 30 \qquad\textbf{(C) } 33 \qquad\textbf{(D) } 81 \qquad\textbf{(E) } 90&lt;/math&gt;<br /> <br /> ==Solution==<br /> ===Solution 1===<br /> To test if a number is divisible by three, you add up the digits of the number. If the sum is divisible by three, then the original number is a multiple of three. If the sum is too large, you can repeat the process until you can tell whether it is a multiple of three. This is the basis for the solution. Since the last two digits are &lt;math&gt; 23 &lt;/math&gt;, the sum of the digits is &lt;math&gt; 2+3 = 5 &lt;/math&gt; (therefore it is not divisible by three). However certain numbers can be added to make the sum of the digits a multiple of three.<br /> <br /> &lt;math&gt; 5+1 = 6 &lt;/math&gt;,<br /> <br /> &lt;math&gt; 5+4 = 9 &lt;/math&gt;, and so on. <br /> <br /> However since the largest four-digit number ending with &lt;math&gt; 23 &lt;/math&gt; is &lt;math&gt; 9923 &lt;/math&gt;, the maximum sum is<br /> <br /> &lt;math&gt; 5+18 = 23 &lt;/math&gt;.<br /> <br /> Using that process we can fairly quickly compile a list of the sum of the first two digits of the number. <br /> <br /> &lt;cmath&gt; \{1, 4, 7, 10, 13, 16\} &lt;/cmath&gt;<br /> <br /> Now we find all the two-digit numbers that have any of the sums shown above. We can do this by listing all the two digit numbers &lt;math&gt; xy &lt;/math&gt; in separate cases.<br /> <br /> &lt;cmath&gt; I. x+y = 1, \{10\} = 1 &lt;/cmath&gt;<br /> &lt;cmath&gt; II. x+y = 4, \{13, 22, 31, 40\} = 4 &lt;/cmath&gt;<br /> &lt;cmath&gt; III. x+y = 7, \{16, 25, 34, 43, 52, 61, 70\} = 7 &lt;/cmath&gt;<br /> &lt;cmath&gt; IV. x+y = 10, \{19, 28, 37, 46, 55, 64, 73, 82, 91\} = 9 &lt;/cmath&gt;<br /> &lt;cmath&gt; V. x+y = 13, \{49, 58, 67, 76, 85, 94\} = 6 &lt;/cmath&gt;<br /> &lt;cmath&gt; VI. x+y = 16, \{79, 88, 97\} = 3 &lt;/cmath&gt;<br /> <br /> And finally, we add the number of elements in each set.<br /> <br /> &lt;cmath&gt; 1+4+7+9+6+3 = \boxed{\textbf{(B)}\ 30} &lt;/cmath&gt;<br /> <br /> ===Solution 2===<br /> A number divisible by &lt;math&gt;3&lt;/math&gt; has all its digits add to a multiple of &lt;math&gt;3.&lt;/math&gt; The last two digits are &lt;math&gt;2&lt;/math&gt; and &lt;math&gt;3&lt;/math&gt; and add up to &lt;math&gt;5 \equiv 2\ (\text{mod}\ 3).&lt;/math&gt; Therefore the first two digits must add up to &lt;math&gt;1\ (\text{mod}\ 3).&lt;/math&gt; &lt;math&gt;4&lt;/math&gt; digits (including &lt;math&gt;0&lt;/math&gt;) are &lt;math&gt;0\ (\text{mod}\ 3),&lt;/math&gt; &lt;math&gt;3&lt;/math&gt; are &lt;math&gt;1\ (\text{mod}\ 3),&lt;/math&gt; and &lt;math&gt;3&lt;/math&gt; are &lt;math&gt;2\ (\text{mod}\ 3).&lt;/math&gt; The following combinations are equivalent to &lt;math&gt;1\ (\text{mod}\ 3)&lt;/math&gt;:<br /> <br /> &lt;cmath&gt; 0\ (\text{mod}\ 3)+1\ (\text{mod}\ 3) \equiv 1\ (\text{mod}\ 3)+0\ (\text{mod}\ 3) \equiv 2\ (\text{mod}\ 3) +2\ (\text{mod}\ 3)&lt;/cmath&gt;<br /> <br /> Let the first term in each combination represent the thousands digit and the second term represent the hundreds digit. We can use this to find the total amount of four-digit numbers.<br /> <br /> &lt;cmath&gt;3\cdot3 + 3\cdot4 + 3\cdot3 = 9 + 12 + 9 = \boxed{\textbf{(B)}\ 30}&lt;/cmath&gt;<br /> <br /> ===Solution 3 ===<br /> <br /> We have the following: &lt;math&gt; n \equiv 0 \pmod{3}&lt;/math&gt; and &lt;math&gt;n \equiv 23\pmod{100}&lt;/math&gt;. Then &lt;math&gt;n = 3a = 100b+23&lt;/math&gt; for some integers &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt;. Taking mod 3 gives:<br /> &lt;math&gt;0 \equiv b+2 \pmod 3 \implies b\equiv 1\pmod 3&lt;/math&gt; so &lt;math&gt;b=3c+1 &lt;/math&gt; for some integer &lt;math&gt;c&lt;/math&gt;. But &lt;math&gt;N = 100b+23 = 100(3c+1)+23&lt;/math&gt; so &lt;math&gt;n = 300c+123&lt;/math&gt;. Bounding this gives us:<br /> &lt;math&gt;999 &lt; 300c+123 &lt; 10000&lt;/math&gt; so &lt;math&gt;876 &lt; 300c &lt; 9877&lt;/math&gt;. Dividing by &lt;math&gt;300&lt;/math&gt; gives &lt;math&gt;2.92 &lt; c &lt; 32.9222&lt;/math&gt; so &lt;math&gt;3\le c \le 32&lt;/math&gt;. This gives &lt;math&gt;32-3+1 = \boxed{\textbf{(B)} \ 30 }&lt;/math&gt;<br /> <br /> ===Solution 4 (Extremely Fast)===<br /> <br /> The number is in the form &lt;math&gt;xy23&lt;/math&gt;<br /> <br /> Notice that the number is divisible by &lt;math&gt;3&lt;/math&gt; if the sum of the digits of &lt;math&gt;xy23&lt;/math&gt; is divisible by &lt;math&gt;3&lt;/math&gt;<br /> <br /> Our first number that is divisible by &lt;math&gt;3&lt;/math&gt; is &lt;math&gt;1023&lt;/math&gt;, next is &lt;math&gt;1323&lt;/math&gt;..<br /> Notice &lt;math&gt;xy&lt;/math&gt; goes from &lt;math&gt;10\implies 97&lt;/math&gt;<br /> <br /> Hence, there are &lt;math&gt;\frac{97-10}{3}+1=30&lt;/math&gt; distinct four digit numbers.<br /> <br /> === Solution 5 (Even Faster than Extremely Fast) ===<br /> <br /> Following the form &lt;math&gt;xy23&lt;/math&gt; in Solution 4, notice that &lt;math&gt;x+y \equiv 1 \pmod 3&lt;/math&gt; to satisfy our condition. Choose a value of &lt;math&gt;x&lt;/math&gt; from 1 to 9. For &lt;math&gt;x=1, 4, 7&lt;/math&gt;, there are exactly 4 values of &lt;math&gt;y: 0, 3, 6, 9&lt;/math&gt;. For the remaining 6 digits, there are 3 choices for y. So our answer is &lt;math&gt;(3)(4)+(6)(3)=30&lt;/math&gt;.<br /> <br /> === Solution 6 (Even Faster than Even Faster than Extremely Fast)===<br /> There are &lt;math&gt;9 \cdot 10 = 90&lt;/math&gt; possible values for the first two digits. One-third of them yield a multiple of &lt;math&gt;3&lt;/math&gt;, so the answer is &lt;math&gt;\frac{90}{3} = \boxed{\textbf{(B)}\ 30}&lt;/math&gt;<br /> <br /> == See also ==<br /> {{AMC10 box|year=2003|num-b=24|after=Last problem|ab=B}}<br /> <br /> <br /> {{MAA Notice}}</div> Ctw0611 https://artofproblemsolving.com/wiki/index.php?title=2003_AMC_10B_Problems/Problem_25&diff=113971 2003 AMC 10B Problems/Problem 25 2020-01-01T20:49:50Z <p>Ctw0611: /* Solution 4 (extremely fast) */</p> <hr /> <div>==Problem 25==<br /> <br /> How many distinct four-digit numbers are divisible by &lt;math&gt;3&lt;/math&gt; and have &lt;math&gt;23&lt;/math&gt; as their last two digits?<br /> <br /> &lt;math&gt;\textbf{(A) } 27 \qquad\textbf{(B) } 30 \qquad\textbf{(C) } 33 \qquad\textbf{(D) } 81 \qquad\textbf{(E) } 90&lt;/math&gt;<br /> <br /> ==Solution==<br /> ===Solution 1===<br /> To test if a number is divisible by three, you add up the digits of the number. If the sum is divisible by three, then the original number is a multiple of three. If the sum is too large, you can repeat the process until you can tell whether it is a multiple of three. This is the basis for the solution. Since the last two digits are &lt;math&gt; 23 &lt;/math&gt;, the sum of the digits is &lt;math&gt; 2+3 = 5 &lt;/math&gt; (therefore it is not divisible by three). However certain numbers can be added to make the sum of the digits a multiple of three.<br /> <br /> &lt;math&gt; 5+1 = 6 &lt;/math&gt;,<br /> <br /> &lt;math&gt; 5+4 = 9 &lt;/math&gt;, and so on. <br /> <br /> However since the largest four-digit number ending with &lt;math&gt; 23 &lt;/math&gt; is &lt;math&gt; 9923 &lt;/math&gt;, the maximum sum is<br /> <br /> &lt;math&gt; 5+18 = 23 &lt;/math&gt;.<br /> <br /> Using that process we can fairly quickly compile a list of the sum of the first two digits of the number. <br /> <br /> &lt;cmath&gt; \{1, 4, 7, 10, 13, 16\} &lt;/cmath&gt;<br /> <br /> Now we find all the two-digit numbers that have any of the sums shown above. We can do this by listing all the two digit numbers &lt;math&gt; xy &lt;/math&gt; in separate cases.<br /> <br /> &lt;cmath&gt; I. x+y = 1, \{10\} = 1 &lt;/cmath&gt;<br /> &lt;cmath&gt; II. x+y = 4, \{13, 22, 31, 40\} = 4 &lt;/cmath&gt;<br /> &lt;cmath&gt; III. x+y = 7, \{16, 25, 34, 43, 52, 61, 70\} = 7 &lt;/cmath&gt;<br /> &lt;cmath&gt; IV. x+y = 10, \{19, 28, 37, 46, 55, 64, 73, 82, 91\} = 9 &lt;/cmath&gt;<br /> &lt;cmath&gt; V. x+y = 13, \{49, 58, 67, 76, 85, 94\} = 6 &lt;/cmath&gt;<br /> &lt;cmath&gt; VI. x+y = 16, \{79, 88, 97\} = 3 &lt;/cmath&gt;<br /> <br /> And finally, we add the number of elements in each set.<br /> <br /> &lt;cmath&gt; 1+4+7+9+6+3 = \boxed{\textbf{(B)}\ 30} &lt;/cmath&gt;<br /> <br /> ===Solution 2===<br /> A number divisible by &lt;math&gt;3&lt;/math&gt; has all its digits add to a multiple of &lt;math&gt;3.&lt;/math&gt; The last two digits are &lt;math&gt;2&lt;/math&gt; and &lt;math&gt;3&lt;/math&gt; and add up to &lt;math&gt;5 \equiv 2\ (\text{mod}\ 3).&lt;/math&gt; Therefore the first two digits must add up to &lt;math&gt;1\ (\text{mod}\ 3).&lt;/math&gt; &lt;math&gt;4&lt;/math&gt; digits (including &lt;math&gt;0&lt;/math&gt;) are &lt;math&gt;0\ (\text{mod}\ 3),&lt;/math&gt; &lt;math&gt;3&lt;/math&gt; are &lt;math&gt;1\ (\text{mod}\ 3),&lt;/math&gt; and &lt;math&gt;3&lt;/math&gt; are &lt;math&gt;2\ (\text{mod}\ 3).&lt;/math&gt; The following combinations are equivalent to &lt;math&gt;1\ (\text{mod}\ 3)&lt;/math&gt;:<br /> <br /> &lt;cmath&gt; 0\ (\text{mod}\ 3)+1\ (\text{mod}\ 3) \equiv 1\ (\text{mod}\ 3)+0\ (\text{mod}\ 3) \equiv 2\ (\text{mod}\ 3) +2\ (\text{mod}\ 3)&lt;/cmath&gt;<br /> <br /> Let the first term in each combination represent the thousands digit and the second term represent the hundreds digit. We can use this to find the total amount of four-digit numbers.<br /> <br /> &lt;cmath&gt;3\cdot3 + 3\cdot4 + 3\cdot3 = 9 + 12 + 9 = \boxed{\textbf{(B)}\ 30}&lt;/cmath&gt;<br /> <br /> ===Solution 3 ===<br /> <br /> We have the following: &lt;math&gt; n \equiv 0 \pmod{3}&lt;/math&gt; and &lt;math&gt;n \equiv 23\pmod{100}&lt;/math&gt;. Then &lt;math&gt;n = 3a = 100b+23&lt;/math&gt; for some integers &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt;. Taking mod 3 gives:<br /> &lt;math&gt;0 \equiv b+2 \pmod 3 \implies b\equiv 1\pmod 3&lt;/math&gt; so &lt;math&gt;b=3c+1 &lt;/math&gt; for some integer &lt;math&gt;c&lt;/math&gt;. But &lt;math&gt;N = 100b+23 = 100(3c+1)+23&lt;/math&gt; so &lt;math&gt;n = 300c+123&lt;/math&gt;. Bounding this gives us:<br /> &lt;math&gt;999 &lt; 300c+123 &lt; 10000&lt;/math&gt; so &lt;math&gt;876 &lt; 300c &lt; 9877&lt;/math&gt;. Dividing by &lt;math&gt;300&lt;/math&gt; gives &lt;math&gt;2.92 &lt; c &lt; 32.9222&lt;/math&gt; so &lt;math&gt;3\le c \le 32&lt;/math&gt;. This gives &lt;math&gt;32-3+1 = \boxed{\textbf{(B)} \ 30 }&lt;/math&gt;<br /> <br /> ===Solution 4 (Extremely Fast)===<br /> <br /> The number is in the form &lt;math&gt;xy23&lt;/math&gt;<br /> <br /> Notice that the number is divisible by &lt;math&gt;3&lt;/math&gt; if the sum of the digits of &lt;math&gt;xy23&lt;/math&gt; is divisible by &lt;math&gt;3&lt;/math&gt;<br /> <br /> Our first number that is divisible by &lt;math&gt;3&lt;/math&gt; is &lt;math&gt;1023&lt;/math&gt;, next is &lt;math&gt;1323&lt;/math&gt;..<br /> Notice &lt;math&gt;xy&lt;/math&gt; goes from &lt;math&gt;10\implies 97&lt;/math&gt;<br /> <br /> Hence, there are &lt;math&gt;\frac{97-10}{3}+1=30&lt;/math&gt; distinct four digit numbers.<br /> <br /> === Solution 5 (Even Faster than extremely fast) ===<br /> <br /> Following the form &lt;math&gt;xy23&lt;/math&gt; in Solution 4, notice that &lt;math&gt;x+y \equiv 1 \pmod 3&lt;/math&gt; to satisfy our condition. Choose a value of &lt;math&gt;x&lt;/math&gt; from 1 to 9. For &lt;math&gt;x=1, 4, 7&lt;/math&gt;, there are exactly 4 values of &lt;math&gt;y: 0, 3, 6, 9&lt;/math&gt;. For the remaining 6 digits, there are 3 choices for y. So our answer is &lt;math&gt;(3)(4)+(6)(3)=30&lt;/math&gt;.<br /> <br /> === Solution 6 (Even Faster than Even Faster than Extremely Fast)===<br /> There are &lt;math&gt;9 \cdot 10 = 90&lt;/math&gt; possible values for the first two digits. One-third of them yield a multiple of &lt;math&gt;3&lt;/math&gt;, so the answer is &lt;math&gt;\frac{90}{3} = \boxed{\textbf{(B)}\ 30}&lt;/math&gt;<br /> <br /> == See also ==<br /> {{AMC10 box|year=2003|num-b=24|after=Last problem|ab=B}}<br /> <br /> <br /> {{MAA Notice}}</div> Ctw0611 https://artofproblemsolving.com/wiki/index.php?title=2003_AMC_10B_Problems/Problem_25&diff=113970 2003 AMC 10B Problems/Problem 25 2020-01-01T20:49:25Z <p>Ctw0611: /* Solution 4(extremely fast) */</p> <hr /> <div>==Problem 25==<br /> <br /> How many distinct four-digit numbers are divisible by &lt;math&gt;3&lt;/math&gt; and have &lt;math&gt;23&lt;/math&gt; as their last two digits?<br /> <br /> &lt;math&gt;\textbf{(A) } 27 \qquad\textbf{(B) } 30 \qquad\textbf{(C) } 33 \qquad\textbf{(D) } 81 \qquad\textbf{(E) } 90&lt;/math&gt;<br /> <br /> ==Solution==<br /> ===Solution 1===<br /> To test if a number is divisible by three, you add up the digits of the number. If the sum is divisible by three, then the original number is a multiple of three. If the sum is too large, you can repeat the process until you can tell whether it is a multiple of three. This is the basis for the solution. Since the last two digits are &lt;math&gt; 23 &lt;/math&gt;, the sum of the digits is &lt;math&gt; 2+3 = 5 &lt;/math&gt; (therefore it is not divisible by three). However certain numbers can be added to make the sum of the digits a multiple of three.<br /> <br /> &lt;math&gt; 5+1 = 6 &lt;/math&gt;,<br /> <br /> &lt;math&gt; 5+4 = 9 &lt;/math&gt;, and so on. <br /> <br /> However since the largest four-digit number ending with &lt;math&gt; 23 &lt;/math&gt; is &lt;math&gt; 9923 &lt;/math&gt;, the maximum sum is<br /> <br /> &lt;math&gt; 5+18 = 23 &lt;/math&gt;.<br /> <br /> Using that process we can fairly quickly compile a list of the sum of the first two digits of the number. <br /> <br /> &lt;cmath&gt; \{1, 4, 7, 10, 13, 16\} &lt;/cmath&gt;<br /> <br /> Now we find all the two-digit numbers that have any of the sums shown above. We can do this by listing all the two digit numbers &lt;math&gt; xy &lt;/math&gt; in separate cases.<br /> <br /> &lt;cmath&gt; I. x+y = 1, \{10\} = 1 &lt;/cmath&gt;<br /> &lt;cmath&gt; II. x+y = 4, \{13, 22, 31, 40\} = 4 &lt;/cmath&gt;<br /> &lt;cmath&gt; III. x+y = 7, \{16, 25, 34, 43, 52, 61, 70\} = 7 &lt;/cmath&gt;<br /> &lt;cmath&gt; IV. x+y = 10, \{19, 28, 37, 46, 55, 64, 73, 82, 91\} = 9 &lt;/cmath&gt;<br /> &lt;cmath&gt; V. x+y = 13, \{49, 58, 67, 76, 85, 94\} = 6 &lt;/cmath&gt;<br /> &lt;cmath&gt; VI. x+y = 16, \{79, 88, 97\} = 3 &lt;/cmath&gt;<br /> <br /> And finally, we add the number of elements in each set.<br /> <br /> &lt;cmath&gt; 1+4+7+9+6+3 = \boxed{\textbf{(B)}\ 30} &lt;/cmath&gt;<br /> <br /> ===Solution 2===<br /> A number divisible by &lt;math&gt;3&lt;/math&gt; has all its digits add to a multiple of &lt;math&gt;3.&lt;/math&gt; The last two digits are &lt;math&gt;2&lt;/math&gt; and &lt;math&gt;3&lt;/math&gt; and add up to &lt;math&gt;5 \equiv 2\ (\text{mod}\ 3).&lt;/math&gt; Therefore the first two digits must add up to &lt;math&gt;1\ (\text{mod}\ 3).&lt;/math&gt; &lt;math&gt;4&lt;/math&gt; digits (including &lt;math&gt;0&lt;/math&gt;) are &lt;math&gt;0\ (\text{mod}\ 3),&lt;/math&gt; &lt;math&gt;3&lt;/math&gt; are &lt;math&gt;1\ (\text{mod}\ 3),&lt;/math&gt; and &lt;math&gt;3&lt;/math&gt; are &lt;math&gt;2\ (\text{mod}\ 3).&lt;/math&gt; The following combinations are equivalent to &lt;math&gt;1\ (\text{mod}\ 3)&lt;/math&gt;:<br /> <br /> &lt;cmath&gt; 0\ (\text{mod}\ 3)+1\ (\text{mod}\ 3) \equiv 1\ (\text{mod}\ 3)+0\ (\text{mod}\ 3) \equiv 2\ (\text{mod}\ 3) +2\ (\text{mod}\ 3)&lt;/cmath&gt;<br /> <br /> Let the first term in each combination represent the thousands digit and the second term represent the hundreds digit. We can use this to find the total amount of four-digit numbers.<br /> <br /> &lt;cmath&gt;3\cdot3 + 3\cdot4 + 3\cdot3 = 9 + 12 + 9 = \boxed{\textbf{(B)}\ 30}&lt;/cmath&gt;<br /> <br /> ===Solution 3 ===<br /> <br /> We have the following: &lt;math&gt; n \equiv 0 \pmod{3}&lt;/math&gt; and &lt;math&gt;n \equiv 23\pmod{100}&lt;/math&gt;. Then &lt;math&gt;n = 3a = 100b+23&lt;/math&gt; for some integers &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt;. Taking mod 3 gives:<br /> &lt;math&gt;0 \equiv b+2 \pmod 3 \implies b\equiv 1\pmod 3&lt;/math&gt; so &lt;math&gt;b=3c+1 &lt;/math&gt; for some integer &lt;math&gt;c&lt;/math&gt;. But &lt;math&gt;N = 100b+23 = 100(3c+1)+23&lt;/math&gt; so &lt;math&gt;n = 300c+123&lt;/math&gt;. Bounding this gives us:<br /> &lt;math&gt;999 &lt; 300c+123 &lt; 10000&lt;/math&gt; so &lt;math&gt;876 &lt; 300c &lt; 9877&lt;/math&gt;. Dividing by &lt;math&gt;300&lt;/math&gt; gives &lt;math&gt;2.92 &lt; c &lt; 32.9222&lt;/math&gt; so &lt;math&gt;3\le c \le 32&lt;/math&gt;. This gives &lt;math&gt;32-3+1 = \boxed{\textbf{(B)} \ 30 }&lt;/math&gt;<br /> <br /> ===Solution 4 (extremely fast)===<br /> <br /> The number is in the form &lt;math&gt;xy23&lt;/math&gt;<br /> <br /> Notice that the number is divisible by &lt;math&gt;3&lt;/math&gt; if the sum of the digits of &lt;math&gt;xy23&lt;/math&gt; is divisible by &lt;math&gt;3&lt;/math&gt;<br /> <br /> Our first number that is divisible by &lt;math&gt;3&lt;/math&gt; is &lt;math&gt;1023&lt;/math&gt;, next is &lt;math&gt;1323&lt;/math&gt;..<br /> Notice &lt;math&gt;xy&lt;/math&gt; goes from &lt;math&gt;10\implies 97&lt;/math&gt;<br /> <br /> Hence, there are &lt;math&gt;\frac{97-10}{3}+1=30&lt;/math&gt; distinct four digit numbers.<br /> <br /> === Solution 5 (Even Faster than extremely fast) ===<br /> <br /> Following the form &lt;math&gt;xy23&lt;/math&gt; in Solution 4, notice that &lt;math&gt;x+y \equiv 1 \pmod 3&lt;/math&gt; to satisfy our condition. Choose a value of &lt;math&gt;x&lt;/math&gt; from 1 to 9. For &lt;math&gt;x=1, 4, 7&lt;/math&gt;, there are exactly 4 values of &lt;math&gt;y: 0, 3, 6, 9&lt;/math&gt;. For the remaining 6 digits, there are 3 choices for y. So our answer is &lt;math&gt;(3)(4)+(6)(3)=30&lt;/math&gt;.<br /> <br /> === Solution 6 (Even Faster than Even Faster than Extremely Fast)===<br /> There are &lt;math&gt;9 \cdot 10 = 90&lt;/math&gt; possible values for the first two digits. One-third of them yield a multiple of &lt;math&gt;3&lt;/math&gt;, so the answer is &lt;math&gt;\frac{90}{3} = \boxed{\textbf{(B)}\ 30}&lt;/math&gt;<br /> <br /> == See also ==<br /> {{AMC10 box|year=2003|num-b=24|after=Last problem|ab=B}}<br /> <br /> <br /> {{MAA Notice}}</div> Ctw0611 https://artofproblemsolving.com/wiki/index.php?title=2001_AMC_12_Problems/Problem_11&diff=113488 2001 AMC 12 Problems/Problem 11 2019-12-26T23:40:10Z <p>Ctw0611: /* Solution 2 */</p> <hr /> <div>{{duplicate|[[2001 AMC 12 Problems|2001 AMC 12 #11]] and [[2001 AMC 10 Problems|2001 AMC 10 #23]]}}<br /> <br /> == Problem ==<br /> <br /> A box contains exactly five chips, three red and two white. Chips are randomly removed one at a time without replacement until all the red chips are drawn or all the white chips are drawn. What is the probability that the last chip drawn is white?<br /> <br /> &lt;math&gt;<br /> \text{(A) }\frac {3}{10}<br /> \qquad<br /> \text{(B) }\frac {2}{5}<br /> \qquad<br /> \text{(C) }\frac {1}{2}<br /> \qquad<br /> \text{(D) }\frac {3}{5}<br /> \qquad<br /> \text{(E) }\frac {7}{10}<br /> &lt;/math&gt;<br /> <br /> == Solution 1 ==<br /> <br /> Imagine that we draw all the chips in random order, i.e., we do not stop when the last chip of a color is drawn. To draw out all the white chips first, the last chip left must be red, and all previous chips can be drawn in any order. Since there are 3 red chips, the probability that the last chip of the five is red (and so also the probability that the last chip drawn is white) is &lt;math&gt;\boxed{(\text{D}) \frac {3}{5}}&lt;/math&gt;.<br /> <br /> ==Solution 2 ==<br /> <br /> Let's assume we don't stop picking until all of the chips are picked. To satisfy this condition, we have to arrange the letters: ''W, W, R, R, R'' such that both ''R's'' appear in the first 4. We find the number of ways to arrange the red chips in the first 4 and divide that by the total ways to choose all the chips. The probability of this occurring is &lt;math&gt;\dfrac{\dbinom{4}{2}}{\dbinom{5}{2}} = \boxed{\text{(D)}\dfrac35}&lt;/math&gt;<br /> <br /> == See Also ==<br /> {{AMC10 box|year=2001|num-b=22|num-a=24}}<br /> {{AMC12 box|year=2001|num-b=10|num-a=12}}<br /> {{MAA Notice}}</div> Ctw0611 https://artofproblemsolving.com/wiki/index.php?title=2000_AMC_10_Problems/Problem_17&diff=113426 2000 AMC 10 Problems/Problem 17 2019-12-25T21:41:46Z <p>Ctw0611: /* Solution */</p> <hr /> <div>==Problem==<br /> <br /> Boris has an incredible coin changing machine. When he puts in a quarter, it returns five nickels; when he puts in a nickel, it returns five pennies; and when he puts in a penny, it returns five quarters. Boris starts with just one penny. Which of the following amounts could Boris have after using the machine repeatedly?<br /> <br /> &lt;math&gt;\mathrm{(A)}&lt;/math&gt; &lt;math&gt;\$3.63&lt;/math&gt; <br /> <br /> &lt;math&gt;\mathrm{(B)}&lt;/math&gt; &lt;math&gt;\$5.13&lt;/math&gt;<br /> <br /> &lt;math&gt;\mathrm{(C)}&lt;/math&gt; &lt;math&gt;\$6.30&lt;/math&gt; <br /> <br /> &lt;math&gt;\mathrm{(D)}&lt;/math&gt; &lt;math&gt;\$7.45&lt;/math&gt; <br /> <br /> &lt;math&gt;\mathrm{(E)}&lt;/math&gt; &lt;math&gt;\$9.07&lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> Consider what happens each time he puts a coin in. If he puts in a quarter, he gets five nickels back, so the amount of money he has doesn't change. Similarly, if he puts a nickel in the machine, he gets five pennies back and the money value doesn't change. However, if he puts a penny in, he gets five quarters back, increasing the amount of money he has by &lt;math&gt;124&lt;/math&gt; cents. <br /> <br /> This implies that the only possible values, in cents, he can have are the ones one more than a multiple of &lt;math&gt;124&lt;/math&gt;. Of the choices given, the only one is &lt;math&gt;\boxed{\text{D}}&lt;/math&gt;<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2000|num-b=16|num-a=18}}<br /> {{MAA Notice}}</div> Ctw0611 https://artofproblemsolving.com/wiki/index.php?title=2000_AMC_10_Problems/Problem_17&diff=113425 2000 AMC 10 Problems/Problem 17 2019-12-25T21:38:37Z <p>Ctw0611: /* Solution */</p> <hr /> <div>==Problem==<br /> <br /> Boris has an incredible coin changing machine. When he puts in a quarter, it returns five nickels; when he puts in a nickel, it returns five pennies; and when he puts in a penny, it returns five quarters. Boris starts with just one penny. Which of the following amounts could Boris have after using the machine repeatedly?<br /> <br /> &lt;math&gt;\mathrm{(A)}&lt;/math&gt; &lt;math&gt;\$3.63&lt;/math&gt; <br /> <br /> &lt;math&gt;\mathrm{(B)}&lt;/math&gt; &lt;math&gt;\$5.13&lt;/math&gt;<br /> <br /> &lt;math&gt;\mathrm{(C)}&lt;/math&gt; &lt;math&gt;\$6.30&lt;/math&gt; <br /> <br /> &lt;math&gt;\mathrm{(D)}&lt;/math&gt; &lt;math&gt;\$7.45&lt;/math&gt; <br /> <br /> &lt;math&gt;\mathrm{(E)}&lt;/math&gt; &lt;math&gt;\$9.07&lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> Consider what happens each time he puts a coin in. If he puts in a quarter, he gets five nickels back, so the amount of money he has doesn't change. Similarly, if he puts a nickel in the machine, he gets five pennies back and the money value doesn't change. However, if he puts a penny in, he gets five quarters back, increasing the amount of money he has by &lt;math&gt;24&lt;/math&gt; cents. <br /> <br /> This implies that the only possible values, in cents, he can have are the ones one more than a multiple of &lt;math&gt;24&lt;/math&gt;. Of the choices given, the only one is &lt;math&gt;\boxed{\text{D}}&lt;/math&gt;<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2000|num-b=16|num-a=18}}<br /> {{MAA Notice}}</div> Ctw0611 https://artofproblemsolving.com/wiki/index.php?title=2000_AMC_10_Problems/Problem_17&diff=113424 2000 AMC 10 Problems/Problem 17 2019-12-25T21:36:36Z <p>Ctw0611: /* Solution */</p> <hr /> <div>==Problem==<br /> <br /> Boris has an incredible coin changing machine. When he puts in a quarter, it returns five nickels; when he puts in a nickel, it returns five pennies; and when he puts in a penny, it returns five quarters. Boris starts with just one penny. Which of the following amounts could Boris have after using the machine repeatedly?<br /> <br /> &lt;math&gt;\mathrm{(A)}&lt;/math&gt; &lt;math&gt;\$3.63&lt;/math&gt; <br /> <br /> &lt;math&gt;\mathrm{(B)}&lt;/math&gt; &lt;math&gt;\$5.13&lt;/math&gt;<br /> <br /> &lt;math&gt;\mathrm{(C)}&lt;/math&gt; &lt;math&gt;\$6.30&lt;/math&gt; <br /> <br /> &lt;math&gt;\mathrm{(D)}&lt;/math&gt; &lt;math&gt;\$7.45&lt;/math&gt; <br /> <br /> &lt;math&gt;\mathrm{(E)}&lt;/math&gt; &lt;math&gt;\\$9.07&lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> Consider what happens each time he puts a coin in. If he puts in a quarter, he gets five nickels back, so the amount of money he has doesn't change. Similarly, if he puts a nickel in the machine, he gets five pennies back and the money value doesn't change. However, if he puts a penny in, he gets five quarters back, increasing the amount of money he has by &lt;math&gt;124&lt;/math&gt; cents. <br /> <br /> This implies that the only possible values, in cents, he can have are the ones one more than a multiple of &lt;math&gt;24&lt;/math&gt;. Of the choices given, the only one is &lt;math&gt;\boxed{\text{D}}&lt;/math&gt;<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2000|num-b=16|num-a=18}}<br /> {{MAA Notice}}</div> Ctw0611 https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_10B_Problems/Problem_24&diff=108359 2015 AMC 10B Problems/Problem 24 2019-08-07T02:01:43Z <p>Ctw0611: /* Problem */</p> <hr /> <div>==Problem==<br /> Erin the ant walks on the coordinate plane according to the following rules. He starts at the origin &lt;math&gt;p_0=(0,0)&lt;/math&gt; facing to the east and walks one unit, arriving at &lt;math&gt;p_1=(1,0)&lt;/math&gt;. For &lt;math&gt;n=1,2,3,\dots&lt;/math&gt;, right after arriving at the point &lt;math&gt;p_n&lt;/math&gt;, if Aaron can turn &lt;math&gt;90^\circ&lt;/math&gt; left and walk one unit to an unvisited point &lt;math&gt;p_{n+1}&lt;/math&gt;, he does that. Otherwise, he walks one unit straight ahead to reach &lt;math&gt;p_{n+1}&lt;/math&gt;. Thus the sequence of points continues &lt;math&gt;p_2=(1,1), p_3=(0,1), p_4=(-1,1), p_5=(-1,0)&lt;/math&gt;, and so on in a counterclockwise spiral pattern. What is &lt;math&gt;p_{2015}&lt;/math&gt;?<br /> <br /> &lt;math&gt; \textbf{(A) } (-22,-13)\qquad\textbf{(B) } (-13,-22)\qquad\textbf{(C) } (-13,22)\qquad\textbf{(D) } (13,-22)\qquad\textbf{(E) } (22,-13) &lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> ===Solution 1===<br /> The first thing we would do is track Aaron's footsteps:<br /> <br /> He starts by taking &lt;math&gt;1&lt;/math&gt; step East and &lt;math&gt;1&lt;/math&gt; step North, ending at &lt;math&gt;(1,1)&lt;/math&gt; after &lt;math&gt;2&lt;/math&gt; steps and about to head West.<br /> <br /> Then he takes &lt;math&gt;2&lt;/math&gt; steps West and &lt;math&gt;2&lt;/math&gt; steps South, ending at &lt;math&gt;(-1,-1&lt;/math&gt;) after &lt;math&gt;2+4&lt;/math&gt; steps, and about to head East.<br /> <br /> Then he takes &lt;math&gt;3&lt;/math&gt; steps East and &lt;math&gt;3&lt;/math&gt; steps North, ending at &lt;math&gt;(2,2)&lt;/math&gt; after &lt;math&gt;2+4+6&lt;/math&gt; steps, and about to head West.<br /> <br /> Then he takes &lt;math&gt;4&lt;/math&gt; steps West and &lt;math&gt;4&lt;/math&gt; steps South, ending at &lt;math&gt;(-2,-2)&lt;/math&gt; after &lt;math&gt;2+4+6+8&lt;/math&gt; steps, and about to head East.<br /> <br /> From this pattern, we can notice that for any integer &lt;math&gt;k \ge 1&lt;/math&gt; he's at &lt;math&gt;(-k, -k)&lt;/math&gt; after &lt;math&gt;2 + 4 + 6 + ... + 4k&lt;/math&gt; steps, and about to head East. There are &lt;math&gt;2k&lt;/math&gt; terms in the sum, with an average value of &lt;math&gt;(2 + 4k)/2 = 2k + 1&lt;/math&gt;, so:<br /> <br /> &lt;cmath&gt;2 + 4 + 6 + ... + 4k = 2k(2k + 1)&lt;/cmath&gt;<br /> <br /> If we substitute &lt;math&gt;k = 22&lt;/math&gt; into the equation: &lt;math&gt;44(45) = 1980 &lt; 2015&lt;/math&gt;. So he has &lt;math&gt;35&lt;/math&gt; moves to go. This makes him end up at &lt;math&gt;(-22+35,-22) = (13,-22) \implies \boxed{\textbf{(D)} (13, -22)}&lt;/math&gt;.<br /> <br /> (Like: 2)<br /> <br /> ===Solution 2===<br /> We are given that Aaron starts at &lt;math&gt;(0, 0)&lt;/math&gt;, and we note that his net steps follow the pattern of &lt;math&gt;+1&lt;/math&gt; in the &lt;math&gt;x&lt;/math&gt;-direction, &lt;math&gt;+1&lt;/math&gt; in the &lt;math&gt;y&lt;/math&gt;-direction, &lt;math&gt;-2&lt;/math&gt; in the &lt;math&gt;x&lt;/math&gt;-direction, &lt;math&gt;-2&lt;/math&gt; in the &lt;math&gt;y&lt;/math&gt;-direction, &lt;math&gt;+3&lt;/math&gt; in the &lt;math&gt;x&lt;/math&gt;-direction, &lt;math&gt;+3&lt;/math&gt; in the &lt;math&gt;y&lt;/math&gt;-direction, and so on, where we add odd and subtract even.<br /> <br /> We want &lt;math&gt;2 + 4 + 6 + 8 + ... + 2n = 2015&lt;/math&gt;, but it does not work out cleanly. Instead, we get that &lt;math&gt;2 + 4 + 6 + ... + 2(44) = 1980&lt;/math&gt;, which means that there are &lt;math&gt;35&lt;/math&gt; extra steps past adding &lt;math&gt;-44&lt;/math&gt; in the &lt;math&gt;x&lt;/math&gt;-direction (and the final number we add in the &lt;math&gt;y&lt;/math&gt;-direction is &lt;math&gt;-44&lt;/math&gt;).<br /> <br /> So &lt;math&gt;p_{2015} = (0+1-2+3-4+5...-44+35, 0+1-2+3-4+5...-44)&lt;/math&gt;.<br /> <br /> We can group &lt;math&gt;1-2+3-4+5...-44&lt;/math&gt; as &lt;math&gt;(1-2)+(3-4)+(5-6)+...+(43-44) = -22&lt;/math&gt;.<br /> <br /> Thus &lt;math&gt;p_{2015} = \boxed{\textbf{(D)}\; (13, -22)}&lt;/math&gt;.<br /> <br /> ===Solution 3===<br /> Looking at his steps, we see that he walks in a spiral shape. At the &lt;math&gt;8&lt;/math&gt;th step, he is on the bottom right corner of the &lt;math&gt;3\times 3&lt;/math&gt; square centered on the origin. On the &lt;math&gt;24&lt;/math&gt;th step, he is on the bottom right corner of the &lt;math&gt;5\times 5&lt;/math&gt; square centered at the origin. It seems that the &lt;math&gt;p_{n^2-1}&lt;/math&gt; is the bottom right corner of the &lt;math&gt;n\times n&lt;/math&gt; square. This makes sense since, after &lt;math&gt;n^2-1&lt;/math&gt;, he has been on &lt;math&gt;n^2&lt;/math&gt; dots, including the point &lt;math&gt;p_0&lt;/math&gt;. Also, this is only for odd &lt;math&gt;n&lt;/math&gt;, because starting with the &lt;math&gt;1\times 1&lt;/math&gt; square, we can only add one extra set of dots to each side, so we cannot get even &lt;math&gt;n&lt;/math&gt;. Since &lt;math&gt;45^2=2025&lt;/math&gt;, &lt;math&gt;p_{2024}&lt;/math&gt; is the bottom right corner of the &lt;math&gt;45\times 45&lt;/math&gt; square. This point is &lt;math&gt;\frac{45-1}{2}=22&lt;/math&gt; over to the right, and therefore &lt;math&gt;22&lt;/math&gt; down, so &lt;math&gt;p_{2024}=(22, -22)&lt;/math&gt;. Since &lt;math&gt;p_{2024}&lt;/math&gt; is &lt;math&gt;9&lt;/math&gt; ahead of &lt;math&gt;p_{2015}&lt;/math&gt;, we go back &lt;math&gt;9&lt;/math&gt; spaces to &lt;math&gt;\boxed{\textbf{(D)}\; (13, -22)}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2015|ab=B|num-b=23|num-a=25}}<br /> {{MAA Notice}}</div> Ctw0611