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

 

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)