Difference between revisions of "1988 AIME Problems/Problem 9"

Revision as of 23:49, 31 July 2020

Problem

Find the smallest positive integer whose cube ends in $888$ .

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 $(10k + 2)^3$ ; using the binomial theorem gives us $1000k^3 + 600k^2 + 120k + 8$ . Since we are looking for the tens digit, $\mod{100}$ we get $20k + 8 \equiv 88 \pmod{100}$ . This is true if the tens digit is either $4$ or $9$ . Casework:

$4$ : Then our cube must be in the form of $(100k + 42)^3 \equiv 3(100k)(42)^2 + 42^3 \equiv 200k + 88 \pmod{1000}$ . Hence the lowest possible value for the hundreds digit is $4$ , and so $442$ is a valid solution.
$9$ : Then our cube is $(100k + 92)^3 \equiv 3(100k)(92)^2 + 92^3 \equiv 200k + 688 \pmod{1000}$ . The lowest possible value for the hundreds digit is $1$ , and we get $192$ . Hence, since $192 < 442$ , the answer is $\fbox{192}$

Solution 2

$n^3 \equiv 888 \pmod{1000} \implies n^3 \equiv 0 \pmod 8$ and $n^3 \equiv 13 \pmod{125}$ . $n \equiv 2 \pmod 5$ due to the last digit of $n^3$ . Let $n = 5a + 2$ . By expanding, $125a^3 + 150a^2 + 60a + 8 \equiv 13 \pmod{125} \implies 5a^2 + 12a \equiv 1 \pmod{25}$ .

By looking at the last digit again, we see $a \equiv 3 \pmod5$ , so we let $a = 5a_1 + 3$ where $a_1 \in \mathbb{Z^+}$ . Plugging this in to $5a^2 + 12a \equiv 1 \pmod{25}$ gives $10a_1 + 6 \equiv 1 \pmod{25}$ . Obviously, $a_1 \equiv 2 \pmod 5$ , so we let $a_1 = 5a_2 + 2$ where $a_2$ can be any non-negative integer.

Therefore, $n = 2 + 5(3+ 5(2+5a_2)) = 125a_2 + 67$ . $n^3$ must also be a multiple of $8$ , so $n$ must be even. $125a_2 + 67 \equiv 0 \pmod 2 \implies a_2 \equiv 1 \pmod 2$ . Therefore, $a_2 = 2a_3 + 1$ , where $a_3$ is any non-negative integer. The number $n$ has form $125(2a_3+1)+67 = 250a_3+192$ . So the minimum $n = \boxed{192}$ .

Solution 3

Let $x^3 = 1000a + 888$ . We factor an $8$ out of the right hand side, and we note that $x$ must be of the form $x = 2y$ , where $y$ is a positive integer. Then, this becomes $y^3 = 125a + 111$ . Taking mod $5$ , $25$ , and $125$ , we get $y^3 \equiv 1\pmod 5$ , $y^3 \equiv 11\pmod{25}$ , and $y^3 \equiv 111\pmod{125}$ .

We can work our way up, and find that $y\equiv 1\pmod 5$ , $y\equiv 21\pmod{25}$ , and finally $y\equiv 96\pmod{125}$ . This gives us our smallest value, $y = 96$ , so $x = \boxed{192}$ , as desired. - Spacesam

@@ Line 14: / Line 14: @@
 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 5a_2 + 3 \pmod 8 \implies a_2 = 1,9,17 \ldots</math>. Therefore, the minimum of <math>n</math> is <math>125 + 67 = \boxed{192}</math>.
+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 ===

1988 AIME (Problems • Answer Key • Resources)
Preceded by Problem 8	Followed by Problem 10
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