Let $x$ be an integer between $1$ and $10^{12}$. What is the probability that $x^3$ ends with 11?
I started with expressing $x^3=(a+10b)^3$ and applied binomial theorem to get $$x^3 = a^3+300b^2a+30a^2b+1000b^3 $$ How do I proceed from here? Thanks.
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3Which of the terms in the expansion do not influence the last two digits of the cube? – Daniel Fischer Jan 17 '21 at 15:15
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@DanielFischer $300b^2a$ and $1000b^3$ don't affect the last two digits, correct? – ForumWhiner Jan 17 '21 at 15:18
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Right. And $30a^2b$ doesn't affect the last. So we need $a^3$ to end with $1$. Which values of $a$ does that leave? – Daniel Fischer Jan 17 '21 at 15:22
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13 is prime to both 2 and 5, which has implications for how the terminal-digit patterns of cubes are distributed. – Oscar Lanzi Jan 17 '21 at 15:28
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1In the first hundred there is just $71$ who cubed gives $357911$. This repeats every hundred. The probability is $1%$ – Raffaele Jan 17 '21 at 16:52
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You've already noticed that only $a^3$ and $30a^2b$ affect the last two digits.
Notice further that only $a^3$ affects the final digit--namely, the final digit of $x^3$ is the final digit of $a^3.$ Thus, $a=1.$
Consequently, the second-to-last digit of $x^3$ is the last digit of $3b,$ and so we need the last digit of $b$ to be $7.$
Can you take it from there?
Cameron Buie
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@WADon: I appreciate the heads-up! I was thinking of squares, rather than cubes. Should be fixed now! – Cameron Buie Jan 17 '21 at 19:59
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If $x^3 \equiv 1 \bmod 100$, then $x^2 x \equiv 1$ and so $x$ is a unit mod $100$. The group of units mod $100$ has order $\phi(100)=40$. Therefore, since $\gcd(40,3)=1$, the map $x \mapsto x^3$ is a bijection. Thus, there is exactly one $x$ such that $x^3 \equiv 11 \bmod 100$.
lhf
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I don't completely follow: 71^3, 171^3, 271^3, etc all equivalent to 11 mod 100 – zcahgg1 May 12 '23 at 19:08
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1@zcahgg1 that's 1/100 probability right there, i.e. for every 100 numbers, you'll have 1 number hitting that
11 mod 1000, hence the probability is 1/100 – godimedia Jan 13 '24 at 02:47