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I tried to make an equation $20+2x+2x^3+x^5$ which satisfied the above property but after putting 2 in place of $x$ I am not getting the correct answer. The answer is 86 but I want to know how or why 86 is true and mine is false.

Rócherz
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Omkar Singh
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  • How do you know $86$ is correct? – Clayton Oct 15 '23 at 02:22
  • Bcz I have posted this question from a prestigious book and the answer given is 86 – Omkar Singh Oct 15 '23 at 02:34
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    No restriction on $a_0$ except it is integer? Then I don't see how the answer could possibly be unique. Perhaps there's a typo and it should say $0 \le a_0,a_1,a_2,\ldots a_n < 3$. – Robert Israel Oct 15 '23 at 03:42
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    $$p(\sqrt3)=a_0+a_1(\sqrt3)+a_2(\sqrt3)^2+\ldots+a_nx^n$$

    $$p(\sqrt3)=a_0+a_2(3)+a_4(3^2)+\cdots+a_1(\sqrt3)+a_3(\sqrt3)^3+\cdots=20+17\sqrt3$$

    $$p(\sqrt3)=(a_0+a_2(3)+a_4(3^2)+\cdots)+\sqrt3(a_1+a_3(3)+\cdots)=20+17\sqrt3$$

    Compare both sides;

    $$a_0+a_2(3)+a_4(3^2)+\cdots=20$$

    $$a_1+a_3(3)+\cdots=17$$

    You should be getting $p(x)=2+2x+2x^3+2x^4+x^5$ and $p(2)=86$

     – NadiKeUssPar Oct 15 '23 at 04:35
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    @ayan In base $3$, we have $17_{10}=122_3$, so $a_1=2, a_3=2, a_5=1$ are fixed. However we cannot similarly say $20_{10}=202_3$ as $a_0$ is not restricted to be $<3$. For e.g. as the OP noted, $a_0=20, a_{2k}=0$ for $k>0$ works perfectly. Similarly $a_0=3m+2$ works for any non-negative integer $m\leqslant 6$, the remainder is a multiple of $3$ which has a base $3$ representation which will fit for remaining $a_{2k}$. e.g. with $m=4$, we have $a_0=14, , (20-14){10}=6{10}=20_3$, so $p(x)=14+2x+2x^2+2x^3+x^5$ gives $p(\sqrt3)=20+17\sqrt3, p(2)=74$. – Macavity Oct 15 '23 at 09:38
  • Uff, I have made a minute mistake in writing up the question, 0≤a0≤3. Very very sorry guys for the mistake – Omkar Singh Oct 15 '23 at 14:51
  • No worries. That is why I asked. – Clayton Oct 16 '23 at 00:59

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