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.


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Question Number 121012 by mathocean1 last updated on 04/Nov/20

$x; y; z \in N^{*} with x > 3 are$ $numbers .$ $we suppose that y is equal$ $to 121 in base x and z is equal$ $to 110 in base x .$ $1) show that we can write$ $(without knowing x) the$ $product xyz in base x .$ $2) we suppose now that$ $x + y + z is equal to 49 in base$ $10; determinate x and xyz$ $in base 10 .$
Commented bymathocean1 last updated on 19/Nov/20

$t h a n k s$
	Answered by JDamian last updated on 04/Nov/20
	$(1) n_((x) × x is the same that apending a zero as the rightmost digit (n_((x) 0). In fact, z=11_((x) ×x. Then zxy is 11_((x) ×121_((x) with 00 as the rightmost digits: 133100_((x) (2) x+y+z. In any base x the value x is 10_((x) 10_((x) +110_((x) +121_((x) = 120_((x) + 121_((x) = 49 120_((x) + 120_((x) + 1 = 48+1 2∙120_((x) = 48 = 2 ∙ 24 120_((x) =24 x^2 +2x=24 x=((−2±(√(2^2 −4∙1∙(−24))))/2)= { ((x=4 ✓)),((x=−6 ×)) :} x=4$
	$(1) n_{(x} \times x is the same that apending a zero$ $as the rightmost digit (n_{(x} 0) . In fact,$ $z = 11_{(x} \times x . Then z x y is 11_{(x} \times 121_{(x} with 00$ $as the rightmost digits : 133100_{(x}$ $(2) x + y + z . In any base x the value x i s$ $10_{(x}$ $10_{(x} + 110_{(x} + 121_{(x} = 120_{(x} + 121_{(x} = 49$ $120_{(x} + 120_{(x} + 1 = 48 + 1$ $2 \cdot 120_{(x} = 48 = 2 \cdot 24$ $120_{(x} = 24$ $x^{2} + 2 x = 24$ $x = \frac{- 2 \pm \sqrt{2^{2} - 4 \cdot 1 \cdot (- 24)}}{2} = {\begin{cases} x = 4 ✓ \\ x = - 6 \times \end{cases}$ $x = 4$
	Answered by Olaf last updated on 04/Nov/20

	$1)$ $y = 121_{x} = x^{2} + 2 x + 1 = {(x + 1)}^{2}$ $z = 110_{x} = x^{2} + x = x (x + 1)$ $x y z = x \times {(x + 1)}^{2} \times x (x + 1)$ $x y z = x^{2} {(x + 1)}^{3}$ $x y z = x^{2} (x^{3} + 3 x^{2} + 3 x + 1)$ $x y z = x^{5} + 3 x^{4} + 3 x^{3} + x^{2} + 0 x + 0$ $\Rightarrow x y z = 133100_{x}$ $2)$ $If x = 4_{10} :$ $y = 121_{x} = 25_{10}$ $z = 110_{x} = 20_{10}$ $\Rightarrow x + y + z = 49_{10}$ $and x y z = x^{2} {(x + 1)}^{3} = 16 \times 5^{3} = 2000_{10}$ $We can verify that :$ $133100_{x} = 4^{5} + 3 \times 4^{4} + 3 \times 4^{3} + 4^{2} = 2000_{10}$
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