2023/003) Prove that there exists 2023 consecutive natural numbers whose sum is a perect square

 Let the 2023 consecutive numbers be from n-1011 to n + 1011

For all to be natual numbers $n>=1012$ 

The sum of them = 2023n

$2023 =17 ^2 * 7$

So if we choose n to be of the form $7m^2$ then the sum becomes a perect square

Not $n >= 1012$ or $7m^2 >=1012$ of $m > 13$

So the 2023 number starting from $7m^2-1011$ where $m > =13$ satisfy the criteria 

2023/002) A room is empty. Each minute a person enters a room or 2 leave. after exactily $3^{1999}$ minutes can the no of persons be $3^{1000} + 2$

Let persons enter A instances and leave B instances

A+B = $3^{1999}\cdots(1)$

A-B = $3^{1000} + 2\cdots(2)$

Subtracting 2nd from the 1st we get

$3B = 3^{1999} - 3^{1000} - 2$

LHS is multiple of 3 but RHS is not so it is not possible  

2023/001) When dividing a polynomial f(x) by $(x-1)^2$ the remainder is x+1. If f(x) is divided by $x^2$ the remainder is 2x+3 . if the remainder when divided by $x^2(x-1)$ is $ax^2+bx+c$ then find a+b+c

 Dividing a polynomial f(x) by $(x-1)^2$ the remainder is $g(x) = x+1$-

so dividing by (x-1) the remainder is $g(1) = 1 + 1 = 2$

Dividing by $x^2(x-1) $ is $ax^2+bx+c$

so deviding  $ax^2+bx+c$ by $x-1$ remainder must be 2

so $f(1) = a + b+ c = 2$ 

2022/080) Let m and n be positive integers such that gcd(m,n) + lcm (m,n) = m + n. Show that one of the 2 numbers is divisible by the other

Let gcd(m,n) = p.

then m = pq and n = pr for some q and r and gcd(q,r) = 1

gcd(m,n)  = p as we have chosen

lcm(m, n) = pqr as q and r are co-primes

gcd(m,n) + lcm(m,n) = m + n

$=>p + pqr = pq + pr$

$=>1 + qr = q + r$

$=>qr - q -r + 1= 0$

$=>(q-1)(r-1) = 0$

q =1 mean n is divisible by m 

or r =1 meand m is divisible byn

hence proved 

2022/079) Simplify $(\frac{4}{(\sqrt{5}+1)(\sqrt[4]{5}+1)(\sqrt[8]{5}+1)(\sqrt[16]{5}+1)} +1)^{48}$

 as we see that the roots are doubleing in term to term so multiply numeraator and denominator by $(\sqrt[16]{5}-1)$ we get

 $(\frac{4}{(\sqrt{5}+1)(\sqrt[4]{5}+1)(\sqrt[8]{5}+1)(\sqrt[16]{5}+1)} +1)^{48}$

= $(\frac{4*(\sqrt[16]{5}-1)}{(\sqrt{5}+1)(\sqrt[4]{5}+1)(\sqrt[8]{5}+1)(\sqrt[16]{5}+1)(\sqrt[16]{5}-1)}  +1)^{48}$

= $(\frac{4*(\sqrt[16]{5}-1)}{(\sqrt{5}+1)(\sqrt[4]{5}+1)(\sqrt[8]{5}+1)(\sqrt[8]{5}-1)}  +1)^{48}$ using $a^2-b^2$ formula for last 2 terms

= $(\frac{4*(\sqrt[16]{5}-1)}{(\sqrt{5}+1)(\sqrt[4]{5}+1)(\sqrt[4]{5}-1)}  +1)^{48}$ using $a^2-b^2$ formula for last 2 terms-

= $(\frac{4*(\sqrt[16]{5}-1)}{(\sqrt{5}+1)(\sqrt{5}-1)}  +1)^{48}$ using $a^2-b^2$ formula for last 2 terms-

= $(\frac{(4*\sqrt[16]{5}-1)}{4}+1)^{48}$ 

= $(\sqrt[16]{5})^{48} = 5^3 = 125$

2022/078) Given 2 positive nymber a and b with a > b. a-b conatenated with a+b is a 5 digit number c which is prime. find c

 Fir example if  = 5 aand a = 7 then a-b = 2 and a + b = 12. Concatenating we get 212,'

Solution:

because c is prine so $a+b$ is odd so $a-b$ ( as it is $a+b -2b$) is odd. as c is a 5 digit number $a-b$ is 2 digit and$ a+ b$ is 3 digit, mallest $a-b$ is 11 and $a+b$ is odd staring from 101 putting the values we get $a+b = 113$ and $a -b = 11$ and hence c = 11113.

2022/077) If $a^2+b^2+c^2 = 1$ find the range of $ab + bc+ca$

We need to find the mininal and maximal of $ab+bc+ca$

we have $(a+b+c)^2 = a^2 + b^2 + c^2 + 2(ab+ bc+ ca)$

so $ a^2 + b^2 + c^2 + 2(ab+ bc+ ca) > = 0$

puttig $a^2+b^2+c^2 = 1$ we get

$ 1 + 2(ab+ bc+ ca) > = 0$

or $(ab+bc+ca) >=  - \frac{1}{2}$

Further to find tthe maximum we have $(a-b)^2 + (b-c)^2 + (c-a)^2 = 2(a^2 +b^2+c^2 - ab - bc - ca)$

or $2(a^2 +b^2+c^2 - ab - bc - ca) >= 0$

or $ab+bc+ca <= a^2+b^ + c^2$

or $ab+bc+ca <= 1$

so we have $ab+bc + ca \in [-.5 .. 1]$