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TIFR-2011-Maths-20

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Yes,

$2010$

$= 2\times3 \times 5 \times 67$

$= 2\times67 \times 15$$= 134\times 15$

$= (63+70+1)\times 15$

$= 63\times 15+70\times 15+15\times1$

i.e, $x=15$, $y=15$, and $z=1$ satisfies the given equation.

So,$(15,15,1)$ is an Integral solution for equation  $63x+70y+15z=2010$

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