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

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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</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>.
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== Solution 3 ==
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Let <math>x^3 = 1000a + 888</math>. We factor an <math>8</math> out of the right hand side, and we note that <math>x</math> must be of the form <math>x = 2y</math>, where <math>y</math> is a positive integer. Then, this becomes <math>y^3 = 125a + 111</math>. Taking mod <math>5</math>, <math>25</math>, and <math>125</math>, we get <math>y^3 \equiv 1\pmod 5</math>, <math>y^3 \equiv 11\pmod{25}</math>, and <math>y^3 \equiv 111\pmod{125}</math>.
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We can work our way up, and find that <math>y\equiv 1\pmod 5</math>, <math>y\equiv 21\pmod{25}</math>, and finally <math>y\equiv 96\pmod{125}</math>. This gives us our smallest value, <math>y = 96</math>, so <math>x = \boxed{192}</math>, as desired. - Spacesam
  
 
== See also ==
 
== See also ==

Revision as of 18:02, 31 July 2019

Problem

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

Solution

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, $n = 2 + 5(3+ 5(2+5a_2)) = 125a_2 + 67$. $n$ must also be a multiple of $8$, so $125a_2 + 67 \equiv 5a_2 + 3 \pmod 8 \implies a_2 = 1,9,17 \ldots$. Therefore, the minimum of $n$ is $125 + 67 = \boxed{192}$.

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

See also

1988 AIME (ProblemsAnswer KeyResources)
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

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