x-y-z-N-with-x-gt-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-su

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 by mathocean1 last updated on 19/Nov/20

t h a n k s

Answered by JDamian last updated on 04/Nov/20

(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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