1988 AIME Problems/Problem 9

Problem

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

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

solution 5

This number is in the form of $10k+2$ , after binomial expansion, we only want $600k^2+120k\equiv 880 \pmod{1000}$ . We realize that $600,120$ are both multiples of $8$ , we only need that $600k^2+120k \equiv 5\pmod{125}$ , so we write $600k^2+120k=125x+5; 120k^2+24k=25x+1, 24(5k^2+k)=25x+1, 5k^2+k\equiv -1\pmod{25}$

Then, we write $5k^2+k=25m-1, 5k^2+k+1=25m$ so $k+1$ must be a multiple of $5$ at least, so $k\equiv {-1, -6, -11, -16, -21} \pmod {25}$ after checking, when $k=-6, 5k^2+k+1=175=25\cdot 7$ . So $k\equiv -6 \pmod{25}$ , smallest $k=19$ , the number is $\boxed{192}$

~bluesoul

solution 6(NOT RECOMMENDED AT ALL. DO NOT DO UNLESS YOU HAVE NOTHING TO DO)

Let the positive integer whose cuber ends in 888 be x. x must have unit digit 2, that's only possible. So we do trial and error 12 no 22 no 32 no 42 yes 52 no 62 no 72 no 82 no 92 no Continuing on, we get that 192 is the smallest positive integer. ALSO: $192^3=7077888$ ~EthanSpoon

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

Page

Toolbox

Search