2022/076) Find prime p such that $16p+1$ is a cube

We have $16p + 1$ is a cube say $x^3$

So $16p= x^3-1 = (x-1)(x^2+ x + 1)$

As $x^2+x+1 = x(x+1) +1$ is odd we must have

$16 | x-1$

So $x = 16k+1$ for some integer k

so $p = k(x^2 + x + 1)$

the above cannot be prime unless k =1 and this gives x = 17 and p = 307

We need to see if  307 is prime and it is so

Hence the only solution is p = 307 (giving $16 * 307 +1  = 17^3$

 

 

2022/075) Given the two number patterns: 1,5,9,13,17... and 1,4,7,10,13... How do I find their 13th common number?

 in the 2 patterns the 1st number is same.

now 1st pattern has common diffrerence 4 and 2nd one 3

the minimum common number difference shall be LCM(4,3) or 12

so 13th common number shall be 1 * 12 * (13–1) = 145

2022/074) if a and b are roots of $x^2-3x+1=0$ then find the value of $(a^a + b^b)(a^b+ b^a)$

 We are given f a and b are roots of $x^2-3x+1=0$

so $a + b = 3\cdots(1)$
$ab=1\cdots(2)$2

now $ $(a^a + b^b)(a^b+ b^a)$$

= $a^{a+b} + (ab)^a + (ba)^b  b^(b+a)$

= $a^3 + 1 + 1 + b^3$ putting the value of a+b and ab from (1) and (2)

=$a^3 + b^3 + 2$

= $(a+b)^3 - 3ab(ab+b) + 2$ using formula for $a^3+b^3$

= $3^3 - 3 . 1 . 3 + 2 = 20$ putting values from (1) ansd (2) 

2022/073) Maximize $x \sqrt{1-y^2} + \sqrt{1-x^2}$

 In the above expression x  not dependent on y term so let us maximize $\sqrt{1-y^2}$ and this is {maxium at y= 0 and the value is 1.

Putting the same we get $x  + \sqrt{1-x^2}$

As we have $\sqrt{1-x^2}$ we can put $x=\sin\,t$ ( $0 \le t \le \frac{\pi}{4}$ to get

$x  + \sqrt{1-x^2}= \sin\, t + \cos \, t $

We need to maximize $\sin\, t + \cos \, t $ so let us get in form $r \cos\, w \sin \, t + r \sin\, w \cos  \, t$

which is $r \sin ( t+w)$ and maximum value is |sqrt{2}$

So let $r\cos\, t = 1$ and $r \sin\, t = 1$

By squaring both and adding 

$r^2 = 2$ or $r=\sqrt{2}$

Hence largest valus is $\sqrt{2}$

 

 

2022/072) find positive integers x and y such that $\frac{1}{y} - \frac{1}{y+2} = \frac{1}{3 * 2^x}$

 We have multipying by $x(y+2)$ 

$\frac{2}{y(y+2)}= \frac{1}{3 * 2^x}$

or $ 3 * 2^{x+1} = y(y+2)$

let z = x+1 so

 $ 3 * 2^{x+1} = y(y+2)$

LHS is even so y is even because if y is odd then y(y+2) is odd

3 is a factor of y or y + 2

let us atke the 2 cases one by one

case 1 :3 is a factor of y

so $y = 3 * 2^a$ and $y+2 = 2^b$ for some inetger a and b 

$3 *2^a + 2 = 2^b$

or $2(3 * 2^{a-1} + 1) = 2^{b+1}$

as RHS is even so $3*2^{a-1}$ is odd or  a = 2

so y = $3 *2^a = 6$ and $y+2 =  8$ from this $3 * 2^{x +1} = 48 = 3 * 2^4$ ot x = 3

case 2 :3 is a factor of y +2

so $y+ 1 = 3 * 2^a$ and $y-2 = 2^b$ for some inetger a and b 

$3 *2^a - 2 = 2^b$

or $2(3 * 2^{a-1}-1 = 2^{b+1}$

as RHS is even so $3*2^{a-1}$ is odd or  a = 2

so y = $3 *2^a = 6$ and $y -2 =  4$ from this $3 * 2^{x +1} = 24 = 3 * 2^3$ ot x = 2  

So 2 solutions are (3,6) and (2,4)

  

2022/071) Find integers a,b,c such that $a^2-b^2 = 24$ and $b^2-c^2= 16$

We ahve $a^2-b^2 = 24\cdots(1)$

$b^2-c^2 = 16\cdots(2)$ 

Because $a^2-b^2 = 24$ so both a and b are odd or both ae even,

So we have a = b + 2n for some n

To put an upper bound on be a = b+ 2 beause a is at least 2 more than b 

So we have $a^2-b^2 = (b+2)^2 - b^2  = 24$

Or $2b + 4 = 24$ or $b= 5$

For the lower bound $b^2 =17$ so $b ge 5$

So we get b = 5 and putting the values 

So we get $a= 7, b = 5, c= 3$ 

2022/070) 7 dices are rolled. What s the probabilty that the sum of top faces is 10 ?

 For eah dice there are 6 outcomes

for 7 dices there are $6^7$ outcomes

so there are $5^7$ out comes

10 can come in followin ways

6 ones and a 4 this can come in 7 ways(4 at any of 7 places)

5 ones one 2 and one 3 this can come in 7 * 6 = 42 ways( 2 at any of  7 places and and then 3 in any of the next 6 paces)

4 ones and 3 twos = $7 \choose 3$ = 35 places

so total number of places = 7 + 42 + 35 = 84.

so probability = $\frac{84}{6^7}$