2023/010) Given $2x = 3y-5$ find the value of $8x^3-27y^3+90xy +125$

 We have 

$2x = 3y-5$

Cubing both sides as in the required condition we have $8x^3$ and $27y^3$

We get $8x^3 = (3y-5)^3 = (3y)^3) - 5^3 -3(3y)5(3y-5)$ (using $(a -b)^3 = a^3- b^3 -3ab(a-b)$

or  $8x^3 = (3y-5)^3 = (3y)^3 - 5^3 -3(3y)*5 * 2x$ as $3y-5 = 2x$

or $8x^3 = 27y^3- 125 - 90xy$

or $8x^3-27y^3 + 90xy +125 = 0$

2023/009) Find n such that $\lfloor \frac{n}{2} \rfloor + \lfloor \frac{n}{3} \rfloor + \lfloor \frac{n}{6} \rfloor = n$

We have $\lfloor x \rfloor \le x$ and equal ony of x is integer 

so    $\lfloor \frac{n}{2} \rfloor + \lfloor \frac{n}{3} \rfloor  + \lfloor \frac{n}{6} \rfloor \le  \frac{n}{2}  + \frac{n}{3}+ \frac{n}{6} = n$ 

so  $\lfloor \frac{n}{2} \rfloor =  \frac{n}{2}$ 

 $\lfloor \frac{n}{3} \rfloor =  \frac{n}{3}$ 

 $\lfloor \frac{n}{6} \rfloor =  \frac{n}{6}$

the above is so because if any expression above is not true LHS is less that RHS in any of them shall make the sum < n

so  $\frac{n}{2}, \frac{n}{3} ,\frac{n}{6}$ all are ntegers of n is a multiple of 2,3 and 6 that is multiple of LCM(2,3,6) that s 6. so n is of the form 6k   

2023/008) find integer n such that $(n-1)! + 1 = n^2$

We have  $(n-1)! = n^2-1 = (n+1)(n-1)$

So n-1 =0 which gives LHS = 2 RHS = 1 which is contradiction 

Or $(n-2)! = n + 1$

Put n- 2 = k giving $k! =k + 3$

As LHS is divisible by so is RHS so k is a factor of 3 k = 1 or 3

k =1 gives n = 3 which is not a solution as it does not satisfy the criteria

k =3 gives n= 5 and as $4! + 1 = 25 = 5^2$ so n = 5 is a solution 

   

2023/007) FInd integer n such that $n^2+ 19n = n!$

 n = 0 is not a solution so $n > 0$.

deviding both sides by n we get

$n + 19 = (n-1)!$

putting n = x+ 1 we get $x+20 = x!$

for this to be valud x is a factor of 20

so we check with 1 LHS = 21 and RHS = 1

x = 2 LHS = 22 RHS = 2 no

x = 4 LHS = 24 LHS = 24 so x = 4 is a solution

x = 5 LHS = 25 RHS = 125 not a solution

if we take x larger RHS grows larger as compared to LHS so no solution

so solution x = 4 or n = 5.