2022/002) Find the smallest number whose product of digits is 10000

 Because the product of digits is 10000 so this can be represented as product os single numbers.

Let us findone such number and then see if we canmake it smaller

$10000 = 2 ^ 4 * 5 ^ 4$

So if the number has 4 2's and 4 5's then the  product of the digits is 10000. This is an 8 digit number.  this can be mad smaller if we can multiply 2 or more one digit number out of  these and getting one digit number, multiplying 3 2's we get 8 and digits are 2, 8, 5,5,5 ,5 giviing the product 1000. the smallest number that can be formed is 255558 and this is the answer   

2022/001) Solve in integers $x+y+z=3$ and $x^3+y^3 + z^3 = 3$

We are given

$x+y+ z = 3\cdots(1)$

$x^3 + y^3 + z^ 3 = 3\cdots(2)$

cubing (1) we get $(x+y+z)^3 = 3^3$

or $x^3 + y^3 + z^3 + 3(x+y)(y+z)(z+x) = 27$

or $3 + 3(x+y)(y+z) (z+x) = 27$ putting thw vaue of $x^3 + y^3 + z^3 = 3$ from (2)

or  $(x+y)(y+z)(z+ x) = 8$

now $((x+y) + (y+z) + (z+ x)) = 2 ( x + y + z) = 6$ usng (1)

we need to find 3 numbers whose sum is 6 and product is 8. They are (2,2,2) or (4,1,1) and other combinations fail

taking x + y = 2 , y + z = 2, z+ x = 2 and using (2) that is x + y + z = 3 we get x=1,y=1,z = 1

 taking x + y = 4 , y + z = 1, z+ x = 1 and using (2) that is x + y + z = 3 we get x=5,y=z = - 4

takinn n rotaton other combinations we get x=y = -4, z = 5 as one solution and z = z = -4 and y =5 as another solution

so $\{x,y,z\} = \{2,2,2\}$ or $\{5,-4,-4\}$ (that is any permuation of the same)  

2021/116) If the product of two numbers is 2,400 and their LCM is 96, then what is their HCF?

The question is ill formed because

Product = 2400

LCM = 96

So HCF = $\frac{product}{LCM} = \frac{2400}{96}=25$

But LCM has to be a multiple of HCF . ths is so because LCM is a multiple of each of the numbers and each number is multiple of HCF

As 25 does not divide 96. so the problem is incorrect, or this case is not possible 

2021/115) What are the maximum and minimum values of $3x+4y$ on the circle $x^2+y^2=1$

as $x^2 + y^ 2 = 1$ we can choose $x = \sin\, t$, $y = \cos\, t$

$3x + 4 y= 3 \sin\, t + 4 \cos\, t$

to convert $3 x + 4y = 3 \sin\,t + 4 \cos\, t$ to the form $A \sin (x+ t)$

$A \sin (x+t) = A \sin\,t \cos\, x + A \cos\, t \sin\, x$

we can choose $3 = 5 \cos\, x$ and $4 = 5 \sin x$  (as $3^2 + 4^2 = 25 = 5^2$

= $5 \cos\, x \sin\, t + 5 \cos\, t \sin\, x = 5 \sin (x-t)$

it is maximum when $\sin (x-t) = 1$ and maximum value = 5

minumum when $\sin (x-t) = -1$ and minimum value = 5

2021/114 solve in integers $3x^2 + 5y^2 = 345$

working in mod 3 we have $5y^2 = 0 \pmod 3$ or $y =0 \pmod 3$

so y = 3a for some a

similarly x = 5 b for som b

so ge get $75 b^2 + 45 y^2 = 345$ 

deviding by 15 we get $5b^2 + 3a^2 = 23$

we need to check for $5b^2 < 23$ or $b <=2$

putting b = 1 we get  $3a^2 = 18$ or $a^2 = 6$ not an integer

b = 2 gives $3a^3 =3$ or a = 1

so we have a= 1 , b= 2 giving x = 10 and y = 3 

2021/113) prove the following identity: $\binom{n}{k}=\binom{n-2}{k}+2\binom{n-2}{k-1}+\binom{n-2}{k-2}$

We can solve the same in 2 ways. Combinotorics way or alegraic ways

We present here to solve in combinonorics way

From n objects we can choose k objects in   $\binom{n}{k}$ ways

let us group the n objects into (n-2,1,1) ways

for choosing k objects this can be done in 3 ways

k  objects from n-2 objects that is from 1st set 0 from 2nd set and 0 from3rd set in $\binom{n-2}{k}$ ways

k-1  objects from n-2 objects that is from 1st set 1 from 2nd set or  1 from 3rd set in $2 * \binom{n-2}{k-1}$ ways

k-2  objects from n-2 objects that is from 1st set 1 from 2nd set and  1 from 3rd set in $\binom{n-2}{k-2}$ ways

as all above 3 are mutually exclusive so no of ways  =$\binom{n-2}{k} + 2 * \binom{n-2}{k-1} + \binom{n-2}{k-2 }$

in  2 ways we have computed the number of choosing k objects from n obects so they must be same or

$\binom{n}{k} = \binom{n-2}{k} + 2 * \binom{n-2}{k-1} + \binom{n-2}{k-2 }$

2021/112) Let G= $\{a+bi$ in complex: $a^2 +b^2=1\}$. Is G a group under multiplication?

For it to be group folllowing must be tue.

1) it should be closed

that is if  x = a + ib and y = c + id and $(a^2+b^2) = 1$ and $c^2+d^2=1$ 

and xy = m + ni then $m^2+n^2 =1$

we have $xy = m + ni = (a+ib)(c+id) = (ac - bd) + (bc + ad)i$

we have m = ac - bd and n = bc + ad

$m^2 + n^2 = (ac-bd)^2 + (bc + ad)^2 = a^2c^2 - 2abcbd + b^2d^2 + b^2 c^2 + 2abcd + a^2d^2$

 $= a^2c^2 + b^2d^2 + b^2c^2 + a^2d^2 = (a^2+b^2)(c^2 + d^2) = 1$ 

So it is closed 

2) It should have an identity

1 or 1+0i is identity element as $(a+bi)(1+0i) = a+ bi$

3) it should have an inverse  

because $a^2+b^2=1$ so it it not zero and hence it has inverse and we need to show that if

m+in is inverse then $m^2+ n^2 =1$ that is the inverse is in this group

$m + in = \frac{1}{a+ib} = \frac{a-ib}{(a+ib)(a=ib)} = \frac{a-ib}{a^2+b^2} = a - ib$

'so m = a , n = - b and $m^2 + n^2 = a^2 + (-b)^2 = a^2 + b^2 =1$

so it has an inverse

4) assosiativity law holds as unde rcomplex number multiplication assosiativity holds