1988 AIME Problems/Problem 9

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

Solution 4 (Bash)

Let this integer be $x.$ Note that $$x^3 \equiv 888 \pmod{1000} \implies x \equiv 0 \pmod {2}~~ \cap ~~ x \equiv 2 \pmod{5}.$$ We wish to find the residue of $x$ mod $125.$ Note that $$x \equiv 2,7,12,17, \text{ or } 22 \pmod{25}$$ using our congruence in mod $5.$ The residue that works must also satisfy $x^3 \equiv 13 \pmod{25}$ from our original congruence. Noting that $17^3 \equiv (-8)^3 \equiv -512 \equiv 13 \pmod{25}$ (and bashing out the other residues perhaps but they're not that hard), we find that $$x \equiv 17 \pmod{25}.$$ Thus, $$x \equiv 17,42,67,92,117 \pmod{125}.$$ The residue that works must also satisfy $x^3 \equiv 13 \pmod{125}$ from our original congruence. It is easy to memorize that $$17^3 \equiv \mathbf{4913} \equiv 38 \pmod{125}.$$ Also, $$42^3 \equiv 42^2 \cdot 42 \equiv 1764 \cdot 42 \equiv 14 \cdot 42 \equiv 88 \pmod{125}.$$ Finally, $$67^3 \equiv 67^2 \cdot 67 \equiv 4489 \cdot 67 \equiv (-11) \cdot 67 \equiv -737 \equiv 13 \pmod{125},$$ as desired. Thus, $x$ must satisfy $$x \equiv 0 \pmod{2}~~ \cap ~~x \equiv 67 \pmod{125} \implies x \equiv 192 \pmod{250} \implies x=\boxed{192}.$$ ~samrocksnature

See also

 1988 AIME (Problems • Answer Key • Resources) Preceded byProblem 8 Followed byProblem 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.

Invalid username
Login to AoPS