Other Approach.?

Hello friends, I solved this question but I want to know other method which u think will be the shortest method. Find the number of positive integral values of \(x \leq 100\) such that \(3^{x} - x^{2}\) is divisible by 5.I think answer is 20.

Note by Kiran Patel
7 years, 11 months ago

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We are looking for the values of xx such that 3xx2=5k3^x-x^2=5k with kZk\in\mathbb{Z}, or 3xx2(mod5)3^x \equiv x^2 \pmod{5}. The sequence ax=3x(mod5)a_x = 3^x \pmod{5} for 1x1001 \leq x \leq 100 is 3,4,2,1,3,4,2,1,3,4,2,1,3,4,2,1,\dots with a period of 4. The sequence bx=x2(mod5)b_x = x^2 \pmod{5} for 1x1001 \leq x \leq 100 is 1,4,4,1,0,1,4,4,1,0,1,4,4,1,0,1,4,4,1,0,\dots with a period of 5. So, if 3xx2(mod5)    ax=bx3^x \equiv x^2 \pmod{5} \implies a_x=b_x, then either ax=bx=1a_x=b_x=1 or ax=bx=4a_x=b_x=4.

Case 1: ax=bx=1a_x=b_x=1. ax=1a_x = 1 whenever x0(mod4)x \equiv 0 \pmod{4} and bx=1b_x = 1 whenever x1(mod5)x \equiv 1 \pmod{5} or x4(mod5)x \equiv 4 \pmod{5}.

\bullet Case 1a) x0(mod4)x \equiv 0 \pmod{4} and x1(mod5)    x16(mod20)    x{16,36,56,76,96}x \equiv 1 \pmod{5} \implies x \equiv 16 \pmod{20} \implies x \in \{16,36,56,76,96\}.

\bullet Case 1b) x0(mod4)x \equiv 0 \pmod{4} and x4(mod5)    x4(mod20)    x{4,24,44,64,84}x \equiv 4 \pmod{5} \implies x \equiv 4 \pmod{20} \implies x \in \{4,24,44,64,84\}.

Case 2: ax=bx=4a_x=b_x=4. ax=4a_x = 4 whenever x2(mod4)x \equiv 2 \pmod{4} and bx=4b_x = 4 whenever x2(mod5)x \equiv 2 \pmod{5} or x3(mod5)x \equiv 3 \pmod{5}.

\bullet Case 2a) x2(mod4)x \equiv 2 \pmod{4} and x2(mod5)    x2(mod20)    x{2,22,42,62,82}x \equiv 2 \pmod{5} \implies x \equiv 2 \pmod{20} \implies x \in \{2,22,42,62,82\}.

\bullet Case 2b) x2(mod4)x \equiv 2 \pmod{4} and x3(mod5)    x18(mod20)    x{18,38,58,78,98}x \equiv 3 \pmod{5} \implies x \equiv 18 \pmod{20} \implies x \in \{18,38,58,78,98\}.

So, x{2,4,16,18,22,24,36,38,42,44,56,58,62,64,76,78,82,84,96,98}x \in \{ 2,4,16,18,22,24,36,38,42,44,56,58,62,64,76,78,82,84,96,98 \}, which are indeed 20 values.

Tim Vermeulen - 7 years, 11 months ago

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I did the same but I want any other short method.Thanks for ur efforts......

Kiran Patel - 7 years, 11 months ago
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