Menu Close

The-numbers-of-pairs-of-natural-numbers-x-y-with-x-y-33-that-satisfy-5-3-x-y-1-2-y-2x-is-A-295-B-296-C-297-D-298-E-299-

Tinku Tara 0 Comments
Pin




Question Number 212515 by efronzo1 last updated on 16/Oct/24
The numbers of pairs of natural numbers (x,y) with x,y ≤ 33 that satisfy 5 ∣ 3^x^(y−1) + 2^y^(2x) is ... (A) 295 (B) 296 (C) 297 (D) 298 (E) 299
$$\:\:\mathrm{The}\:\mathrm{numbers}\:\mathrm{of}\:\mathrm{pairs}\:\mathrm{of}\:\mathrm{natural}\: \\ $$$$\:\:\:\mathrm{numbers}\:\left(\mathrm{x},\mathrm{y}\right)\:\mathrm{with}\:\mathrm{x},\mathrm{y}\:\leqslant\:\mathrm{33}\:\mathrm{that}\: \\ $$$$\:\:\:\mathrm{satisfy}\:\mathrm{5}\:\mid\:\mathrm{3}^{\mathrm{x}^{\mathrm{y}−\mathrm{1}} } \:+\:\mathrm{2}^{\mathrm{y}^{\mathrm{2x}} } \:\mathrm{is}\:…\: \\ $$$$\:\:\left(\mathrm{A}\right)\:\mathrm{295}\:\:\:\:\left(\mathrm{B}\right)\:\mathrm{296}\:\:\:\:\:\:\:\left(\mathrm{C}\right)\:\mathrm{297}\:\:\:\:\:\:\left(\mathrm{D}\right)\:\mathrm{298}\:\:\:\left(\mathrm{E}\right)\:\mathrm{299} \\ $$
Commented by A5T last updated on 16/Oct/24
I′m getting 313 pairs, do you have any solution to this?
$${I}'{m}\:{getting}\:\mathrm{313}\:{pairs},\:{do}\:{you}\:{have}\:{any}\:{solution} \\ $$$${to}\:{this}? \\ $$
Answered by A5T last updated on 16/Oct/24
3^x^(y−1) +2^y^(2x) ≡(−2)^x^(y−1) +2^y^(2x) ≡0(mod 5) Note that when x(=2k) is even and y≥3 (−2)^x^(y−1) +2^y^(2x) =((−2)^2^(y−1) )^k^(y−1) +2^y^(4k) ≡0(mod 5) ⇒2^y^(4k) ≡4(mod 5) ⇒y^(4k) −2≡0(mod 4)⇒y^(4k) ≡2(mod 4) →← So, x(=2k+1) is odd when y≥3 when y=1;3^x^(y−1) +2^y^(2x) ≡ 3+2≡0(mod 5) for any x ⇒(x,y)=(1,1),(2,1),(3,1),...,(33,1)⇒33 solutions when y=2; 3^x^(y−1) +2^y^(2x) ≡3^x +2^4^x ≡3^x +1≡0(mod 5) ⇒3^x ≡4(mod 5)⇒y=2∧x=4q+2⇒8 solutions when y≥3, we know that x is odd when x=1; 3^x^(y−1) +2^y^(2x) ≡3+2^y^2 ≡0(mod 5) ⇒2^y^2 ≡2(mod 5)⇒y^2 ≡1(mod 4)⇒y=4l+1 or 4l+3≥3 ⇒(x,y)=(1,3),(1,5),...,(1,33)⇒16 solutions Suppose y(=2m+1)≥3 is also odd and x≥3 3^x^(y−1) +2^y^(2x) =3^((2k+1)^(2m) ) +2^((2m+1)^(4k+2) ) ⇒2^((2m+1)^(4k+2) ) ≡2^((2k+1)^(2m) ) (mod 5) ⇒(2m+1)^(4k+2) ≡(2k+1)^(2m) (mod 4) ⇒1+(2m)^1 (4k+2)≡1+(2k)(2m) (mod 4) which is always true for all m and k(≥1) ⇒16×16=256 solutions When y(=2m) is even. ⇒3^((2k+1)^(2m−1) ) +2^((2m)^(4k+2) ) ≡0(mod 5) ⇒(2m)^(4k+2) ≡(2k+1)^(2m−1) (mod 4) →← ⇒Total number of solutions=256+16+8+33 =313
$$\mathrm{3}^{{x}^{{y}−\mathrm{1}} } +\mathrm{2}^{{y}^{\mathrm{2}{x}} } \equiv\left(−\mathrm{2}\right)^{{x}^{{y}−\mathrm{1}} } +\mathrm{2}^{{y}^{\mathrm{2}{x}} } \equiv\mathrm{0}\left({mod}\:\mathrm{5}\right) \\ $$$${Note}\:{that}\:{when}\:{x}\left(=\mathrm{2}{k}\right)\:{is}\:{even}\:{and}\:{y}\geqslant\mathrm{3} \\ $$$$\left(−\mathrm{2}\right)^{{x}^{{y}−\mathrm{1}} } +\mathrm{2}^{{y}^{\mathrm{2}{x}} } =\left(\left(−\mathrm{2}\right)^{\mathrm{2}^{{y}−\mathrm{1}} } \right)^{{k}^{{y}−\mathrm{1}} } +\mathrm{2}^{{y}^{\mathrm{4}{k}} } \equiv\mathrm{0}\left({mod}\:\mathrm{5}\right) \\ $$$$\Rightarrow\mathrm{2}^{{y}^{\mathrm{4}{k}} } \equiv\mathrm{4}\left({mod}\:\mathrm{5}\right) \\ $$$$\Rightarrow{y}^{\mathrm{4}{k}} −\mathrm{2}\equiv\mathrm{0}\left({mod}\:\mathrm{4}\right)\Rightarrow{y}^{\mathrm{4}{k}} \equiv\mathrm{2}\left({mod}\:\mathrm{4}\right)\:\:\:\:\:\rightarrow\leftarrow \\ $$$${So},\:{x}\left(=\mathrm{2}{k}+\mathrm{1}\right)\:{is}\:{odd}\:{when}\:{y}\geqslant\mathrm{3} \\ $$$${when}\:{y}=\mathrm{1};\mathrm{3}^{{x}^{{y}−\mathrm{1}} } +\mathrm{2}^{{y}^{\mathrm{2}{x}} } \equiv\:\mathrm{3}+\mathrm{2}\equiv\mathrm{0}\left({mod}\:\mathrm{5}\right)\:{for}\:{any}\:{x} \\ $$$$\Rightarrow\left({x},{y}\right)=\left(\mathrm{1},\mathrm{1}\right),\left(\mathrm{2},\mathrm{1}\right),\left(\mathrm{3},\mathrm{1}\right),…,\left(\mathrm{33},\mathrm{1}\right)\Rightarrow\mathrm{33}\:{solutions} \\ $$$${when}\:{y}=\mathrm{2};\:\mathrm{3}^{{x}^{{y}−\mathrm{1}} } +\mathrm{2}^{{y}^{\mathrm{2}{x}} } \equiv\mathrm{3}^{{x}} +\mathrm{2}^{\mathrm{4}^{{x}} } \equiv\mathrm{3}^{{x}} +\mathrm{1}\equiv\mathrm{0}\left({mod}\:\mathrm{5}\right) \\ $$$$\Rightarrow\mathrm{3}^{{x}} \equiv\mathrm{4}\left({mod}\:\mathrm{5}\right)\Rightarrow{y}=\mathrm{2}\wedge{x}=\mathrm{4}{q}+\mathrm{2}\Rightarrow\mathrm{8}\:{solutions} \\ $$$${when}\:{y}\geqslant\mathrm{3},\:{we}\:{know}\:{that}\:{x}\:{is}\:{odd} \\ $$$${when}\:{x}=\mathrm{1};\:\mathrm{3}^{{x}^{{y}−\mathrm{1}} } +\mathrm{2}^{{y}^{\mathrm{2}{x}} } \equiv\mathrm{3}+\mathrm{2}^{{y}^{\mathrm{2}} } \equiv\mathrm{0}\left({mod}\:\mathrm{5}\right) \\ $$$$\Rightarrow\mathrm{2}^{{y}^{\mathrm{2}} } \equiv\mathrm{2}\left({mod}\:\mathrm{5}\right)\Rightarrow{y}^{\mathrm{2}} \equiv\mathrm{1}\left({mod}\:\mathrm{4}\right)\Rightarrow{y}=\mathrm{4}{l}+\mathrm{1}\:{or}\:\mathrm{4}{l}+\mathrm{3}\geqslant\mathrm{3} \\ $$$$\Rightarrow\left({x},{y}\right)=\left(\mathrm{1},\mathrm{3}\right),\left(\mathrm{1},\mathrm{5}\right),…,\left(\mathrm{1},\mathrm{33}\right)\Rightarrow\mathrm{16}\:{solutions} \\ $$$${Suppose}\:{y}\left(=\mathrm{2}{m}+\mathrm{1}\right)\geqslant\mathrm{3}\:{is}\:{also}\:{odd}\:{and}\:{x}\geqslant\mathrm{3} \\ $$$$\mathrm{3}^{{x}^{{y}−\mathrm{1}} } +\mathrm{2}^{{y}^{\mathrm{2}{x}} } =\mathrm{3}^{\left(\mathrm{2}{k}+\mathrm{1}\right)^{\mathrm{2}{m}} } +\mathrm{2}^{\left(\mathrm{2}{m}+\mathrm{1}\right)^{\mathrm{4}{k}+\mathrm{2}} } \\ $$$$\Rightarrow\mathrm{2}^{\left(\mathrm{2}{m}+\mathrm{1}\right)^{\mathrm{4}{k}+\mathrm{2}} } \equiv\mathrm{2}^{\left(\mathrm{2}{k}+\mathrm{1}\right)^{\mathrm{2}{m}} } \left({mod}\:\mathrm{5}\right) \\ $$$$\Rightarrow\left(\mathrm{2}{m}+\mathrm{1}\right)^{\mathrm{4}{k}+\mathrm{2}} \equiv\left(\mathrm{2}{k}+\mathrm{1}\right)^{\mathrm{2}{m}} \left({mod}\:\mathrm{4}\right) \\ $$$$\Rightarrow\mathrm{1}+\left(\mathrm{2}{m}\right)^{\mathrm{1}} \left(\mathrm{4}{k}+\mathrm{2}\right)\equiv\mathrm{1}+\left(\mathrm{2}{k}\right)\left(\mathrm{2}{m}\right)\:\left({mod}\:\mathrm{4}\right) \\ $$$${which}\:{is}\:{always}\:{true}\:{for}\:{all}\:{m}\:{and}\:{k}\left(\geqslant\mathrm{1}\right) \\ $$$$\Rightarrow\mathrm{16}×\mathrm{16}=\mathrm{256}\:{solutions} \\ $$$${When}\:{y}\left(=\mathrm{2}{m}\right)\:{is}\:{even}. \\ $$$$\Rightarrow\mathrm{3}^{\left(\mathrm{2}{k}+\mathrm{1}\right)^{\mathrm{2}{m}−\mathrm{1}} } +\mathrm{2}^{\left(\mathrm{2}{m}\right)^{\mathrm{4}{k}+\mathrm{2}} } \equiv\mathrm{0}\left({mod}\:\mathrm{5}\right) \\ $$$$\Rightarrow\left(\mathrm{2}{m}\right)^{\mathrm{4}{k}+\mathrm{2}} \equiv\left(\mathrm{2}{k}+\mathrm{1}\right)^{\mathrm{2}{m}−\mathrm{1}} \left({mod}\:\mathrm{4}\right)\:\:\:\:\:\rightarrow\leftarrow \\ $$$$\Rightarrow{Total}\:{number}\:{of}\:{solutions}=\mathrm{256}+\mathrm{16}+\mathrm{8}+\mathrm{33} \\ $$$$=\mathrm{313} \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *