2022/069) find the pythegorean triplet whose one number 10x+ y and largest number 10y + x

let the 3rd number be t

so $t^2 + (10x+y)^2 = (10y+x)^21$

or $t^2 = (10y+x)^2 - (10y+x)^2 = 99(y^2 - x^2) = 99(y-x)(y+x)= 11 . 3^2(y+x)(y-x)$

y+x is less than 18 and y-x is less than 10 

for the RHS to be a perfect square we must have y +x = 11 and y-x is to be perfect squiare and odd

because 11 cannot be a factor of y-x and if y +x is odd then y-x has to be odd because y-x = y + x - 2x

so y+ x = 11 and y-x = 1 giving y = 6 and x = 5

hence $t^2 = 99 * 11 * 1$ or t = 33

so triple is (33,56,65)

 

2022/068) Find the smallest positive number n such that $3n^3-2019$ is divisible by 2016

 $3n^2-2019$ is divisible by 2016

Both are mutiple of 3 so $n^3-667$ is divisible by 672

Or  $n^3-667+672$ that is $n^3-1$ is divisible by 672

Now $n^3-1= (n-1)(n^2+n+1)$ and 672 = 32 * 21 (product of even and odd)

As $n^2+n+1 = n(n+1) +1 $ is always odd so we have (n-1) multiple of 32.

Futher we must have $n^3 \equiv 1 \pmod 3$ and  $n^3 \equiv 1 \pmod 7$

For $n^3 \equiv 1 \pmod 3$ we take mod 3 and get 0 for 0 1 or 1 and 2 for 2 so n must be 1 mod 3

Or f n-1 must be mutiple of 3

So $n \equiv 1 \pmod {96}$

Futher we must have $n^3 \equiv 1 \pmod 7$ and  $n^3 \equiv 1 \pmod 7$

for $n^3 \equiv 1 \pmod 7$ we take mod 7 and get 0 for 0,1 for 1,2,4,  and 6 for 3,5,6

So we must have n = 1/2/4 mod 7

n = 96k + 1

k =1 gives n = 97 , mod 7 = 6 so not a sulution

k = 2 gives n = 193, mod 7 = 4 so is a solution

So ans smallest n = 193

 

2022/067) Find the no of solutions of equation $\frac{1}{m} + \frac{1}{n} = \frac{1}{143}$ where m , n are distinct positive inetegers

 We have $\frac{1}{m} + \frac{1}{n} = \frac{1}{143}$

or $143n + 143m = mn$

or mn - 143m - 143n = 0

this if of the form $m(n-143) - 143n = 0$

To solve these type of probem add and $143^2$ to both sdes to get

 $m(n-143) - 143(n- 143)= 143^2$

or $(m-143)(n-143) = 143^2= 11^2 * 13^2$

we get the following pairs as solutions (n-143,n-143) = (1,20449), (11, 1859), (13,1573), (121,169),(143,143), (169,121), (1573,13), (1859,11),(20449.1)

or (n,m) = (144, 20592), (11,2002),(13,1716), (264, 312), (286,286), (312,264),(1716,13),(2002,11),(20592.144)

there are 9 pairs out of which 8 are distinct  

2022/066) Let a,b be real numbers (b≠0) and consider the infinite arithmetic sequence a,a+b,a+2b,⋯. How do I show that this sequence contains an infinite geometric subsequence if$\frac{a}{b}$ is rational?

Let $\frac{a}{b}$ be rational and so let it be $\frac{n}{m}$ where n and m are integers

so $\frac{a}{b}= \frac{n}{m}$ or $b= \frac{ma}{n}$

now let $t_n$ the the nth term

so $t_(n+1) = a + ma = a(1+m)$

to get $a(1+m)^2$ we need to add $a(m^2 + 2m)$ or $am(m+2)$ as $nb = ma$ so $nb(m+2) = am(m+2)$ and so we get this term as well

to get $a(1+m)^k$ we need to add $a((1+m)^k -1)$ or $amf(m)$ as $nb = ma$ so $nbf(m) = amf(m)$ and so we get this term as well

so we get the terms in GP

2022/065) Find the number of 3 digit numbers whose GCD with 36 is 2

 Let n ba me number.

now $36 = 2 ^ 2 * 3^2$

$gcd(n, 36) = 2$  means 2 is a facttor bf n but is is not a factor of n. and 3 is not a factor of it

So nwe take the multiple of 2 and remove from this mutipls of 4 and multiples of 6.

because we need to remove multiple of 3 so we need to remove multiple of 6 (as they are multiple of 2) 

To find the number of numbers 

the number of numbers which are mutilple of 2 = $\lfloor\frac{999}{2}\rfloor - \lfloor{99}{2}\rfloor = 449 - 49 = 500$

the number of numbers which are mutilple of 4 = $\lfloor\frac{999}{4}\rfloor - \lfloor{99}{4}\rfloor = 249 - 24 = 225$

the number of numbers which are mutilple of  6 = $\lfloor\frac{999}{6}\rfloor - \lfloor{99}{6}\rfloor 166  - 16 = 150$

multiple of 4 and 6 both contain multiples of 12(LCM of 4 and 6).

mutiples of 12 will be in both the lists that is multiple of 4 and 6.

the number of numbers which are mutilple of  12= $\lfloor\frac{999}{12}\rfloor - \lfloor{99}{12}\rfloor 83  - 8 = 75$

so number of mutiples of 4 and 6 are 225 + 150 -75 = 300

so number of numbers whose GCD with 36 =2 is 450 - 300 = 150