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☞ If 'a' is the actual product of 'b' and 'c', how many solutions are there in the positive integer of the equation a + b + c = 60?

(Answer : 144)

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Answer 1

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Answer 2

$\large\underline{\sf{Solution-}}$

Given that,

a is the actual product of b

Let assume that b = ax where x is a natural number more than 1.

Also, given that,

a is the actual product of c.

Let assume that c = ay where y is a natural number more than 1.

It means

$\boxed{ \bf{ \: a \: \ne \: b \: \ne \: c}}$

Now, its further given that,

$\rm :\longmapsto\:a + b + c = 60$

$\rm :\longmapsto\:a + ax + ay = 60$

$\rm :\longmapsto\:a(1 + x + y) = 60$

$\rm :\longmapsto\:1 + x + y = \dfrac{60}{a}$

$\rm :\longmapsto\:x + y = \dfrac{60}{a} - 1$

As x and y are natural numbers,

a can assume the values 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30.

Now, we know from combinations,

The number of ways in which n identical things can be distributed in to r different groups, no group being blank is

$\rm \:  =  \:  \: \boxed{ \bf{ \: ^{n+r-1}C_{r-1}}}$

So, number of ways in which x + y can assume the values from 60/a - 1 is given by

$\rm \:  =  \:  \: ^{\dfrac{60}{a} - 1-1}C_{2-1}$

$\rm \:  =  \:  \: ^{\dfrac{60}{a} - 2}C_{1}$

$\rm \:  =  \:  \: \dfrac{60}{a} - 2$

Now the number of possible ways are

$\rm :\longmapsto\: \:When \: a = 1, x+y = \:  \: \dfrac{60}{1} - 2 = 58$

$\rm :\longmapsto\: \:When \: a = 2, x+y = \:  \: \dfrac{60}{2} - 2 = 28$

$\rm :\longmapsto\: \:When \: a = 3, x+y = \:  \: \dfrac{60}{3} - 2 = 18$

$\rm :\longmapsto\: \:When \: a = 4, x+y = \:  \: \dfrac{60}{4} - 2 = 13$

$\rm :\longmapsto\: \:When \: a = 5, x+y = \:  \: \dfrac{60}{5} - 2 = 10$

$\rm :\longmapsto\: \:When \: a = 6, x+y = \:  \: \dfrac{60}{6} - 2 = 8$

$\rm :\longmapsto\: \:When \: a = 10, x+y = \:  \: \dfrac{60}{10} - 2 = 4$

$\rm :\longmapsto\: \:When \: a = 12, x+y = \:  \: \dfrac{60}{12} - 2 = 3$

$\rm :\longmapsto\: \:When \: a = 15, \: x+y = \:  \: \dfrac{60}{15} - 2 = 2$

$\rm :\longmapsto\: \:When \: a = 20, \: x+y = \:  \: \dfrac{60}{20} - 2 = 1 \\ \bf\implies \:b = 20 \: and \: c = 20 \: which \: is \: rejected$

$\rm :\longmapsto\: \:When \: a = 30, \: x+y = \:  \: \dfrac{60}{30} - 2 = 0$

So, total number of solutions for which a + b + c = 60 is

$\rm \:  =  \:  \: 58 + 28 + 18 + 13 + 10 + 8 + 4 + 3 + 2 + 0$

$\rm \:  =  \:  \: 144$