Difference between revisions of "2018 AIME I Problems/Problem 11"
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==Problem== | ==Problem== | ||
− | Find the least positive integer <math>n</math> such that when <math>3^n</math> is written in base <math>143</math>, its two right-most digits in base <math>143</math> are <math>01</math>. | + | Find the least positive integer <math>n</math> such that when <math>3^n</math> is written in base <math>143^2</math>, its two right-most digits in base <math>143</math> are <math>01</math>. |
− | == | + | ==Solution 1== |
− | |||
− | |||
Note that the given condition is equivalent to <math>3^n \equiv 1 \pmod{143^2}</math> and <math>143=11\cdot 13</math>. Because <math>\gcd(11^2, 13^2) = 1</math>, the desired condition is equivalent to <math>3^n \equiv 1 \pmod{121}</math> and <math>3^n \equiv 1 \pmod{169}</math>. | Note that the given condition is equivalent to <math>3^n \equiv 1 \pmod{143^2}</math> and <math>143=11\cdot 13</math>. Because <math>\gcd(11^2, 13^2) = 1</math>, the desired condition is equivalent to <math>3^n \equiv 1 \pmod{121}</math> and <math>3^n \equiv 1 \pmod{169}</math>. | ||
If <math>3^n \equiv 1 \pmod{121}</math>, one can see the sequence <math>1, 3, 9, 27, 81, 1, 3, 9...</math> so <math>5|n</math>. | If <math>3^n \equiv 1 \pmod{121}</math>, one can see the sequence <math>1, 3, 9, 27, 81, 1, 3, 9...</math> so <math>5|n</math>. | ||
− | Now if <math>3^n \equiv 1 \pmod{169}</math>, it is harder. But we do observe that <math>3^3 \equiv 1 \pmod{13}</math>, therefore <math>3^3 = 13a + 1</math> for some integer <math>a</math>. So our goal is to find the first number <math>p_1</math> such that <math>(13a+1)^ {p_1} \equiv 1 \pmod{169}</math>. | + | Now if <math>3^n \equiv 1 \pmod{169}</math>, it is harder. But we do observe that <math>3^3 \equiv 1 \pmod{13}</math>, therefore <math>3^3 = 13a + 1</math> for some integer <math>a</math>. So our goal is to find the first number <math>p_1</math> such that <math>(13a+1)^ {p_1} \equiv 1 \pmod{169}</math>. Then, <math>p_1 \equiv 0 \pmod{13}, </math> which follows from the binomial theorem. It is not difficult to see that the smallest <math>p_1=13</math>, so ultimately <math>3^{39} \equiv 1 \pmod{169}</math>. Therefore, <math>39|n</math>. |
The first <math>n</math> satisfying both criteria is thus <math>5\cdot 39=\boxed{195}</math>. | The first <math>n</math> satisfying both criteria is thus <math>5\cdot 39=\boxed{195}</math>. | ||
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Note that Euler's Totient Theorem would not necessarily lead to the smallest <math>n</math> and that in this case that <math>n</math> is greater than <math>1000</math>. | Note that Euler's Totient Theorem would not necessarily lead to the smallest <math>n</math> and that in this case that <math>n</math> is greater than <math>1000</math>. | ||
− | We wish to find the least <math>n</math> such that <math>3^n \equiv 1 \pmod{143^2}</math>. This factors as <math>143^2=11^{2}*13^{2}</math>. Because <math>gcd(121, 169) = 1</math>, we can simply find the least <math>n</math> such that <math>3^n \equiv 1 \pmod{121}</math> and <math>3^n \equiv 1 \pmod{169}</math>. | + | We wish to find the least <math>n</math> such that <math>3^n \equiv 1 \pmod{143^2}</math>. This factors as <math>143^2=11^{2}*13^{2}</math>. Because <math>\gcd(121, 169) = 1</math>, we can simply find the least <math>n</math> such that <math>3^n \equiv 1 \pmod{121}</math> and <math>3^n \equiv 1 \pmod{169}</math>. |
− | Quick inspection yields <math>3^5 \equiv 1 \pmod{121}</math> and <math>3^3 \equiv 1 \pmod{13}</math>. Now we must find the smallest <math>k</math> such that <math>3^{3k} \equiv 1 \pmod{ | + | Quick inspection yields <math>3^5 \equiv 1 \pmod{121}</math> and <math>3^3 \equiv 1 \pmod{13}</math>. Now we must find the smallest <math>k</math> such that <math>3^{3k} \equiv 1 \pmod{169}</math>. Euler's gives <math>3^{156} \equiv 1 \pmod{169}</math>. So <math>3k</math> is a factor of <math>156</math>. This gives <math>k=1,2, 4, 13, 26, 52</math>. Some more inspection yields <math>k=13</math> is the smallest valid <math>k</math>. So <math>3^5 \equiv 1 \pmod{121}</math> and <math>3^{39} \equiv 1 \pmod{169}</math>. The least <math>n</math> satisfying both is <math>lcm(5, 39)=\boxed{195}</math>. (RegularHexagon) |
==Solution 3 (Big Bash)== | ==Solution 3 (Big Bash)== | ||
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\end{array}</cmath> | \end{array}</cmath> | ||
− | The powers of <math>3</math> repeat in cycles of <math>5</math> | + | The powers of <math>3</math> repeat in cycles of <math>5</math> and <math>39</math> in modulo <math>121</math> and modulo <math>169</math>, respectively. The answer is <math>\text{lcm}(5, 39) = \boxed{195}</math>. |
− | ==Solution 4(Order+Bash)== | + | ~Minor edit by Yiyj1 |
+ | |||
+ | ==Solution 4 (Order+Bash)== | ||
We have that <cmath>3^n \equiv 1 \pmod{143^2}.</cmath>Now, <math>3^{110} \equiv 1 \pmod{11^2}</math> so by the Fundamental Theorem of Orders, <math>\text{ord}_{11^2}(3)|110</math> and with some bashing, we get that it is <math>5</math>. Similarly, we get that <math>\text{ord}_{13^2}(3)=39</math>. Now, <math>\text{lcm}(39,5)=\boxed{195}</math> which is our desired solution. | We have that <cmath>3^n \equiv 1 \pmod{143^2}.</cmath>Now, <math>3^{110} \equiv 1 \pmod{11^2}</math> so by the Fundamental Theorem of Orders, <math>\text{ord}_{11^2}(3)|110</math> and with some bashing, we get that it is <math>5</math>. Similarly, we get that <math>\text{ord}_{13^2}(3)=39</math>. Now, <math>\text{lcm}(39,5)=\boxed{195}</math> which is our desired solution. | ||
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-Stormersyle | -Stormersyle | ||
− | ==Solution 6(LTE)== | + | ==Solution 6 (LTE)== |
We can see that <math>3^n-1 = 143^2*x</math>, which means that <math>v_{11}(3^n-1) \geq 2</math>, <math>v_{13}(3^n-1) \geq 2</math>. <math>v_{11}(3^n-1) = v_{11}(242) + v_{11}(\frac{n}{5})</math>, <math>v_{13}(3^n-1) = v_{13}(26) + v_{13}(\frac{n}{3})</math> by the Lifting the Exponent lemma. From the first equation we gather that 5 divides n, while from the second equation we gather that both 13 and 3 divide n as <math>v_{13}(3^n-1) \geq 2</math>. Therefore the minimum possible value of n is <math>3\times5\times13=\boxed{195}</math>. | We can see that <math>3^n-1 = 143^2*x</math>, which means that <math>v_{11}(3^n-1) \geq 2</math>, <math>v_{13}(3^n-1) \geq 2</math>. <math>v_{11}(3^n-1) = v_{11}(242) + v_{11}(\frac{n}{5})</math>, <math>v_{13}(3^n-1) = v_{13}(26) + v_{13}(\frac{n}{3})</math> by the Lifting the Exponent lemma. From the first equation we gather that 5 divides n, while from the second equation we gather that both 13 and 3 divide n as <math>v_{13}(3^n-1) \geq 2</math>. Therefore the minimum possible value of n is <math>3\times5\times13=\boxed{195}</math>. | ||
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The least power of <math>3</math> that is congruent to <math>1</math> modulo <math>13</math> is <math>3^3=27=2\cdot 13+1.</math> It follows that <math>3^n\equiv 1\pmod{13}</math> if and only if <math>n=3k</math> for some positive integer <math>k</math>. Additionally, for some positive integer <math>k</math>, the Binomial Theorem shows that <math>3^{3k}=(26+1)^k=26\cdot k+1 \pmod{13^2}</math>. In particular, <math>3^n=3^{3k}\equiv 1\pmod {13^2}</math> if and only if <math>k=13m</math> for some positive integer <math>m</math>, that is, if and only if <math>n=39m.</math> | The least power of <math>3</math> that is congruent to <math>1</math> modulo <math>13</math> is <math>3^3=27=2\cdot 13+1.</math> It follows that <math>3^n\equiv 1\pmod{13}</math> if and only if <math>n=3k</math> for some positive integer <math>k</math>. Additionally, for some positive integer <math>k</math>, the Binomial Theorem shows that <math>3^{3k}=(26+1)^k=26\cdot k+1 \pmod{13^2}</math>. In particular, <math>3^n=3^{3k}\equiv 1\pmod {13^2}</math> if and only if <math>k=13m</math> for some positive integer <math>m</math>, that is, if and only if <math>n=39m.</math> | ||
− | Because <math>11^2</math> and <math>13^2</math> are relatively prime, <math>3^n\equiv 1\pmod {143^2}</math> if and only if <math>3^n\equiv 1\pmod{11^2}</math> and <math>3^n \equiv 1\pmod {13^2}</math>. This occurs if and only if <math>n</math> is a multiple of both of the relatively prime integers <math>5</math> and <math>39</math>, so the least possible value of <math>n</math> is <math>5\cdot 39=195.</math> | + | Because <math>11^2</math> and <math>13^2</math> are relatively prime, <math>3^n\equiv 1\pmod {143^2}</math> if and only if <math>3^n\equiv 1\pmod{11^2}</math> and <math>3^n \equiv 1\pmod {13^2}</math>. This occurs if and only if <math>n</math> is a multiple of both of the relatively prime integers <math>5</math> and <math>39</math>, so the least possible value of <math>n</math> is <math>5\cdot 39=\boxed{195}.</math> |
− | == | + | |
− | + | ==Solution 9 (Motivation and LTE)== | |
+ | |||
+ | We first note that we wish to find <math>3^n \equiv 1 \pmod{11^2}</math> and <math>3^n \equiv 1 \pmod{13^2}.</math> Not thinking of anything else, we try a few numbers for the first condition to get that <math>5 \mid n.</math> For the second condition, upon testing up to 729, we find that it doesn't yield a solution readily, so we use Lifting the Exponent from our toolkit to get that | ||
+ | <cmath>v_{13}(27^m-1^m)=v_{13}(26)+v(m)=1+v(m), 3m = n</cmath> which clearly implies <math>m=13</math> and <math>39 | n.</math> Our answer is then obviously <math>39 \cdot 5 = \boxed{195}.</math> | ||
+ | |||
+ | ~Dhillonr25 | ||
==See Also== | ==See Also== | ||
{{AIME box|year=2018|n=I|num-b=10|num-a=12}} | {{AIME box|year=2018|n=I|num-b=10|num-a=12}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 06:30, 21 April 2024
Contents
Problem
Find the least positive integer such that when
is written in base
, its two right-most digits in base
are
.
Solution 1
Note that the given condition is equivalent to and
. Because
, the desired condition is equivalent to
and
.
If , one can see the sequence
so
.
Now if , it is harder. But we do observe that
, therefore
for some integer
. So our goal is to find the first number
such that
. Then,
which follows from the binomial theorem. It is not difficult to see that the smallest
, so ultimately
. Therefore,
.
The first satisfying both criteria is thus
.
-expiLnCalc
Solution 2
Note that Euler's Totient Theorem would not necessarily lead to the smallest and that in this case that
is greater than
.
We wish to find the least such that
. This factors as
. Because
, we can simply find the least
such that
and
.
Quick inspection yields and
. Now we must find the smallest
such that
. Euler's gives
. So
is a factor of
. This gives
. Some more inspection yields
is the smallest valid
. So
and
. The least
satisfying both is
. (RegularHexagon)
Solution 3 (Big Bash)
Listing out the powers of , modulo
and modulo
, we have:
The powers of repeat in cycles of
and
in modulo
and modulo
, respectively. The answer is
.
~Minor edit by Yiyj1
Solution 4 (Order+Bash)
We have that Now,
so by the Fundamental Theorem of Orders,
and with some bashing, we get that it is
. Similarly, we get that
. Now,
which is our desired solution.
Solution 5 (Easy Binomial Theorem)
We wish to find the smallest such that
, so we want
and
. Note that
, so
repeats
with a period of
, so
. Now, in order for
, then
. Because
,
repeats with a period of
, so
.
Hence, we have that for some positive integer
,
, so
and
. Thus, we have that
,
, and
, so the smallest possible value of
is
.
-Stormersyle
Solution 6 (LTE)
We can see that , which means that
,
.
,
by the Lifting the Exponent lemma. From the first equation we gather that 5 divides n, while from the second equation we gather that both 13 and 3 divide n as
. Therefore the minimum possible value of n is
.
-bradleyguo
Solution 7
Note that the problem is basically asking for the least positive integer such that
It is easy to see that
where
is the least positive integer satisfying
and
the least positive integer satisfying
. Luckily, finding
is a relatively trivial task, as one can simply notice that
. However, finding
is slightly more nontrivial. The order of
modulo
(which is
) is trivial to find, as one can either bash out a pattern of remainders upon dividing powers of
by
, or one can notice that
(the latter which is the definition of period/orders by FLT). We can thus rewrite
as
. Now suppose that
I claim that
Proof:
To find we can simply multiply
by
which is congruent to
modulo
.
By expanding the product out, we obtain
and since the
on the LHS cancels out, we're left with
. Thus, our claim is proven.
Let
be the second to last digit when
is written in base
.
Using our proof, it is easy to see that
satisfies the recurrence
and
. Since this implies
we just have to find the least positive integer
such that
is a multiple of
, which is trivially obtained as
.
The least integer
such that
is divisible by
is
so our final answer is
-fidgetboss_4000 -minor edits made by srisainandan6
Solution 8 (Official MAA)
The requested positive integer is the least value of such that
Note that
The least power of
that is congruent to
modulo
is
It follows that.
if and only if
for some positive integer
.
The least power of that is congruent to
modulo
is
It follows that
if and only if
for some positive integer
. Additionally, for some positive integer
, the Binomial Theorem shows that
. In particular,
if and only if
for some positive integer
, that is, if and only if
Because and
are relatively prime,
if and only if
and
. This occurs if and only if
is a multiple of both of the relatively prime integers
and
, so the least possible value of
is
Solution 9 (Motivation and LTE)
We first note that we wish to find and
Not thinking of anything else, we try a few numbers for the first condition to get that
For the second condition, upon testing up to 729, we find that it doesn't yield a solution readily, so we use Lifting the Exponent from our toolkit to get that
which clearly implies
and
Our answer is then obviously
~Dhillonr25
See Also
2018 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.