Question

The number of solutions in natural numbers to the equation xyz = 200

Answer 1

The number of solutions are 60

Step-by-step explanation:

Given:

$xyz=200$

To find out:

No. of natural number solutions of the above equation

Solution:

$xyz=200$

$\implies xyz=2\times 2\times 5\times 2\times 5$

$\implies xyz=2^35^2$

We can write

$x=2^{m_1}5^{n_1}$

$x=2^{m_2}5^{n_2}$

$x=2^{m_3}5^{n_3}$

Thus,

$xyz=2^{(m_1+m_2+m_3)}5^{(n_1+n_2+n_3)}$

Therefore

$m_1+m_2+m_3=3$   and  $0\le m_1+m_2+m_3\le 3$

$n_1+n_2+n_3=2$       and $0\le n_1+n_2+n_3\le 2$

Possibilities for $m_1,m_2,m_3$ are

(1,1,1), [(0,1,2) pair] × 3!, [(0,0,3) pair] × (3!/2!)     Total 10 pairs

Possibilities for $n_1,n_2,n_3$ are

(0,1,1) × (3!/2!) and (0,0,2) × (3!/2!)      Total 6 pairs

Therefore total number of solutions of the equation = 10 × 6 = 60

Hope this answer is helpful.

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