Question

please help me....... in solvoing my question

Answer 1

$hi$
Let the H.C.F be x

By the given condition ,

L.C.M = H.C.F + 57
L.C.M = x + 57

By the 2nd given condition ,

$( {x})^{2} + {(x + 57)}^{2} = 3609$

$x ^{2} + ( {x}^{2} + 114x + 3249) \\ = 3609$

$2 {x}^{2} + 114x + 3249 = 3609$

$2 {x}^{2} + 114x - 360 = 0$

$(dividing \: through \: by \: 2)$

${x}^{2} + 57x - 180 = 0$

${x}^{2} + 60x - 3x - 360 = 0$

$x(x + 60) - 3(x + 60) = 0$

$(x + 60)(x - 3) = 0$

$therefore..$

$x + 60 = 0 \: \: or \: \: x - 3 = 0$

$x = - 60 \: \: or \: \: x = 3$

$x = - 60 \: isn't \: acceptable$
So ...
$x = 3$
The H.C.F of two numbers is 3

Now,
$l.c.m = x + 57$
$l.c.m = 3 + 57$
$l.c.m = 60$
We have got
$l.c.m = 60 \: \: and \: \: h.c.f = 3$
We know that ,
L.C.M × H.C.F = Product of numbers

60 × 3 = Product of numbers

Thus ,
$product \: of \: numbers = 180$
___________________________
$hope \: \: this \: \: helps..$