Difference between revisions of "2022 AIME I Problems/Problem 2"
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Find the three-digit positive integer <math>\underline{a}\,\underline{b}\,\underline{c}</math> whose representation in base nine is <math>\underline{b}\,\underline{c}\,\underline{a}_{\,\text{nine}},</math> where <math>a,</math> <math>b,</math> and <math>c</math> are (not necessarily distinct) digits. | Find the three-digit positive integer <math>\underline{a}\,\underline{b}\,\underline{c}</math> whose representation in base nine is <math>\underline{b}\,\underline{c}\,\underline{a}_{\,\text{nine}},</math> where <math>a,</math> <math>b,</math> and <math>c</math> are (not necessarily distinct) digits. | ||
− | == Solution == | + | == Solution 1 == |
We are given that <cmath>100a + 10b + c = 81b + 9c + a,</cmath> which rearranges to <cmath>99a = 71b + 8c.</cmath> | We are given that <cmath>100a + 10b + c = 81b + 9c + a,</cmath> which rearranges to <cmath>99a = 71b + 8c.</cmath> | ||
Line 16: | Line 16: | ||
~MRENTHUSIASM | ~MRENTHUSIASM | ||
+ | |||
+ | == Solution 2 == | ||
+ | |||
+ | As shown in Solution 1, we get <math>99a = 71b+8c</math>. | ||
+ | |||
+ | Note that <math>99</math> and <math>71</math> are large numbers comparatively to <math>8</math>, so we hypothesize that <math>a</math> and <math>b</math> are equal and <math>8c</math> fills the gap between them. The difference between <math>99</math> and <math>71</math> is <math>28</math>, which is a multiple of <math>4</math>. So, if we multiply this by <math>2</math>, it will be a multiple of <math>8</math> and thus the gap can be filled. Therefore, the only solution is <math>(a,b,c)=(2,2,7)</math>, and the answer is <math>\underline{a}\,\underline{b}\,\underline{c}=\boxed{227}</math>. | ||
+ | |||
+ | ~KingRavi | ||
+ | |||
+ | == Solution 3 == | ||
+ | |||
+ | As shown in Solution 1, we get <math>99a = 71b+8c.</math> | ||
+ | |||
+ | We list a few multiples of <math>99</math> out: <cmath>99,198,297,396.</cmath> | ||
+ | Of course, <math>99</math> can't be made of just <math>8</math>'s. If we use one <math>71</math>, we get a remainder of <math>28</math>, which can't be made of <math>8</math>'s either. So <math>99</math> doesn't work. | ||
+ | <math>198</math> can't be made up of just <math>8</math>'s. If we use one <math>71</math>, we get a remainder of <math>127</math>, which can't be made of <math>8</math>'s. If we use two <math>71</math>'s, we get a remainder of <math>56</math>, which can be made of <math>8</math>'s. | ||
+ | Therefore we get <math>99\cdot2=71\cdot2+8\cdot7</math> so <math>a=2,b=2,</math> and <math>c=7</math>. Plugging this back into the original problem shows that this answer is indeed correct. Therefore, <math>\underline{a}\,\underline{b}\,\underline{c}=\boxed{227}.</math> | ||
+ | |||
+ | ~Technodoggo | ||
+ | |||
+ | == Solution 4 == | ||
+ | As shown in Solution 1, we get <math>99a = 71b+8c</math>. | ||
+ | |||
+ | We can see that <math>99</math> is <math>28</math> larger than <math>71</math>, and we have an <math>8c</math>. We can clearly see that <math>56</math> is a multiple of <math>8</math>, and any larger than <math>56</math> would result in <math>c</math> being larger than <math>9</math>. Therefore, our only solution is <math>a = 2, b = 2, c = 7</math>. Our answer is <math>\underline{a}\,\underline{b}\,\underline{c}=\boxed{227}</math>. | ||
+ | |||
+ | ~Arcticturn | ||
+ | |||
+ | == Solution 5 == | ||
+ | As shown in Solution 1, we get <math>99a = 71b+8c,</math> which rearranges to <cmath>99(a – b) = 8c – 28 b = 4(2c – 7b) \le 4(2\cdot 9 - 0 ) = 72.</cmath> | ||
+ | So <math>a=b, 2c = 7b \implies c=7, b=2,a=2.</math> | ||
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | |||
+ | == Solution 6 == | ||
+ | |||
+ | As shown in Solution 1, we have that <math>99a = 71b + 8c</math>. | ||
+ | |||
+ | Note that by the divisibility rule for <math>9</math>, we have <math>a+b+c \equiv a \pmod{9}</math>. Since <math>b</math> and <math>c</math> are base-<math>9</math> digits, we can say that <math>b+c = 0</math> or <math>b+c=9</math>. The former possibility can be easily eliminated, and thus <math>b+c=9</math>. Next, we write the equation from Solution 1 as <math>99a = 63b + 8(b+c)</math>, and dividing this by <math>9</math> gives <math>11a = 7b+8</math>. Taking both sides modulo <math>7</math>, we have <math>4a \equiv 1 \pmod{7}</math>. Multiplying both sides by <math>2</math> gives <math>a\equiv 2 \pmod{7}</math>, which implies <math>a=2</math>. From here, we can find that <math>b=2</math> and <math>c=7</math>, giving an answer of <math>\boxed{227}</math>. | ||
+ | |||
+ | ~Sedro | ||
+ | |||
+ | == Video Solution by OmegaLearn == | ||
+ | https://youtu.be/SCGzEOOICr4?t=340 | ||
+ | |||
+ | ~ pi_is_3.14 | ||
==Video Solution (Mathematical Dexterity)== | ==Video Solution (Mathematical Dexterity)== | ||
https://www.youtube.com/watch?v=z5Y4bT5rL-s | https://www.youtube.com/watch?v=z5Y4bT5rL-s | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://www.youtube.com/watch?v=CwSkAHR3AcM | ||
+ | |||
+ | ~Steven Chen (www.professorchenedu.com) | ||
+ | |||
+ | == Video Solution == | ||
+ | |||
+ | https://youtu.be/MJ_M-xvwHLk?t=392 | ||
+ | |||
+ | ~ThePuzzlr | ||
+ | |||
+ | ==Video Solution by MRENTHUSIASM (English & Chinese)== | ||
+ | https://www.youtube.com/watch?v=v4tHtlcD9ww&t=360s&ab_channel=MRENTHUSIASM | ||
+ | |||
+ | ~MRENTHUSIASM | ||
+ | |||
+ | == Video Solution == | ||
+ | https://youtu.be/YcZzxez-j-c | ||
+ | |||
+ | ~AMC & AIME Training | ||
==See Also== | ==See Also== | ||
{{AIME box|year=2022|n=I|num-b=1|num-a=3}} | {{AIME box|year=2022|n=I|num-b=1|num-a=3}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 15:05, 2 February 2024
Contents
Problem
Find the three-digit positive integer whose representation in base nine is where and are (not necessarily distinct) digits.
Solution 1
We are given that which rearranges to Taking both sides modulo we have The only solution occurs at from which
Therefore, the requested three-digit positive integer is
~MRENTHUSIASM
Solution 2
As shown in Solution 1, we get .
Note that and are large numbers comparatively to , so we hypothesize that and are equal and fills the gap between them. The difference between and is , which is a multiple of . So, if we multiply this by , it will be a multiple of and thus the gap can be filled. Therefore, the only solution is , and the answer is .
~KingRavi
Solution 3
As shown in Solution 1, we get
We list a few multiples of out: Of course, can't be made of just 's. If we use one , we get a remainder of , which can't be made of 's either. So doesn't work. can't be made up of just 's. If we use one , we get a remainder of , which can't be made of 's. If we use two 's, we get a remainder of , which can be made of 's. Therefore we get so and . Plugging this back into the original problem shows that this answer is indeed correct. Therefore,
~Technodoggo
Solution 4
As shown in Solution 1, we get .
We can see that is larger than , and we have an . We can clearly see that is a multiple of , and any larger than would result in being larger than . Therefore, our only solution is . Our answer is .
~Arcticturn
Solution 5
As shown in Solution 1, we get which rearranges to So
vladimir.shelomovskii@gmail.com, vvsss
Solution 6
As shown in Solution 1, we have that .
Note that by the divisibility rule for , we have . Since and are base- digits, we can say that or . The former possibility can be easily eliminated, and thus . Next, we write the equation from Solution 1 as , and dividing this by gives . Taking both sides modulo , we have . Multiplying both sides by gives , which implies . From here, we can find that and , giving an answer of .
~Sedro
Video Solution by OmegaLearn
https://youtu.be/SCGzEOOICr4?t=340
~ pi_is_3.14
Video Solution (Mathematical Dexterity)
https://www.youtube.com/watch?v=z5Y4bT5rL-s
Video Solution
https://www.youtube.com/watch?v=CwSkAHR3AcM
~Steven Chen (www.professorchenedu.com)
Video Solution
https://youtu.be/MJ_M-xvwHLk?t=392
~ThePuzzlr
Video Solution by MRENTHUSIASM (English & Chinese)
https://www.youtube.com/watch?v=v4tHtlcD9ww&t=360s&ab_channel=MRENTHUSIASM
~MRENTHUSIASM
Video Solution
~AMC & AIME Training
See Also
2022 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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