Difference between revisions of "1988 AIME Problems/Problem 9"
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Find the smallest positive integer whose [[perfect cube|cube]] ends in <math>888</math>. | Find the smallest positive integer whose [[perfect cube|cube]] ends in <math>888</math>. | ||
− | == Solution == | + | ==Solution 1== |
A little bit of checking tells us that the units digit must be 2. Now our cube must be in the form of <math>(10k + 2)^3</math>; using the [[binomial theorem]] gives us <math>1000k^3 + 600k^2 + 120k + 8</math>. Since we are looking for the tens digit, <math>\mod{100}</math> we get <math>20k + 8 \equiv 88 \pmod{100}</math>. This is true if the tens digit is either <math>4</math> or <math>9</math>. Casework: | A little bit of checking tells us that the units digit must be 2. Now our cube must be in the form of <math>(10k + 2)^3</math>; using the [[binomial theorem]] gives us <math>1000k^3 + 600k^2 + 120k + 8</math>. Since we are looking for the tens digit, <math>\mod{100}</math> we get <math>20k + 8 \equiv 88 \pmod{100}</math>. This is true if the tens digit is either <math>4</math> or <math>9</math>. Casework: | ||
*<math>4</math>: Then our cube must be in the form of <math>(100k + 42)^3 \equiv 3(100k)(42)^2 + 42^3 \equiv 200k + 88 \pmod{1000}</math>. Hence the lowest possible value for the hundreds digit is <math>4</math>, and so <math>442</math> is a valid solution. | *<math>4</math>: Then our cube must be in the form of <math>(100k + 42)^3 \equiv 3(100k)(42)^2 + 42^3 \equiv 200k + 88 \pmod{1000}</math>. Hence the lowest possible value for the hundreds digit is <math>4</math>, and so <math>442</math> is a valid solution. | ||
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By looking at the last digit again, we see <math>a \equiv 3 \pmod5</math>, so we let <math>a = 5a_1 + 3</math> where <math>a_1 \in \mathbb{Z^+}</math>. Plugging this in to <math>5a^2 + 12a \equiv 1 \pmod{25}</math> gives <math>10a_1 + 6 \equiv 1 \pmod{25}</math>. Obviously, <math>a_1 \equiv 2 \pmod 5</math>, so we let <math>a_1 = 5a_2 + 2</math> where <math>a_2</math> can be any non-negative integer. | By looking at the last digit again, we see <math>a \equiv 3 \pmod5</math>, so we let <math>a = 5a_1 + 3</math> where <math>a_1 \in \mathbb{Z^+}</math>. Plugging this in to <math>5a^2 + 12a \equiv 1 \pmod{25}</math> gives <math>10a_1 + 6 \equiv 1 \pmod{25}</math>. Obviously, <math>a_1 \equiv 2 \pmod 5</math>, so we let <math>a_1 = 5a_2 + 2</math> where <math>a_2</math> can be any non-negative integer. | ||
− | Therefore, <math>n = 2 + 5(3+ 5(2+5a_2)) = 125a_2 + 67</math>. <math>n</math> must also be a multiple of <math>8</math>, so <math>125a_2 + 67 \equiv | + | Therefore, <math>n = 2 + 5(3+ 5(2+5a_2)) = 125a_2 + 67</math>. <math>n^3</math> must also be a multiple of <math>8</math>, so <math>n</math> must be even. <math>125a_2 + 67 \equiv 0 \pmod 2 \implies a_2 \equiv 1 \pmod 2</math>. Therefore, <math>a_2 = 2a_3 + 1</math>, where <math>a_3</math> is any non-negative integer. The number <math>n</math> has form <math>125(2a_3+1)+67 = 250a_3+192</math>. So the minimum <math>n = \boxed{192}</math>. |
== Solution 3 == | == Solution 3 == | ||
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We can work our way up, and find that <math>y\equiv 1\pmod 5</math>, <math>y\equiv 21\pmod{25}</math>, and finally <math>y\equiv 96\pmod{125}</math>. This gives us our smallest value, <math>y = 96</math>, so <math>x = \boxed{192}</math>, as desired. - Spacesam | We can work our way up, and find that <math>y\equiv 1\pmod 5</math>, <math>y\equiv 21\pmod{25}</math>, and finally <math>y\equiv 96\pmod{125}</math>. This gives us our smallest value, <math>y = 96</math>, so <math>x = \boxed{192}</math>, as desired. - Spacesam | ||
+ | |||
+ | == Solution 4 (Bash) == | ||
+ | |||
+ | Let this integer be <math>x.</math> Note that <cmath>x^3 \equiv 888 \pmod{1000} \implies x \equiv 0 \pmod {2}~~ \cap ~~ x \equiv 2 \pmod{5}.</cmath> We wish to find the residue of <math>x</math> mod <math>125.</math> Note that <cmath>x \equiv 2,7,12,17, \text{ or } 22 \pmod{25}</cmath> using our congruence in mod <math>5.</math> The residue that works must also satisfy <math>x^3 \equiv 13 \pmod{25}</math> from our original congruence. Noting that <math>17^3 \equiv (-8)^3 \equiv -512 \equiv 13 \pmod{25}</math> (and bashing out the other residues perhaps but they're not that hard), we find that <cmath>x \equiv 17 \pmod{25}.</cmath> Thus, <cmath>x \equiv 17,42,67,92,117 \pmod{125}.</cmath> The residue that works must also satisfy <math>x^3 \equiv 13 \pmod{125}</math> from our original congruence. It is easy to memorize that <cmath>17^3 \equiv \mathbf{4913} \equiv 38 \pmod{125}.</cmath> Also, <cmath>42^3 \equiv 42^2 \cdot 42 \equiv 1764 \cdot 42 \equiv 14 \cdot 42 \equiv 88 \pmod{125}.</cmath> Finally, <cmath>67^3 \equiv 67^2 \cdot 67 \equiv 4489 \cdot 67 \equiv (-11) \cdot 67 \equiv -737 \equiv 13 \pmod{125},</cmath> as desired. Thus, <math>x</math> must satisfy <cmath>x \equiv 0 \pmod{2}~~ \cap ~~x \equiv 67 \pmod{125} \implies x \equiv 192 \pmod{250} \implies x=\boxed{192}.</cmath> ~samrocksnature | ||
+ | |||
+ | |||
+ | ==solution 5== | ||
+ | This number is in the form of <math>10k+2</math>, after binomial expansion, we only want <math>600k^2+120k\equiv 880 \pmod{1000}</math>. We realize that <math>600,120</math> are both multiples of <math>8</math>, we only need that <math>600k^2+120k \equiv 5\pmod{125}</math>, so we write <math>600k^2+120k=125x+5; 120k^2+24k=25x+1, 24(5k^2+k)=25x+1, 5k^2+k\equiv -1\pmod{25}</math> | ||
+ | |||
+ | Then, we write <math>5k^2+k=25m-1, 5k^2+k+1=25m</math> so <math>k+1</math> must be a multiple of <math>5</math> at least, so <math>k\equiv {-1, -6, -11, -16, -21} \pmod {25}</math> after checking, when <math>k=-6, 5k^2+k+1=175=25\cdot 7</math>. So <math>k\equiv -6 \pmod{25}</math>, smallest <math>k=19</math>, the number is <math>\boxed{192}</math> | ||
+ | |||
+ | ~bluesoul | ||
+ | |||
+ | ==Solution 6 (A bit of brute force using basic knowledge.)== | ||
+ | |||
+ | We do know the unit digit has to be 2, So lets consider the number of the form of <math>x2</math>. | ||
+ | |||
+ | On cubing <math>x2</math>, we get a number of the form <math>x^3 \,6x^2 \,12x \,8</math> where the unit digit of <math>12x</math> must be 8, therefore x can be 4 or 9. But for this value of <math>x</math> the hundreds digit won't be 8. | ||
+ | |||
+ | Thus our number must be of the form <math>x42</math> / <math>x92</math>. Repeating the above process we get values of <math>x</math> as 4 and 1 respectively. | ||
+ | |||
+ | Therefore the smallest number is 192.~ Dubey619 | ||
== See also == | == See also == |
Latest revision as of 08:15, 2 February 2023
Contents
Problem
Find the smallest positive integer whose cube ends in .
Solution 1
A little bit of checking tells us that the units digit must be 2. Now our cube must be in the form of ; using the binomial theorem gives us
. Since we are looking for the tens digit,
we get
. This is true if the tens digit is either
or
. Casework:
: Then our cube must be in the form of
. Hence the lowest possible value for the hundreds digit is
, and so
is a valid solution.
: Then our cube is
. The lowest possible value for the hundreds digit is
, and we get
. Hence, since
, the answer is
Solution 2
and
.
due to the last digit of
. Let
. By expanding,
.
By looking at the last digit again, we see , so we let
where
. Plugging this in to
gives
. Obviously,
, so we let
where
can be any non-negative integer.
Therefore, .
must also be a multiple of
, so
must be even.
. Therefore,
, where
is any non-negative integer. The number
has form
. So the minimum
.
Solution 3
Let . We factor an
out of the right hand side, and we note that
must be of the form
, where
is a positive integer. Then, this becomes
. Taking mod
,
, and
, we get
,
, and
.
We can work our way up, and find that ,
, and finally
. This gives us our smallest value,
, so
, as desired. - Spacesam
Solution 4 (Bash)
Let this integer be Note that
We wish to find the residue of
mod
Note that
using our congruence in mod
The residue that works must also satisfy
from our original congruence. Noting that
(and bashing out the other residues perhaps but they're not that hard), we find that
Thus,
The residue that works must also satisfy
from our original congruence. It is easy to memorize that
Also,
Finally,
as desired. Thus,
must satisfy
~samrocksnature
solution 5
This number is in the form of , after binomial expansion, we only want
. We realize that
are both multiples of
, we only need that
, so we write
Then, we write so
must be a multiple of
at least, so
after checking, when
. So
, smallest
, the number is
~bluesoul
Solution 6 (A bit of brute force using basic knowledge.)
We do know the unit digit has to be 2, So lets consider the number of the form of .
On cubing , we get a number of the form
where the unit digit of
must be 8, therefore x can be 4 or 9. But for this value of
the hundreds digit won't be 8.
Thus our number must be of the form /
. Repeating the above process we get values of
as 4 and 1 respectively.
Therefore the smallest number is 192.~ Dubey619
See also
1988 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 8 |
Followed by Problem 10 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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