Difference between revisions of "1988 AIME Problems/Problem 9"
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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(\mod 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> | 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(\mod 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 | + | 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 | ~bluesoul |
Revision as of 12:00, 8 August 2022
Contents
Problem
Find the smallest positive integer whose cube ends in .
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 ; 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
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.