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Question Number 121012 by mathocean1 last updated on 04/Nov/20
x;y;z ∈ N^(∗ ) with x>3 are numbers. we suppose that y is equal to 121 in base x and z is equal to 110 in base x. 1) show that we can write (without knowing x) the product xyz in base x. 2) we suppose now that x+y+z is equal to 49 in base 10; determinate x and xyz in base 10.
$$\mathrm{x};\mathrm{y};\mathrm{z}\:\in\:\mathbb{N}^{\ast\:} \mathrm{with}\:\mathrm{x}>\mathrm{3}\:\mathrm{are}\: \\ $$$$\mathrm{numbers}. \\ $$$$\mathrm{we}\:\mathrm{suppose}\:\mathrm{that}\:\mathrm{y}\:\mathrm{is}\:\mathrm{equal} \\ $$$$\mathrm{to}\:\mathrm{121}\:\mathrm{in}\:\mathrm{base}\:\mathrm{x}\:\mathrm{and}\:\mathrm{z}\:\mathrm{is}\:\mathrm{equal} \\ $$$$\mathrm{to}\:\mathrm{110}\:\mathrm{in}\:\mathrm{base}\:\mathrm{x}. \\ $$$$\left.\mathrm{1}\right)\:\mathrm{show}\:\mathrm{that}\:\mathrm{we}\:\mathrm{can}\:\mathrm{write} \\ $$$$\left(\mathrm{without}\:\mathrm{knowing}\:\mathrm{x}\right)\:\mathrm{the}\: \\ $$$$\mathrm{product}\:\mathrm{xyz}\:\mathrm{in}\:\mathrm{base}\:\mathrm{x}. \\ $$$$\left.\mathrm{2}\right)\:\mathrm{we}\:\mathrm{suppose}\:\mathrm{now}\:\mathrm{that} \\ $$$$\mathrm{x}+\mathrm{y}+\mathrm{z}\:\:\:\mathrm{is}\:\mathrm{equal}\:\mathrm{to}\:\mathrm{49}\:\mathrm{in}\:\mathrm{base} \\ $$$$\mathrm{10};\:\mathrm{determinate}\:\mathrm{x}\:\mathrm{and}\:\mathrm{xyz} \\ $$$$\mathrm{in}\:\mathrm{base}\:\mathrm{10}. \\ $$
Commented by mathocean1 last updated on 19/Nov/20
thanks
$${thanks} \\ $$
Answered by JDamian last updated on 04/Nov/20
(1) n_((x) × x is the same that apending a zero as the rightmost digit (n_((x) 0). In fact, z=11_((x) ×x. Then zxy is 11_((x) ×121_((x) with 00 as the rightmost digits: 133100_((x) (2) x+y+z. In any base x the value x is 10_((x) 10_((x) +110_((x) +121_((x) = 120_((x) + 121_((x) = 49 120_((x) + 120_((x) + 1 = 48+1 2∙120_((x) = 48 = 2 ∙ 24 120_((x) =24 x^2 +2x=24 x=((−2±(√(2^2 −4∙1∙(−24))))/2)= { ((x=4 ✓)),((x=−6 ×)) :} x=4
$$\left(\mathrm{1}\right)\:{n}_{\left({x}\right.} ×\:{x}\:\:\:\:\mathrm{is}\:\mathrm{the}\:\mathrm{same}\:\mathrm{that}\:\mathrm{apending}\:\mathrm{a}\:\mathrm{zero} \\ $$$$\mathrm{as}\:\mathrm{the}\:\mathrm{rightmost}\:\mathrm{digit}\:\left({n}_{\left({x}\right.} \mathrm{0}\right).\:\mathrm{In}\:\mathrm{fact}, \\ $$$${z}=\mathrm{11}_{\left({x}\right.} ×{x}.\:\mathrm{Then}\:{zxy}\:\mathrm{is}\:\mathrm{11}_{\left({x}\right.} ×\mathrm{121}_{\left({x}\right.} \:\mathrm{with}\:\mathrm{00} \\ $$$$\mathrm{as}\:\mathrm{the}\:\mathrm{rightmost}\:\mathrm{digits}:\:\mathrm{133100}_{\left({x}\right.} \\ $$$$\left(\mathrm{2}\right)\:{x}+{y}+{z}.\:\mathrm{In}\:\mathrm{any}\:\mathrm{base}\:{x}\:\mathrm{the}\:\mathrm{value}\:{x}\:{is} \\ $$$$\mathrm{10}_{\left({x}\right.} \\ $$$$\mathrm{10}_{\left({x}\right.} +\mathrm{110}_{\left({x}\right.} +\mathrm{121}_{\left({x}\right.} =\:\mathrm{120}_{\left({x}\right.} +\:\mathrm{121}_{\left({x}\right.} =\:\mathrm{49} \\ $$$$\mathrm{120}_{\left({x}\right.} +\:\mathrm{120}_{\left({x}\right.} +\:\mathrm{1}\:=\:\mathrm{48}+\mathrm{1} \\ $$$$\mathrm{2}\centerdot\mathrm{120}_{\left({x}\right.} =\:\mathrm{48}\:=\:\mathrm{2}\:\centerdot\:\mathrm{24} \\ $$$$\mathrm{120}_{\left({x}\right.} =\mathrm{24} \\ $$$${x}^{\mathrm{2}} +\mathrm{2}{x}=\mathrm{24} \\ $$$${x}=\frac{−\mathrm{2}\pm\sqrt{\mathrm{2}^{\mathrm{2}} −\mathrm{4}\centerdot\mathrm{1}\centerdot\left(−\mathrm{24}\right)}}{\mathrm{2}}=\begin{cases}{{x}=\mathrm{4}\:\checkmark}\\{{x}=−\mathrm{6}\:×}\end{cases} \\ $$$${x}=\mathrm{4} \\ $$
Answered by Olaf last updated on 04/Nov/20
1) y = 121_x = x^2 +2x+1 = (x+1)^2 z = 110_x = x^2 +x = x(x+1) xyz = x×(x+1)^2 ×x(x+1) xyz = x^2 (x+1)^3 xyz = x^2 (x^3 +3x^2 +3x+1) xyz = x^5 +3x^4 +3x^3 +x^2 +0x+0 ⇒ xyz = 133100_x 2) If x = 4_(10) : y = 121_x = 25_(10) z = 110_x = 20_(10) ⇒ x+y+z = 49_(10) and xyz = x^2 (x+1)^3 = 16×5^3 = 2000_(10) We can verify that : 133100_x = 4^5 +3×4^4 +3×4^3 +4^2 = 2000_(10)
$$\left.\mathrm{1}\right) \\ $$$${y}\:=\:\mathrm{121}_{{x}} \:=\:{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{1}\:=\:\left({x}+\mathrm{1}\right)^{\mathrm{2}} \\ $$$${z}\:=\:\mathrm{110}_{{x}} \:=\:{x}^{\mathrm{2}} +{x}\:=\:{x}\left({x}+\mathrm{1}\right) \\ $$$$ \\ $$$${xyz}\:=\:{x}×\left({x}+\mathrm{1}\right)^{\mathrm{2}} ×{x}\left({x}+\mathrm{1}\right) \\ $$$${xyz}\:=\:{x}^{\mathrm{2}} \left({x}+\mathrm{1}\right)^{\mathrm{3}} \\ $$$${xyz}\:=\:{x}^{\mathrm{2}} \left({x}^{\mathrm{3}} +\mathrm{3}{x}^{\mathrm{2}} +\mathrm{3}{x}+\mathrm{1}\right) \\ $$$${xyz}\:=\:{x}^{\mathrm{5}} +\mathrm{3}{x}^{\mathrm{4}} +\mathrm{3}{x}^{\mathrm{3}} +{x}^{\mathrm{2}} +\mathrm{0}{x}+\mathrm{0} \\ $$$$\Rightarrow\:{xyz}\:=\:\mathrm{133100}_{{x}} \\ $$$$ \\ $$$$\left.\mathrm{2}\right) \\ $$$$\mathrm{If}\:{x}\:=\:\mathrm{4}_{\mathrm{10}} \:: \\ $$$${y}\:=\:\mathrm{121}_{{x}} \:=\:\mathrm{25}_{\mathrm{10}} \\ $$$${z}\:=\:\mathrm{110}_{{x}} \:=\:\mathrm{20}_{\mathrm{10}} \\ $$$$\Rightarrow\:{x}+{y}+{z}\:=\:\mathrm{49}_{\mathrm{10}} \\ $$$$\mathrm{and}\:{xyz}\:=\:{x}^{\mathrm{2}} \left({x}+\mathrm{1}\right)^{\mathrm{3}} \:=\:\mathrm{16}×\mathrm{5}^{\mathrm{3}} \:=\:\mathrm{2000}_{\mathrm{10}} \\ $$$$ \\ $$$$\mathrm{We}\:\mathrm{can}\:\mathrm{verify}\:\mathrm{that}\:: \\ $$$$\mathrm{133100}_{{x}} \:=\:\mathrm{4}^{\mathrm{5}} +\mathrm{3}×\mathrm{4}^{\mathrm{4}} +\mathrm{3}×\mathrm{4}^{\mathrm{3}} +\mathrm{4}^{\mathrm{2}} \:=\:\mathrm{2000}_{\mathrm{10}} \\ $$

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