2023/007) FInd integer n such that $n^2+ 19n = n!$

 n = 0 is not a solution so $n > 0$.

deviding both sides by n we get

$n + 19 = (n-1)!$

putting n = x+ 1 we get $x+20 = x!$

for this to be valud x is a factor of 20

so we check with 1 LHS = 21 and RHS = 1

x = 2 LHS = 22 RHS = 2 no

x = 4 LHS = 24 LHS = 24 so x = 4 is a solution

x = 5 LHS = 25 RHS = 125 not a solution

if we take x larger RHS grows larger as compared to LHS so no solution

so solution x = 4 or n = 5.

2023/006) Find 2 digit numbers that have exactly 5 factors

 If the umber is of the form $\prod_{n=1}^{k}p_n^{q_n}$ then the number of factors = $\prod_{n=1}^{k}(q_n+1)$

Because 5 is prime the number must be of the form $p^4$ where p is prime

We have the number of digits =2 and we should find p such that $ 9 \lt p^4 \lt 100$

The only number that satsfies the condition is n = 2 and n is a prime and $2^4= 16$

So 16 is the only 2 digit number having 5 factors

 

2023/005) What is the difference of numbers if the HCF of two numbers is 9 and their LCM is 270 if the sum of the numbers is 99?

HCF is 9 so the 2 numbers are 9x, 9y where HCF(x,y) = 1 and x and y >0 and without loss of generality let x > y

product of the 2 numbers are 9xy = 270

or $xy = 30\cdots(1)$

sum of the 2 numbers are 9(x+y) = 99 or $x + y = 11\cdots(2)

we have $(x-y)^2 = (x+y)^2 - 4 * xy = 11^2 - 120 = 1$

so $x - y = 1\cdots(3)$

so we get x = 6 and y = 5 and numbers are $54, 45$ 

2023/004) Solve in integers $4^x-5^y= 39$

we have $4^x = 2^{2x}$

now working in mod 3 we get $1-5^y \equiv 0 \pmod 3$

or $5^y \equiv 1 \pmod 3$

as we know $5 \equiv -1 \pmod 3$ so y has to be even say 2m

now $4^x - 5^y = 2^{2x} - 5^{2m} = 39$

Farctoring  we get $(2^x + 5^m)(2^x-5^m) = 39 = 39 *1 = 13 * 3$ (39 can be factored in 2 ways)

so we have 2 cases

$2^x+5^m= 39$ and $2^x - 5^m=1$ adding we get $2^x *2 = 40$ and this does not have integer

or 

$2^x+5^m= 13$ and $2^x - 5^m=3$ adding we get $2^x *2 = 16$ or x = 3 and subtracting $2 * 5^m = 10$ and m = 1

so x = 3 and y = 2  

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 $f(1) = a + b+ c = 2$