Difference between revisions of "2017 AIME I Problems/Problem 5"
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==Solution 1== | ==Solution 1== | ||
− | First, note that the first two digits will always be a positive number. We will start with base twelve because of its repetition. List all the positive numbers in base | + | First, note that the first two digits will always be a positive number. We will start with base twelve because of its repetition. List all the positive numbers in base eight that have equal tens and ones digits in base 12. |
<math>11_{12}=15_8</math> | <math>11_{12}=15_8</math> | ||
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==Solution 4 (modular congruency/nt bash)== | ==Solution 4 (modular congruency/nt bash)== | ||
− | In the problem, we are given that < | + | In the problem, we are given that <cmath>8a+b+\dfrac c8+\dfrac d{64}=12b+b+\dfrac b{12}+\dfrac a{144}.</cmath> We multiply by the LCM of the denominators, which is <math>576</math> to get <cmath>4608a+576b+72c+9d=6912b+576b+48b+4a.</cmath> We then group like terms and factor to get <cmath>4604a+72c+9d=48b(144+1)=145\cdot48b=24\cdot290b.</cmath> |
− | Observe that the coefficients of <math>a</math>, <math>c</math>, and <math>b</math> are all divisible by <math>4</math>. Therefore, we know that <math>d</math> must also be divisible by <math>4</math> to compensate. (To observe this, one could rearrange the terms to see <math>9d=24\cdot290b-4604a-72c=4(6\cdot290b-1151a-18c)</math>. At this point, it is obvious that <math>9d</math> must be divisible by <math>4</math>, so <math>d</math> must be divisible by <math>4</math>.) Thus, we let <math>d=4y</math> to see <math>4604a+72c+9\cdot4y=24\cdot290b</math>. We divide by <math>4</math> in the equation to get < | + | Observe that the coefficients of <math>a</math>, <math>c</math>, and <math>b</math> are all divisible by <math>4</math>. Therefore, we know that <math>d</math> must also be divisible by <math>4</math> to compensate. (To observe this, one could rearrange the terms to see <math>9d=24\cdot290b-4604a-72c=4(6\cdot290b-1151a-18c)</math>. At this point, it is obvious that <math>9d</math> must be divisible by <math>4</math>, so <math>d</math> must be divisible by <math>4</math>.) Thus, we let <math>d=4y</math> to see <math>4604a+72c+9\cdot4y=24\cdot290b</math>. We divide by <math>4</math> in the equation to get <cmath>1151a+18c+9y=6\cdot290b.</cmath> |
− | Observe that the coefficients of <math>c</math>, <math>y</math>, and <math>b</math> are all divisible by <math>3</math>. By similar reasoning, <math>a</math> must be divisible by <math>3</math>, so we let <math>a=3x</math>. Substituting and dividing by <math>3</math>, we get < | + | Observe that the coefficients of <math>c</math>, <math>y</math>, and <math>b</math> are all divisible by <math>3</math>. By similar reasoning, <math>a</math> must be divisible by <math>3</math>, so we let <math>a=3x</math>. Substituting and dividing by <math>3</math>, we get <cmath>1151x+6c+3y=2\cdot290b.</cmath> We observe that this can no longer be reduced by similar means. |
− | We know that <math>0<a,b,c<8\implies0<3x,b,4y<8</math>. We examine <math>0<4y<8</math>; this becomes <math>0<y<2</math>. It is apparent that <math>y=1\implies d=4</math>. However, the problem does not even ask for <math>d</math>, so it may appear that this find is meaningless. However, we substitute our value of <math>y</math> in to get < | + | We know that <math>0<a,b,c<8\implies0<3x,b,4y<8</math>. We examine <math>0<4y<8</math>; this becomes <math>0<y<2</math>. It is apparent that <math>y=1\implies d=4</math>. However, the problem does not even ask for <math>d</math>, so it may appear that this find is meaningless. However, we substitute our value of <math>y</math> in to get <cmath>1151x+6c+3=580b.</cmath> |
− | We know that <math>x</math> is either <math>1</math> or <math>2</math>, since <math>0<3x<8</math>. We take our equation modulo <math>6</math> to find < | + | We know that <math>x</math> is either <math>1</math> or <math>2</math>, since <math>0<3x<8</math>. We take our equation modulo <math>6</math> to find <cmath>5x+3\equiv4b\pmod{6}\implies3\equiv4b+x\pmod{6}.</cmath> If <math>x=2</math>, then <math>1\equiv4b\pmod{6}</math> - this is equivalent to saying that <math>1+6n=4b\implies 1=4b-6n</math>. However, <math>4b-6n</math> is always even, and <math>1</math> is odd. Thus, this case is not possible, so <math>x=1\implies a=3</math>. We know that <math>x=1</math>; thus, <math>3\equiv4b+1\pmod{6}\implies2\equiv4b\pmod{6}\implies1\equiv2b\pmod{3}</math>. Obviously, <math>b=2</math> and <math>b=5</math> work; however, if <math>b=5</math>, then the RHS of our original equation (<math>2\cdot290b</math>) is much too large to be equal to the LHS (the maximum possible value of the LHS is <math>1151\cdot1+6\cdot7+3\cdot1</math>, which is less than <math>1200</math> while the RHS would become <math>2\cdot290\cdot5=2900</math>), so we have <math>b=2</math>. |
We recall our equation of <math>1151x+6c+3=580b</math>. Plugging in what we know, we have <math>1151\cdot1+6c+3=580\cdot2\implies6c=6\implies c=1</math>. | We recall our equation of <math>1151x+6c+3=580b</math>. Plugging in what we know, we have <math>1151\cdot1+6c+3=580\cdot2\implies6c=6\implies c=1</math>. |
Latest revision as of 22:03, 18 January 2024
Contents
Problem 5
A rational number written in base eight is , where all digits are nonzero. The same number in base twelve is
. Find the base-ten number
.
Solution 1
First, note that the first two digits will always be a positive number. We will start with base twelve because of its repetition. List all the positive numbers in base eight that have equal tens and ones digits in base 12.
We stop because we only can have two-digit numbers in base 8 and 101 is not a 2 digit number.
Compare the ones places to check if they are equal. We find that they are equal if or
.
Evaluating the places to the right side of the decimal point gives us
or
.
When the numbers are converted into base 8, we get
and
. Since
, the first value is correct. Compiling the necessary digits leaves us a final answer of
Solution 2
The parts before and after the decimal points must be equal. Therefore and
. Simplifying the first equation gives
. Plugging this into the second equation gives
. Multiplying both sides by 64 gives
.
and
are both digits between 1 and 7 (they must be a single non-zero digit in base eight) so using
,
or
. Testing these gives that
doesn't work, and
gives
, and
. Therefore
Solution 3
Converting to base we get
Since and
are much smaller than the other two terms, dividing by
and approximating we get
Writing out the first few values of and
, the first possible tuple is
and the second possible tuple is
Note that can not be
, therefore the answer is
By maxamc
Solution 4 (modular congruency/nt bash)
In the problem, we are given that We multiply by the LCM of the denominators, which is
to get
We then group like terms and factor to get
Observe that the coefficients of ,
, and
are all divisible by
. Therefore, we know that
must also be divisible by
to compensate. (To observe this, one could rearrange the terms to see
. At this point, it is obvious that
must be divisible by
, so
must be divisible by
.) Thus, we let
to see
. We divide by
in the equation to get
Observe that the coefficients of ,
, and
are all divisible by
. By similar reasoning,
must be divisible by
, so we let
. Substituting and dividing by
, we get
We observe that this can no longer be reduced by similar means.
We know that . We examine
; this becomes
. It is apparent that
. However, the problem does not even ask for
, so it may appear that this find is meaningless. However, we substitute our value of
in to get
We know that is either
or
, since
. We take our equation modulo
to find
If
, then
- this is equivalent to saying that
. However,
is always even, and
is odd. Thus, this case is not possible, so
. We know that
; thus,
. Obviously,
and
work; however, if
, then the RHS of our original equation (
) is much too large to be equal to the LHS (the maximum possible value of the LHS is
, which is less than
while the RHS would become
), so we have
.
We recall our equation of . Plugging in what we know, we have
.
Therefore, .
~~this insane timesink solution was composed by Technodoggo~~
Video Solution
https://youtu.be/Mk-MCeVjSGc?t=298 ~Shreyas S
See Also
2017 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 4 |
Followed by Problem 6 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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