2020/029) Solve in positive integers $ab - 4a -4b = -8$

Because it is symmetric in a ,b without loss of generality we can take $a>b$ and find a solution for the same
then permutation shall be another solution.

We need to factor the same in some form and as it is of the form $ab-4a-4b$ this shall have the form $(a-4)(b-4)$
Now $(a-4)(b-4) = ab -4a -4b + 16$
Or $(a-4)(b-4) = -8 + 16$ Putting the value of ab-4a-4b from given condition
Or $(a-4)(b-4) = 8$
For a and b to be integers we need to take the factors of 8 giving 2 sets 8 = 8 * 1 or 4 * 2
$a-4=8,b=4=>a=12, b= 5$

and
$a-4=4,b=2=>a=8, b= 6$

 

2020/028) Compare $\sqrt[n+1]{(n+1)!}$ and $\sqrt[n]{(n)!}$ where n is integer

 We have $(n+1)!= (n+1)n!$

Taking $n^{th}$ power we have $((n+1)!)^n = ((n+1)n!)^n = (n+1)^n (n!)^n\cdots(1)$
Now $(n+1)^n > n!\cdots(2)$
From (1) and (2) we get $((n+1)!)^n > (n!)^{n+1}$
Taking $(n(n+1))^{th}$ root we get $\sqrt[n+1]{(n+1)!} > \sqrt[n]{(n)!}$

2020/027) Find the smallest positive prime that divides $n^2 + 5n + 23$ for some integer n.

1st we see that $n^2 + 5n + 23 = n(n+5) + 23$ which is odd so

it is not divisible by 2

We have $4n^2 + 20n + 92 = (2n + 5)^2 + 67$
So we have for x = 1 we get 67 + 1 = 68 = 4 * 17
So we need to check for odd numbers upto 17 and can see if we have any smaller factor less that 17. this is based on the fact that if it divisible by p then we can always have a number $n < p$ for which it is true.
x = 3 67 + 9 = 76 = 4 * 19
x=5 67 + 25 = 92 = 4 * 23
x =7 67 + 49 = 116 = 4 * 29
x = 9 67 + 81 = 148 = 4 * 37
x =11 67 + 121 = 188 = 4 * 47
x = 13 67 + 169 = 236 = 4 * 59
x = 15 67 + 225 = 292 = 4 * 73
from the above we find that 17 is the smallest number for which we have
x = 1
or $2n + 5 \equiv 1 \pmod {17}$
solving this n = 15 and for n = 15 we have

If it divides $n^2 + 5n + 23$ then if divides $4n^2 + 20n + 92$ and no smaller prime can divide the same as the number is coprime to 4

Now let us take square of odd number add to 67 that is $x^2+67$

So the smallest prime is 17

$n^2 + 5n + 23 = 15^2 + 5 * 15 + 23 = 323 = 17 * 19$    

2020/026) Call a 3-digit number geometric if it has 3 distinct digits which, when read from left to right, form a geometric sequence. Find the difference between the largest and smallest geometric number.

 Let the common ratio of the digits of smallest number be x.

The smallest digit is 1 and largest digit is 9 and the ratio of them is 9 which is $3^2$

So we have a number 139 starting with 1 and let us check if can have a smaller number than this. if take the ratio 2 then we get a smaller number 124.
Now that larger number is 931 with the common ratio with $\frac{1}{3}$ and as we do not have a factor of 9 more than 3( excluding 9 it self) the largest number is 931.
So the difference is $931-124= 807$