Difference between revisions of "1988 AIME Problems/Problem 9"
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*<math>9</math>: Then our cube is <math>(100k + 92)^3 \equiv 3(100k)(92)^2 + 92^3 \equiv 200k + 688 \pmod{1000}</math>. The lowest possible value for the hundreds digit is <math>1</math>, and we get <math>192</math>. Hence, since <math>192 < 442</math>, the answer is <math>\fbox{192}</math> | *<math>9</math>: Then our cube is <math>(100k + 92)^3 \equiv 3(100k)(92)^2 + 92^3 \equiv 200k + 688 \pmod{1000}</math>. The lowest possible value for the hundreds digit is <math>1</math>, and we get <math>192</math>. Hence, since <math>192 < 442</math>, the answer is <math>\fbox{192}</math> | ||
− | ==Solution 2== | + | ===Solution 2=== |
<math>n^3 \equiv 888 \pmod{1000} \implies n^3 \equiv 0 \pmod 8</math> and <math>n^3 \equiv 13 \pmod{125}</math>. | <math>n^3 \equiv 888 \pmod{1000} \implies n^3 \equiv 0 \pmod 8</math> and <math>n^3 \equiv 13 \pmod{125}</math>. | ||
<math>n \equiv 2 \pmod 5</math> due to the last digit of <math>n^3</math>. Let <math>n = 5a + 2</math>. By expanding, <math>125a^3 + 150a^2 + 60a + 8 \equiv 13 \pmod{125} \implies 5a^2 + 12a \equiv 1 \pmod{25}</math>. | <math>n \equiv 2 \pmod 5</math> due to the last digit of <math>n^3</math>. Let <math>n = 5a + 2</math>. By expanding, <math>125a^3 + 150a^2 + 60a + 8 \equiv 13 \pmod{125} \implies 5a^2 + 12a \equiv 1 \pmod{25}</math>. |
Revision as of 15:04, 16 May 2020
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
. Therefore, the minimum of
is
.
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
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.