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For a prime p and a natural number n we shall denote by v_p(n) the power of p nbsp;dividing n. Then it is enough to show that v_p(a_2) = v_p nbsp ...

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Byju's Answer

Standard X

Mathematics

Applications of HCF & LCM in Real-World Problems

Let a_1,...

Question

Let $a_{1}, b_{1}, c_{1}$ be natural numbers. We define
$a_{2} = g c d (b_{1}, c_{1}), b_{2} = g c d (c_{1}, a_{1}), c_{2} = g c d (a_{1}, b_{1}),$
and
$a_{3} = l c m (b_{2}, c_{2}), b_{3} = l c m (c_{2}, a_{2}), c_{3} = l c m (a_{2}, b_{2}) .$
Show that $g c d (b_{3}, c_{3}) = a_{2}$ .

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Solution

For a prime p and a natural number n we shall denote by $v_{p} (n)$ the power of p dividing n.
Then it is enough to show that $v_{p} (a_{2}) = v_{p} (g c d (b_{3}, c_{3}))$ for all primes p.
Let p be a prime and let $α = v_{p} (a_{1}), β = v_{p} (b_{1})$ and $γ = v_{p} (c_{1})$ .
Bysymmetry, we assume that $α \leq β \leq γ$ .
$∴ v_{p} (a_{2}) = m i n$ { $β, γ$ } = $β$ and similarly $v_{p} (b_{2}) = v_{p} (c_{2}) = α$ .
$∴ v_{p} (b_{3}) = m a x$ { $α, β$ } = $β$ and similarly $v_{p} (c_{3}) = m a x$ { $α, β$ } = $β$ .
$∴ v_{p} (g c d (b_{3}, c_{3})) = v_{p} (a_{2}) = β$

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Similar questions

Q. If each pair of equations

a_{1} x^{2} + b_{1} x + c_{1} = 0, a_{2} x^{2} + b_{2} x + c_{2} = 0

and

a_{3} x^{2} + b_{3} x + c_{3} = 0

has a common root, then show that
(i)

\frac{c_{1} a_{2} + c_{2} a_{1}}{c_{1} a_{2} - c_{2} a_{1}} + \frac{a_{1} a_{2} b_{3}}{a_{3} (a_{1} b_{2} - a_{2} b_{1})} = 0

(ii)

{(\frac{a_{1} b_{2} - a_{2} b_{1}}{a_{1} c_{2} - a_{2} c_{1}})}^{2} = \frac{a_{1} a_{2} c_{3}}{c_{1} c_{2} a_{3}}

Q. If

| a_{1} | > | a_{2} | + | a_{3} |, | b_{2} | > | b_{1} | + | b_{3} |

and

| c_{3} | > | c_{1} | + | c_{2} |

, then show

∣ ∣ ∣ ∣ \begin{matrix} a_{1} & a_{2} & a_{3} b_{1} & b_{2} & b_{3} c_{1} & c_{2} & c_{3} \end{matrix} ∣ ∣ ∣ ∣ = 0

Q. If

(a_{1}, a_{2}, a_{3})

(b_{1}, b_{2}, b_{3})

and

(c_{1}, c_{2}, c_{3})

are linearly independent and

x (a_{1}, a_{2}, a_{3}) + y (b_{1}, b_{2}, b_{3}) + z (c_{1}, c_{2}, c_{3}) = 0

then

If A1, B1, C1 … are the cofactors of the elements a1, b1, c1... of the matrix

$⎡ ⎢ ⎣ \begin{matrix} a 1 & b 1 & c 1 a 2 & b 2 & c 2 a 3 & b 3 & c 3 \end{matrix} ⎤ ⎥ ⎦$ then $∣ ∣ ∣ \begin{matrix} B 2 & C 2 B 3 & C 3 \end{matrix} ∣ ∣ ∣ =$

A set of vectors

{(a_{1}, a_{2}, a_{3}), (b_{1}, b_{2}, b_{3}), (c_{1}, c_{2}, c_{3})}

is said to be linearly independent if and only if
$\begin{vmatrix}
a_{1} & a_{2} & a_{3}\\
b_{1} & b_{2} & b_{3}\\
c_{1} & c_{2} & c_{3}
\end{vmatrix}\neq 0$
otherwise the set is said to be linearly dependent. A similar result holds for

{(a_{1}, a_{2}), (b_{1}, b_{2})}

(a_{1}, a_{2}, a_{3})

(b_{1}, b_{2}, b_{3})

and

(c_{1}, c_{2}, c_{3})

are linearly independent and

x, y, z ϵ R

, then

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Let a1,b1,c1 be natural numbers. We definea2=gcd(b1,c1),b2=gcd(c1,a1),c2=gcd(a1,b1),anda3=lcm(b2,c2),b3=lcm(c2,a2),c3=lcm(a2,b2).Show that gcd(b3,c3)=a2.

Let $a_{1}, b_{1}, c_{1}$ be natural numbers. We define
$a_{2} = g c d (b_{1}, c_{1}), b_{2} = g c d (c_{1}, a_{1}), c_{2} = g c d (a_{1}, b_{1}),$
and
$a_{3} = l c m (b_{2}, c_{2}), b_{3} = l c m (c_{2}, a_{2}), c_{3} = l c m (a_{2}, b_{2}) .$
Show that $g c d (b_{3}, c_{3}) = a_{2}$ .