2024/057) The LCM of three numbers is 72 and their GCF is 6. If the first and second number is 12 and 18 respectively, what is the third number?

We have GCD = $6 = 2 * 3$

$1^{st}   number = $12 = 2^2  *3$

$2^{nd}$  number =  $18= 2 * 3^2$

LCM = $72 = 2^3 * 3^2$

So the $3^{rd}$ number has to of the form $2^a3^b$

Taking the lowest powers of 2 and 3 we get the HCF 

So HCF = $2^{min(1,a)}3^{min(1,b)}= 2 * 3$ which gives $a >=1$ and $b >=1$

Taking the highest  powers of 2 and 3 we get the LCM  

 LCM=  $2^{max(2,a)}3^{max(2,b)}= 2^3 * 3^2$ so $a=3$ and $ b <=2$

Giving $a=3,b=1$ that is number = $2^3*3=24$ or $a=3,b=3$ that is number = $2^3*3^2=72$

  so $3^{rd}$ number can be 24 or 72

2024/056) How do I find the minimum positive integer $n>114$, such that $2001 \equiv 114(\mod n)$

It is 114 mod n so 2001–114 should be divisible by n and $n > 114$

Factors of 1887 are 1 3 17 37 51 111 629 1887. any number $<= 111$ shall give a smaller remainder and 2 numbers 629 and 1187 shall give a remainder 114 so the number is lower of the 2 that is 629.

2024/055) if $2n+ 1$ is a perferct square show that $n+1$ is sum of 2 squares when n is an integer

 We know $2n+1$ is odd so it is square of an odd number say $(2k+1)^2$

So $2n+1 = (2k+1)^2 = 4k^2 + 4k +1$

or $2n= 4k^2+ 4k$

or $n = 2k^2 + 2k$

or $n + 1=  2k^2+2k + 1 = k^2+(k^2+2k+1) = k^2+(k+1)^2$

Proved 

2024/054) Find x and y such that $LCM(x,y) = LCM(12x,5y) = 720$

 As $LCM(x,y) = 720$ we need to find prime factors of 720

$720=2^4 * 3^2 *5$

So $x=2^a3^b5^c\cdots(1)$

And $y=2^d3^e4^f\cdots(2)$

Where $max(a,d) = 4\cdots(3)$

  $max(b,e) = 2\cdots(4)$

 $max(c,f) = 1\cdots(5)$'

Now

$LCM(12x,5y) = LCM(2^{a+2}3^{b+1}5^c $$, 2^d3^e5^{f+1})= 2^4 * 3^2 *5$

So $max(a+2,d) = 4\cdots(6)$

$max(b+1,e) = 2\cdots(7)$

$max(c,f+1) = 1\cdots(8)$

From (3) and (6) we get $d=4$ and $a \le 2$

From (4) and (7) we get $e=2$ and $b\le 2$

from (5) nd (8) we get $c=1$ and $f=0$

So we get following pairs $(2^0 * 3^0 * 5, 2^4*3^2)$ that is $(5,144)$

$(2^1 * 3^0 * 5, 2^4*3^2)$ that is $(10,144)$

$(2^2 * 3^0 * 5, 2^4*3^2)$ that is $(20,144)$

$(2^0 * 3^1 * 5, 2^4*3^2)$ that is $(15,144)$

$(2^1 * 3^1 * 5, 2^4*3^2)$ that is $(30,144)$

$(2^2 * 3^1 * 5, 2^4*3^2)$ that is $(60,144)$

 

    

 

2024/053) Given $ab + bc + ca = abc$ find the value of $\frac{(b+c)}{bc(a-1)} + \frac{(c+a)}{ca(b-1)} + \frac{(a+b)}{ab(c-1)}$

 We are given

$ab+bc+ca = abc$

Hence $bc(a-1) = abc - bc = ab + ac = a(b+c)$

Hence

 $\frac{(b+c)}{bc(a-1)} =  \frac{(b+c)}{a(b+c)} = \frac{1}{a}$

or 

 $\frac{(b+c)}{bc(a-1)} = \frac{1}{a}\cdots(1)$

 similarly

 $\frac{c+a)}{ca(b-1)} = \frac{1}{b}\cdots(2)$

and

 $\frac{(a+b)}{ab(c-1)} = \frac{1}{c}\cdots(3)$

Adding (1),(2),(3) we get

 $\frac{(b+c)}{bc(a-1)} + \frac{(c+a)}{ca(b-1)} + \frac{(a+b)}{ab(c-1)}$

$=\frac{1}{a} + \frac{1}{b} + \frac{1}{c}$

$=\frac{bc+ca+ab}{abc} = \frac{abc}{abc}=1$

2024/052) Solve in integers $x^3+3367=2^n$

We are given $x^3+3367=2^n$

We know $3367=7 * 13 &*37

So let is work $x^3=2^n \pmod 7$

Working in mod 7 we have $x^3 \in \{1,-1,0\}$ and  $2^n \in \{1,2,4\}$ so we get 1 as common and for that n has to be multiple of 3.

So we get

$x^3= 2^{3k} \pmod 7$

Going back to the original equation we get

$x^3 + 3367 = 2^{3k}$

Or $(2^k)^3 - x^3 = 3367$

Or $y^3 -x^3 = 3367$ where $y = 2^k$

As $(y-x) | y^3-x^3$ so $y-x |  3367\cdots(1)$

Further 

As we know $(y-x)^3 = y^3-3y^2x + 3yx^2 - x^3 = (y^3-x^3) - 3yx(y-x) \lt y^3-x^3$ when $y-x \gt 0$

So $y-x\lt 15\cdots(2)$

So $y-x \in \{1,7\}\cdots(2)$ from (1) and (2)  

 $y-x =1$

gives $x^3 + 3367 = (x+1)^3 = x^3 + 3x^2 + 3x + 1$

Or $3367 = 3x^2+ 3x + 1$

Or $3x^2+3x = 3366$

Or $x^2 + x = 1122$

x = 33 and y = 34 and y is not power of 2 so not a solution

$y-x=7$

gives

$x^3 + 3367 = x^3 + 21x^2 + 147 x + 343$

or $3024 = 21x^2+ 147x

or $x^2+ 7x = 144 $

or $x = 9$ or $x=-16$

only positive value admissible giving x = 9 and y = 16

$2^n = y^3$ giving $2^n = 2^{12}$ or n = 12

Hence x=3 , n = 12

 

 

2024/051) Find integer solutions to $1+2^x+ 2^{2x+1} = y^2$

 We have 

 $1+2^x+ 2^{2x+1} = y^2$

Or  $2^x+ 2^{2x+1} = y^2-1$

Let us check one case x = 0 for which $2^x$ is odd that is 1 and we get $1 + 1+ 2 = 4 = y^2$ or 

(0, 2), (0, -2) are 2 solutions.

 $2^x(1+2^{x+1}) = y^2-1 = (y+1)(y-1)\cdots(1)$ in this x is positive

As LHS is even  number so RHS is even.

Now y+1 and y-1 both must be even and one has to be an odd multiple of power of 2 and other one odd multiple of 2.

So $y = 2^{x-1}p\pm 1$  where p is odd.

Or  $y = 2^{x-1}p +q$ where p is odd and q is $\pm 1$

Putting in the equation (1) we get

$2^x(1+2^{x+1}) = y^2-1 = (2^{x-1}p +q)^2-1 = 2^{2x-2}p^2 + 2^x pq$

Or $1+2^{x+1} = 2^{x-2} p^2 + pq$

Or  $1-pq = 2^{x-2}(p^2-8)$

q =1 gives $1-p = 2^{x-1}(p^2-8)=>(p^2-8)<0$ or $p=1$ which does not satisfy the condtion.

If q = -1 then we have

$1+p = 2^{x-2}(p^2-8)$ giving $1 + p > p^2 -8$

Or    $p^2-p -7 <=1$ this has $p  <= 3$ as p = 3 giving x = 4 and $y = \pm 23$

So 4 solutions  are $(0,2),(0,-2),(4,23), (4,-23)$