Difference between revisions of "2020 AMC 10A Problems/Problem 24"
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Alternatively, we could have said <math>\alpha = 40k+14 \equiv{0} \pmod{3}</math> for <math>k \equiv{1} \pmod{3}</math> only, so <math>k \equiv{0,2} \pmod{3}</math>, giving us our answer. Since the problem asks for the sum of the digits of <math>n</math>, our answer is <math>1+9+1+7=\boxed{\textbf{(C) } 18}</math>. | Alternatively, we could have said <math>\alpha = 40k+14 \equiv{0} \pmod{3}</math> for <math>k \equiv{1} \pmod{3}</math> only, so <math>k \equiv{0,2} \pmod{3}</math>, giving us our answer. Since the problem asks for the sum of the digits of <math>n</math>, our answer is <math>1+9+1+7=\boxed{\textbf{(C) } 18}</math>. | ||
− | ~Prabh1512 | + | ~Prabh1512 |
==Solution 2== | ==Solution 2== | ||
The conditions of the problem reduce to the following: <math>n+120 = 21k</math> and <math>n+63 = 60\ell</math>, where <math>\gcd(k,3) = 1</math> and <math>\gcd(\ell,2) = 1</math>. From these equations, we see that <math>21k - 60\ell = 57</math>. Solving this Diophantine equation gives us that <math>k = 20a + 17</math> and <math>\ell = 7a + 5</math>. Since, <math>n>1000</math>, we can do some bounding and get that <math>k > 53</math> and <math>\ell > 17</math>. Now we start bashing by plugging in numbers that satisfy these conditions: <math>a=4</math> is the first number that works so we get <math>\ell = 33</math>, <math>k = 97</math>. Therefore, we have <math>n=21(97)-120=60(33)-63=1917</math>, and our answer is <math>1+9+1+7=\boxed{\textbf{(C) } 18}</math>. | The conditions of the problem reduce to the following: <math>n+120 = 21k</math> and <math>n+63 = 60\ell</math>, where <math>\gcd(k,3) = 1</math> and <math>\gcd(\ell,2) = 1</math>. From these equations, we see that <math>21k - 60\ell = 57</math>. Solving this Diophantine equation gives us that <math>k = 20a + 17</math> and <math>\ell = 7a + 5</math>. Since, <math>n>1000</math>, we can do some bounding and get that <math>k > 53</math> and <math>\ell > 17</math>. Now we start bashing by plugging in numbers that satisfy these conditions: <math>a=4</math> is the first number that works so we get <math>\ell = 33</math>, <math>k = 97</math>. Therefore, we have <math>n=21(97)-120=60(33)-63=1917</math>, and our answer is <math>1+9+1+7=\boxed{\textbf{(C) } 18}</math>. | ||
+ | |||
+ | ==Solution 3== | ||
+ | We first find that <math>n\equiv6\pmod{21}</math> and <math>n\equiv57\pmod{60}</math>, then we get <math>n=21x+6</math> and <math>n=60y+57</math> by definitions, where <math>x</math> and <math>y</math> are integers. It follows that <math>y</math> must be odd, since the GCD will be <math>120</math> instead of <math>60</math> if <math>y</math> is even. Also, the unit digit of <math>n</math> must be <math>7</math>, since the unit digit of <math>60y</math> is always <math>0</math> and the unit digit of <math>57</math> is <math>7</math>. Therefore, we find that <math>x</math> must end in <math>1</math> to satisfy <math>n</math> having a unit digit of <math>7</math>. Also, we find that <math>x</math> must not be a multiple of <math>3</math> or else the GCD will be <math>63</math>. Therefore, we test values for <math>x</math> and find that <math>x=91</math> satisfies all these conditions. Therefore, <math>n=1917</math> and <math>1+9+1+7 = \boxed{\textbf{(C) } 18}</math>. | ||
+ | |||
+ | ~happykeeper | ||
==Solution 4== | ==Solution 4== | ||
− | + | We are given that <math>\gcd(63, n+120) =21</math> and <math>\gcd(n+63, 120)=60.</math> By applying the Euclidean algorithm in reverse, we have <cmath>\gcd(63, n+120) = \gcd(63, n+120 + 63) = \gcd(63, n+183) = 21</cmath> and <cmath>\gcd(n+63, 120) = \gcd(n+63 + 120, 120) = \gcd(n+183, 120) = 60.</cmath> | |
− | |||
− | |||
− | We are given that <math>\gcd(63, n+120) =21</math> and <math>\gcd(n+63, 120)=60.</math> By applying the Euclidean algorithm | ||
− | We now know that <math>n+183</math> must be divisible by <math>21</math> and <math>60,</math> so it is divisible by <math>\text{lcm}(21, 60) = 420.</math> Therefore, <math>n+183 = 420k</math> for some integer <math>k.</math> We know that <math>3 \nmid k,</math> or else the first condition won't hold (<math>\gcd</math> will be <math>63</math>) and <math>2 \nmid k,</math> or else the second condition won't hold (<math>\gcd</math> will be <math>120</math>). Since <math>k = 1</math> gives us too small of an answer, then <math>k=5 | + | We now know that <math>n+183</math> must be divisible by <math>21</math> and <math>60,</math> so it is divisible by <math>\text{lcm}(21, 60) = 420.</math> Therefore, <math>n+183 = 420k</math> for some integer <math>k.</math> We know that <math>3 \nmid k,</math> or else the first condition won't hold (<math>\gcd</math> will be <math>63</math>) and <math>2 \nmid k,</math> or else the second condition won't hold (<math>\gcd</math> will be <math>120</math>). Since <math>k = 1</math> gives us too small of an answer, then <math>k=5,</math> from which <math>n = 1917.</math> So, the answer is <math>1+9+1+7 = \boxed{\textbf{(C) } 18}.</math> |
− | ==Solution | + | ==Solution 5== |
− | <math>\gcd(n+63,120)=60</math> tells us <math>n+63\equiv60\pmod {120}</math>. The smallest <math>n+63</math> that satisfies the previous condition and <math>n>1000</math> is <math>1140</math>, so we start from there. If <math>n+63=1140</math>, then <math>n+120=1197</math>. Because <math>\gcd(n+120,63)=21</math>, <math>n+120\equiv21\pmod {63}</math> or <math>n+120\equiv42\pmod {63}</math>. We see that <math>1197\equiv0\pmod {63}</math>, which does not fulfill the requirement for <math>n+120</math>, so we continue by keep on adding <math>120</math> to <math>1197</math>, in order to also fulfill the requirement for <math>n+63</math>. Soon, we see that <math>n+120\pmod {63}</math> decreases by <math>6</math> every time we add <math>120</math>, so we can quickly see that <math>n=1917</math> because at that point <math>n+120\equiv21\pmod {63}</math>. We add up all digits of <math>1917</math> | + | <math>\gcd(n+63,120)=60</math> tells us <math>n+63\equiv60\pmod {120}</math>. The smallest <math>n+63</math> that satisfies the previous condition and <math>n>1000</math> is <math>1140</math>, so we start from there. If <math>n+63=1140</math>, then <math>n+120=1197</math>. Because <math>\gcd(n+120,63)=21</math>, <math>n+120\equiv21\pmod {63}</math> or <math>n+120\equiv42\pmod {63}</math>. We see that <math>1197\equiv0\pmod {63}</math>, which does not fulfill the requirement for <math>n+120</math>, so we continue by keep on adding <math>120</math> to <math>1197</math>, in order to also fulfill the requirement for <math>n+63</math>. Soon, we see that <math>n+120\pmod {63}</math> decreases by <math>6</math> every time we add <math>120</math>, so we can quickly see that <math>n=1917</math> because at that point <math>n+120\equiv21\pmod {63}</math>. We add up all digits of <math>1917</math> to get <math>\boxed{\textbf{(C) } 18}</math>. |
~SmileKat32 | ~SmileKat32 | ||
− | ==Solution | + | ==Solution 6== |
We are able to set up the following system of congruences: | We are able to set up the following system of congruences: | ||
<cmath>\begin{align*} | <cmath>\begin{align*} | ||
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~nikenissan ~Midnight | ~nikenissan ~Midnight | ||
− | == Solution | + | == Solution 7== |
First, we find <math>n</math>. We know that it is greater than <math>1000</math>, so we first input <math>n = 1000</math>. From the first equation, <math>\gcd(63, n + 120) = 21</math>, we know that if <math>n</math> is correct, after we add <math>120</math> to it, it should be divisible by <math>21</math>, but not <math>63</math>: | First, we find <math>n</math>. We know that it is greater than <math>1000</math>, so we first input <math>n = 1000</math>. From the first equation, <math>\gcd(63, n + 120) = 21</math>, we know that if <math>n</math> is correct, after we add <math>120</math> to it, it should be divisible by <math>21</math>, but not <math>63</math>: | ||
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~primegn | ~primegn | ||
− | ==Solution | + | ==Solution 8 (Euclidean Algorithm)== |
By the Euclidean Algorithm, we have | By the Euclidean Algorithm, we have | ||
<cmath>\begin{alignat*}{8} | <cmath>\begin{alignat*}{8} | ||
− | \gcd(63,n+120)&=\hspace{1mm}&&\gcd(63,\phantom{ }\underbrace{n+120-63k_1}_{(n+120) \ \mathrm{mod} \ 63}\phantom{ })&&=21, | + | \gcd(63,n+120)&=\hspace{1mm}&&\gcd(63,\phantom{ }\underbrace{n+120-63k_1}_{(n+120) \ \mathrm{mod} \ 63}\phantom{ })&&=21, \\ |
− | \gcd(n+63,120)&=&&\gcd(\phantom{ }\underbrace{n+63-120k_2}_{(n+63) \ \mathrm{mod} \ 120}\phantom{ },120)&&=60. | + | \gcd(n+63,120)&=&&\gcd(\phantom{ }\underbrace{n+63-120k_2}_{(n+63) \ \mathrm{mod} \ 120}\phantom{ },120)&&=60. |
\end{alignat*}</cmath> | \end{alignat*}</cmath> | ||
Clearly, <math>n+120-63k_1</math> must be either <math>21</math> or <math>42,</math> and <math>n+63-120k_2</math> must be <math>60.</math> | Clearly, <math>n+120-63k_1</math> must be either <math>21</math> or <math>42,</math> and <math>n+63-120k_2</math> must be <math>60.</math> | ||
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More generally, let <math>t\in\{1,2\},</math> so we get | More generally, let <math>t\in\{1,2\},</math> so we get | ||
<cmath>\begin{align*} | <cmath>\begin{align*} | ||
− | n+120-63k_1&=21t, &\hspace{55.5mm}(1 | + | n+120-63k_1&=21t, &\hspace{55.5mm}(1) \\ |
− | n+63-120k_2&=60. &\hspace{55.5mm}(2 | + | n+63-120k_2&=60. &\hspace{55.5mm}(2) |
\end{align*}</cmath> | \end{align*}</cmath> | ||
− | Subtracting <math>(2 | + | Subtracting <math>(2)</math> from <math>(1)</math> and then simplifying give |
<cmath>\begin{align*} | <cmath>\begin{align*} | ||
57-63k_1+120k_2&=21t-60 \\ | 57-63k_1+120k_2&=21t-60 \\ | ||
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k_2&\equiv2\pmod{7}. | k_2&\equiv2\pmod{7}. | ||
\end{align*}</cmath> | \end{align*}</cmath> | ||
− | Recall that <math>n>1000.</math> From <math>(2 | + | Recall that <math>n>1000.</math> From <math>(2),</math> it follows that <cmath>1063-120k_2<n+63-120k_2=60,</cmath> from which <math>k_2>8.</math> Therefore, the possible values for <math>k_2</math> are <math>9,16,23,\ldots.</math> |
− | We need to check whether positive integers <math>k_1</math> and <math>t</math> (where <math>t\in\{1,2\}</math>) exist in <math>(1 | + | We need to check whether positive integers <math>k_1</math> and <math>t</math> (where <math>t\in\{1,2\}</math>) exist in <math>(1):</math> |
<ul style="margin-left: 1.5em, list-style-type: square;"> | <ul style="margin-left: 1.5em, list-style-type: square;"> | ||
− | <li>If <math>k_2=9,</math> then substituting into <math>(2 | + | <li>If <math>k_2=9,</math> then substituting into <math>(2)</math> gives <math>n=1077.</math> Next, substituting into <math>(1)</math> produces <math>1197-63k_1=21t,</math> or <math>57-3k_1=t.</math> <p> |
There are no solutions <math>(k_1,t).</math></li><p> | There are no solutions <math>(k_1,t).</math></li><p> | ||
− | <li>If <math>k_2=16,</math> then substituting into <math>(2 | + | <li>If <math>k_2=16,</math> then substituting into <math>(2)</math> gives <math>n=1917.</math> Next, substituting into <math>(1)</math> produces <math>2037-63k_1=21t,</math> or <math>97-3k_1=t.</math> <p> |
The solution is <math>(k_1,t)=(32,1).</math></li><p> | The solution is <math>(k_1,t)=(32,1).</math></li><p> | ||
</ul> | </ul> | ||
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~MRENTHUSIASM | ~MRENTHUSIASM | ||
− | == Solution | + | == Solution 9 (Euclidean Algorithm) == |
Because we are finding value of <math>n</math> for <math>n > 1000</math>, let <math>n = 1000 + k</math>. | Because we are finding value of <math>n</math> for <math>n > 1000</math>, let <math>n = 1000 + k</math>. | ||
− | Using the | + | Using the Euclidean Algorithm, |
<cmath>\begin{align*} | <cmath>\begin{align*} | ||
− | \gcd(63, n+120) &= \gcd(63, 1000 + k + 120) = \gcd(63, k + 1120 - 63 \cdot 18) = \gcd(63, k-14) = 21, \\ | + | \gcd(63, n+120) &= \gcd(63, 1000 + k + 120) \\ |
− | \gcd(n+63, 120) &= \gcd(k + 1000 + 63, 120) = \gcd(k + 1000 + 63 - 120 \cdot 9, 120) = \gcd(k-17, 120) = 60. | + | &= \gcd(63, k + 1120 - 63 \cdot 18) \\ |
+ | &= \gcd(63, k-14) \\ | ||
+ | &= 21, \\ | ||
+ | \gcd(n+63, 120) &= \gcd(k + 1000 + 63, 120) \\ | ||
+ | &= \gcd(k + 1000 + 63 - 120 \cdot 9, 120) \\ | ||
+ | &= \gcd(k-17, 120) \\ | ||
+ | &= 60. | ||
\end{align*}</cmath> | \end{align*}</cmath> | ||
So, we have | So, we have | ||
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k &\equiv 17 \pmod{60}. | k &\equiv 17 \pmod{60}. | ||
\end{align*}</cmath> | \end{align*}</cmath> | ||
− | Let <math>k = 21p + 14 = 60q + 17</math>, <math>7p= 20q + 1</math>, <math>7p = 21q - (q - 1)</math>, <math>q - 1</math> is a multiple of <math>7</math>. | + | Let <math>k = 21p + 14 = 60q + 17</math>, <math>7p= 20q + 1</math> (by the Division Algorithm), <math>7p = 21q - (q - 1)</math>, <math>q - 1</math> is a multiple of <math>7</math>. |
Let <math>q-1 = 7r</math>, <math>q = 7r+1</math>, <math>k = 60(7r+1) + 17 = 420r + 77</math>. | Let <math>q-1 = 7r</math>, <math>q = 7r+1</math>, <math>k = 60(7r+1) + 17 = 420r + 77</math>. | ||
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By substituting <math>1077</math>, <math>1497</math>, <math>1917</math> into <math>\gcd(63, n+120) =21</math> and <math>\gcd(n+63, 120)=60</math>, <math>1077</math> and <math>1497</math> aren't valid answers, only <math>1917</math> is. Therefore, the answer is <math>1+9+1+7=\boxed{\textbf{(C) } 18}</math>. | By substituting <math>1077</math>, <math>1497</math>, <math>1917</math> into <math>\gcd(63, n+120) =21</math> and <math>\gcd(n+63, 120)=60</math>, <math>1077</math> and <math>1497</math> aren't valid answers, only <math>1917</math> is. Therefore, the answer is <math>1+9+1+7=\boxed{\textbf{(C) } 18}</math>. | ||
− | ~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen] | + | ~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen] (Solution) |
+ | |||
+ | ~[https://artofproblemsolving.com/wiki/index.php/User:Plainoldnumbertheory Plainoldnumbertheory] (Minor Edits) | ||
+ | |||
+ | ~[https://artofproblemsolving.com/wiki/index.php/User:Plainoldnumbertheory Puck_0] (Formatting) | ||
− | == Solution | + | == Solution 10 (Diophantine Equations)== |
We know <math>63 = 3 \cdot 21</math>, and <math>\gcd(63, n + 120) = 21 \Longrightarrow n + 120 = 21a</math>, where <math>a</math> is not a multiple of <math>3</math>. | We know <math>63 = 3 \cdot 21</math>, and <math>\gcd(63, n + 120) = 21 \Longrightarrow n + 120 = 21a</math>, where <math>a</math> is not a multiple of <math>3</math>. | ||
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~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen] | ~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen] | ||
− | == Solution | + | == Solution 11 (Chinese Remainder Theorem) == |
We have | We have | ||
<cmath>\begin{align*} | <cmath>\begin{align*} | ||
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~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen] | ~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen] | ||
+ | |||
+ | == Solution 12 (does not require ANY guess and check, Chinese Remainder Theorem) == | ||
+ | From <math>\gcd(63, n + 120) = 21</math>, we know that | ||
+ | <cmath> | ||
+ | \begin{align} | ||
+ | n + 120 &\equiv 0 \pmod{7} \Rightarrow n \equiv 6 \pmod{7} \tag{1} \\ | ||
+ | n + 120 &\equiv 0 \pmod{3} \Rightarrow n \equiv 0 \pmod{3} \Rightarrow n \equiv 0, 3, 6 \pmod{9} \notag | ||
+ | \end{align} | ||
+ | </cmath> | ||
+ | Since | ||
+ | <cmath> | ||
+ | \begin{align} | ||
+ | n + 120 &\not\equiv 0 \pmod{9} \Rightarrow n \not\equiv 6 \pmod{9} \notag | ||
+ | \end{align} | ||
+ | </cmath> | ||
+ | We must have | ||
+ | <cmath> | ||
+ | \begin{align} | ||
+ | n &\equiv 0 \pmod{9} \tag{2} | ||
+ | \end{align} | ||
+ | </cmath> | ||
+ | Or | ||
+ | <cmath> | ||
+ | \begin{align} | ||
+ | n &\equiv 3 \pmod{9} \tag{3} | ||
+ | \end{align} | ||
+ | </cmath> | ||
+ | (Note that <math>n + 120 \equiv 0 \pmod{9}</math> is impossible since then <math>9 | \gcd(63, n+120)</math>.) | ||
+ | |||
+ | From <math>\gcd(n+63, 120) = 60</math>, we have | ||
+ | <cmath> | ||
+ | \begin{align} | ||
+ | n + 63 &\equiv 0 \pmod{15} \Rightarrow n \equiv 12 \pmod {15} \tag{4} \\ | ||
+ | n + 63 &\equiv 0 \pmod{4} \Rightarrow n \equiv 1 \pmod {4} \Rightarrow n \equiv 1, 5 \pmod{8} \notag | ||
+ | \end{align} | ||
+ | </cmath> | ||
+ | Since | ||
+ | <cmath> | ||
+ | \begin{align} | ||
+ | n + 63 &\not\equiv 0 \pmod{8} \Rightarrow n \not\equiv 1 \pmod{8} \notag | ||
+ | \end{align} | ||
+ | </cmath> | ||
+ | We must have | ||
+ | <cmath> | ||
+ | \begin{align} | ||
+ | n &\equiv 5 \pmod {8} \tag{5} | ||
+ | \end{align} | ||
+ | </cmath> | ||
+ | Solving equations <math>(1)</math>, <math>(4)</math>, and <math>(5)</math>, we get | ||
+ | <cmath> | ||
+ | \begin{align} | ||
+ | n &\equiv 237 \pmod{840} \tag{6} | ||
+ | \end{align} | ||
+ | </cmath> | ||
+ | Solving <math>(6)</math> with <math>(2)</math>, we have | ||
+ | <cmath> | ||
+ | \begin{align} | ||
+ | n &\equiv 1917 \pmod{2520} \tag{7} | ||
+ | \end{align} | ||
+ | </cmath> | ||
+ | Solving <math>(6)</math> with <math>(3)</math>, we have | ||
+ | <cmath> | ||
+ | \begin{align} | ||
+ | n &\equiv 237 \pmod{2520} \tag{8} | ||
+ | \end{align} | ||
+ | </cmath> | ||
+ | From <math>(7)</math>, we obtain <math>1917</math>, and from <math>(8)</math>, we obtain <math>237 + 2520 = 2757</math> as the smallest solutions above <math>1000</math>. Thus, we get <math>1 + 9 + 1 + 7 = \boxed{\textbf{(C) } 18}</math>. | ||
==Video Solutions== | ==Video Solutions== | ||
===Video Solution 1 (Richard Rusczyk)=== | ===Video Solution 1 (Richard Rusczyk)=== | ||
− | https:// | + | https://www.youtube.com/watch?v=tk3yOGG2K-s& |
===Video Solution 2=== | ===Video Solution 2=== | ||
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===Video Solution 5=== | ===Video Solution 5=== | ||
https://youtu.be/R220vbM_my8?t=899 ~ amritvignesh0719062.0 | https://youtu.be/R220vbM_my8?t=899 ~ amritvignesh0719062.0 | ||
+ | |||
+ | ==Video Solution 6== | ||
+ | https://youtu.be/YcXsdSKVywk | ||
+ | |||
+ | ~savannahsolver | ||
==See Also== | ==See Also== |
Revision as of 12:09, 1 September 2023
Contents
- 1 Problem
- 2 Solution 1
- 3 Solution 2
- 4 Solution 3
- 5 Solution 4
- 6 Solution 5
- 7 Solution 6
- 8 Solution 7
- 9 Solution 8 (Euclidean Algorithm)
- 10 Solution 9 (Euclidean Algorithm)
- 11 Solution 10 (Diophantine Equations)
- 12 Solution 11 (Chinese Remainder Theorem)
- 13 Solution 12 (does not require ANY guess and check, Chinese Remainder Theorem)
- 14 Video Solutions
- 15 Video Solution 6
- 16 See Also
Problem
Let be the least positive integer greater than
for which
What is the sum of the digits of ?
Solution 1
We know that and
by the Euclidean Algorithm. Hence, let
and
, where
and
. Subtracting the two equations,
. Letting
, we get
. Taking modulo
, we have
. We are given that
, so
. Notice that if
then the condition
is violated. The next possible value of
satisfies the given condition, giving us
.
Alternatively, we could have said for
only, so
, giving us our answer. Since the problem asks for the sum of the digits of
, our answer is
.
~Prabh1512
Solution 2
The conditions of the problem reduce to the following: and
, where
and
. From these equations, we see that
. Solving this Diophantine equation gives us that
and
. Since,
, we can do some bounding and get that
and
. Now we start bashing by plugging in numbers that satisfy these conditions:
is the first number that works so we get
,
. Therefore, we have
, and our answer is
.
Solution 3
We first find that and
, then we get
and
by definitions, where
and
are integers. It follows that
must be odd, since the GCD will be
instead of
if
is even. Also, the unit digit of
must be
, since the unit digit of
is always
and the unit digit of
is
. Therefore, we find that
must end in
to satisfy
having a unit digit of
. Also, we find that
must not be a multiple of
or else the GCD will be
. Therefore, we test values for
and find that
satisfies all these conditions. Therefore,
and
.
~happykeeper
Solution 4
We are given that and
By applying the Euclidean algorithm in reverse, we have
and
We now know that must be divisible by
and
so it is divisible by
Therefore,
for some integer
We know that
or else the first condition won't hold (
will be
) and
or else the second condition won't hold (
will be
). Since
gives us too small of an answer, then
from which
So, the answer is
Solution 5
tells us
. The smallest
that satisfies the previous condition and
is
, so we start from there. If
, then
. Because
,
or
. We see that
, which does not fulfill the requirement for
, so we continue by keep on adding
to
, in order to also fulfill the requirement for
. Soon, we see that
decreases by
every time we add
, so we can quickly see that
because at that point
. We add up all digits of
to get
.
~SmileKat32
Solution 6
We are able to set up the following system of congruences:
Therefore, by definition, we are able to set-up the following system of equations:
Thus, we have
from which
We know
and since
therefore
Simplifying this congruence further, we have
Thus, by definition,
Substituting this back into our original equation, we get
By definition, we are able to set up the following congruence:
Thus,
, so our answer is simply
.
Remark
Since we conclude that
Since we conclude that
Remember that
Lastly, the reason why is
would be divisible by
, which is not possible due to the certain condition.
~nikenissan ~Midnight
Solution 7
First, we find . We know that it is greater than
, so we first input
. From the first equation,
, we know that if
is correct, after we add
to it, it should be divisible by
, but not
:
This does not work. To get to the nearest number divisible by
, we have to add
to cancel out the remainder. (Note that we don't subtract
to get to
;
is already at its lowest possible value!)
Adding
to
gives us
. (Note:
is currently divisible by 63, but that's fine since we'll be changing it in the next step.)
Now using the second equation, , we know that if
is correct, then
is divisible by
but not
:
Again, this does not work. This requires some guessing and checking. We can add
over and over again until
is valid. This changes
while also maintaining that
has no remainders.
After adding
once, we get
. By pure luck, adding
two more times gives us
with no remainders.
We now have
. However, this number is divisible by
. To get the next possible number, we add the LCM of
and
(once again, to maintain divisibility), which is
. Unfortunately,
is still divisible by
. Adding
again gives us
, which is valid. However, remember that this is equal to
, so subtracting
from
gives us
, which is
.
The sum of its digits are .
~primegn
Solution 8 (Euclidean Algorithm)
By the Euclidean Algorithm, we have
Clearly,
must be either
or
and
must be
More generally, let so we get
Subtracting
from
and then simplifying give
Taking
modulo
produces
Recall that
From
it follows that
from which
Therefore, the possible values for
are
We need to check whether positive integers and
(where
) exist in
- If
then substituting into
gives
Next, substituting into
produces
or
There are no solutions
- If
then substituting into
gives
Next, substituting into
produces
or
The solution is
Finally, the least such positive integer is
The sum of its digits is
~MRENTHUSIASM
Solution 9 (Euclidean Algorithm)
Because we are finding value of for
, let
.
Using the Euclidean Algorithm,
So, we have
Let
,
(by the Division Algorithm),
,
is a multiple of
.
Let ,
,
.
,
.
By substituting ,
,
into
and
,
and
aren't valid answers, only
is. Therefore, the answer is
.
~isabelchen (Solution)
~Plainoldnumbertheory (Minor Edits)
~Puck_0 (Formatting)
Solution 10 (Diophantine Equations)
We know , and
, where
is not a multiple of
.
Also, , and
, where
is not a multiple of
.
Let ,
,
.
Now the problem becomes and
.
Meaning has to be a multiple of
but not
, and
is a multiple of
but not
.
Using trial and error, the least values are and
.
Therefore, the answer is .
Solution 11 (Chinese Remainder Theorem)
We have
So, we conclude that
and
, respectively.
Because the moduli
and
are not relatively prime, namely
,
, and
, we convert the system of
linear congruences with non-coprime moduli into a system of
linear congruences with coprime moduli:
By Chinese Remainder Theorem, the general solution of system of
linear congruences is
We construct the following table:
Only
satisfies
and
. Therefore, the answer is
.
Solution 12 (does not require ANY guess and check, Chinese Remainder Theorem)
From , we know that
Since
We must have
Or
(Note that
is impossible since then
.)
From , we have
Since
We must have
Solving equations
,
, and
, we get
Solving
with
, we have
Solving
with
, we have
From
, we obtain
, and from
, we obtain
as the smallest solutions above
. Thus, we get
.
Video Solutions
Video Solution 1 (Richard Rusczyk)
https://www.youtube.com/watch?v=tk3yOGG2K-s&
Video Solution 2
https://youtu.be/8mNMKH0T9W0 - Happytwin
Video Solution 3 (Quick & Simple)
Education The Study of Everything
Video Solution 4
https://www.youtube.com/watch?v=gdGmSyzR908&list=PLLCzevlMcsWNcTZEaxHe8VaccrhubDOlQ&index=5 ~ MathEx
Video Solution 5
https://youtu.be/R220vbM_my8?t=899 ~ amritvignesh0719062.0
Video Solution 6
~savannahsolver
See Also
2020 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 23 |
Followed by Problem 25 | |
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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