We have
$x^2-2x + 2 \equiv 0 \pmod 5$
$=>x^2 -2x + 1 + 1 \equiv 0 \pmod 5$ to make square on left side
$=>(x-1)^2 + 1 \equiv 0 \pmod 5$
$=>(x-1)^2 + 1 \equiv 0 \pmod 5$
$=>(x-1)^2 \equiv 4 \pmod 5$
Working in mod 5 we have
$0^2 \equiv 0 \pmod 5$
$1^2 \equiv 1 \pmod 5$
$2^2 \equiv 4 \pmod 5$
$3 \equiv 4 \pmod 5$
$4^2 \equiv 1 \pmod 5$
So we have
$x-1 \equiv 2\pmod 5$ or $x-1 \equiv 3 \pmod 5$
Or $ x \equiv 3 \pmod 5 $ or $x \equiv 4 \pmod 5$
The number shall be 10000x +2020 and we need it to be divisible by 77 and we need to solve it for smallest x
Or 10(1000x + 202) should be divisible by 77
As 77 is co-prime to 10 so 1000 x + 202 is divisible by 77
As 77 = 7 * 11 and 7 and 11 are co-primes
1000 x + 202 is divisible by 7 and also 11
So we need to solve
$1000x + 202 \equiv 0 \pmod 7\cdots(1)$
And
$1000x + 202 \equiv 0 \pmod {11}\cdots(2)$
Let us solve both the equations and then combine the 2
Now the1st equation is
$1000x + 202 \equiv 0 \pmod 7$
Or $6x + 6 \equiv 0 \pmod 7$
Or $x + 1 \equiv= 0 \pmod 7$
Or $x \equiv 6 \pmod 7\cdots(3)$
Now second second equation is
$1000x + 202 \equiv 0 \pmod {11}$
or $10x + 4 \equiv 0 \pmod {11}$
or $- x + 4 \equiv 0 \pmod {11}$
or $x \equiv 4 \pmod {11}\cdots(4)$
So we need to solve
$x \equiv 4 \pmod {11}\cdots(4)$
$x \equiv 6 \pmod {7}\cdots(4)$
Tthis can be solved by Chinese Remainder Theorem but as the numbers are very small we solve as below
$x + 1$ that is $5 \pmod 11$ is divisible by 7
So we see the numbers of the $5 \pmod {11}$ and see it divisible by 7
we find 16,17,38,49 and see that 49 id divisible by 7
So $x = 48$
And number is $482020$
This is the smallest number
Shorter Method
We need 1000n + 202 divisible by 77
As
$1000 \equiv -1 \pmod {77}$
We need
$202 - n$ divisible by 77
So n = $202 \pmod {77}$
or n = 48
and smallest number is $482020$
The expressions are cyclic so WLOG we can assume
a +b < a +c < b+c
From given conditions
$a+b = 2n -1\cdots(1)$ as the lowest number is odd
$a+c = 2n\cdots(2)$
$b+c = 2n +1\cdots(3)$
Adding these 3 we get
$2a + 2b + 2c = 6n$
or $a +b +c = 3n\cdots(4)$
so $a = (a +b + c)-(b+c) = 3n -(2n+1) = n-1$
$b= a+b -a = (2n)-(2n-1) = n$
$c = a+c -a = n+1$
so all are consecutive integers
I shall extend it get all number less than 1000. Let us check upto 3 digit number because for 4 digit number another approach shall be taken( I shall mention the approach at the end)
Now the largest 3 digit number is 999 and sum of digits is 27. so the sum of digits is 17 as it is divisible by 17. Of course it cannot be 0.
Let the number be n
n is divisible by 17
So
$n \equiv 0 \pmod {17}\cdots(1)$
Again sum of digits is 17 so
$n \equiv 8 \pmod 9 \cdots(2)$
We need to solve (1) and (2) to get the solution set from which we shall find the solution.The result shall be mod 153. (17 * 9 = 153)
Because one of the mod is zero we can provide a simpler solution
We have $n = 17k$ (from (1))
Putting in (2) we get
$17k \equiv 8 \pmod 9$
Or $8k \equiv 8 \pmod 9$
Or $k \equiv 1 \pmod 9$
So n = 17 is a solution
So we have general solution is 153m + 17
By putting m = 0 to 6 we get solution 17,170,323,476,629,782,935
Out of which some have digit sum 8 and other have digit sum 17 they are 476,629,782,935 with the lowest number 476
If we need 4 digit number sum of 4 digits can be 34 also . So we have another set of solution to look This is of the form 153k + 34.
This shall include another set besides 153k+ 17 that is 153k+ 34 all are not solution .this is the set from which we need to find solution.
I shall prove the same using modular arithmetic keeping arithmetic simple
We have $2^5 = 32 \equiv -15 \pmod {47}$
$=> 2^{10} \equiv (-15) ^2 \pmod {47}$
$=> 2^{10} \equiv 225 \pmod {47}$
$=> 2^{10} \equiv -10 \pmod {47}$ (as 225 = 5 * 47 -10)
$=> 2^{20} \equiv (-10)^2 \pmod {47}$
$=> 2^{20} \equiv 100 \pmod {47}$
$=> 2^{20} \equiv 6 \pmod {47}$ as 100 =2 * 47 + 6
$=> 2^{23} \equiv 6 * 2^3 \pmod {47}$
$=> 2^{23} \equiv 48 \pmod {47}$
$=> 2^{23} \equiv 1 \pmod {47}$
$=> 2^{23} -1 \equiv 0 \pmod {47}$
Hence $2^{23} -1$ is divisible by 47
The 2 numbers are even so let the 2 numbers be 2x and 2y where x and y are co-primes
So lcm = 2xy
As product of numbers is same of product of LCM and HCF so we have
2xy = 90
Or xy = 45
x and y are co-primes
Factor $45 = 3^2 * 5$
As x and y are co-primes so no prime shall divide both and highest power will be present of a prime is a factor
So $(x,y)$ are $(1,45)$ or $(9,5)$ so the 2 even numbers are $(2,90)$ or $(18,10)$
Let us take the mod 9 of the expression
$A_n=7^{2n} - 48n -1 \mod 9$
$=(7^2)^n - 48n -1 \mod 9$
$=49^n - 48n -1 \mod 9$
$=4^n -3n -1 \mod 9$ reducing to mod 9
This can be proved by principle of mathematical induction also binomial expansion
Method 1 Principle of Mathematical Induction
To prove by principle of mathematical induction we first need to check for initial condition then induction step
We have
$A_1= 4^1 - 3 *1 -1 = 0$ is divisible by 9
We have checked the basic step
So we proceed to check the induction step
Let us assume that it is true for some n say k
So $A_k= 4^k - 3k -1 \equiv 0 \mod 9$
We need to prove
$A_{k+1}= 4^{k+1} - 3 *(k+1) -1 \equiv 0 \mod 9\cdots(2)$
We have
$A_k= 4^k - 3k -1$
$A_{k+1}= 4^{k+1} - 3(k+1) -1$
Now $A_{k+1}- 4 * A_k = 4^{k+1} - 3(k+1) -1- 4 (4^k - 3k -1)$
$= -3k -4 + 12k + 4 = 9k$
Hence Or $A_{k+1} - 4 * A_k+9k$
or $A_{k+1} = 4 * A_k+9k$
So if $A_k$ is divisible by 9 the $A_{k+1}$ is divisible by 9
We proved the induction step
Hence Proved
Method 2 :Binomial expansion
We shall prove the same by Binomial expansionLet us use for 1 and 2 for simple cases and the binomial theorem for $n>2$
$A_k = 4^k - 3k -1$
So we have
$A_1 = 4 - 3 - 1 =0$ or $A_1$ is divisible by 9
and $A_2 = 4^2 - 3 * 2 -1 = 9$ or $A_2$ is divisible by 9
for $n>2$ we have (it is rue for all n but we shall use for n>2)
$4^n= (3+1)^n = \sum_{k=0}^n \binom {n}{k} 3^{n-k}1^k$
$=\sum_{k=0}^{n-2} \binom {n}{k} 3^{n-k}1^k+3n + 1$ by grouping 1st n-2 terms and separating last 2 terms
As the sum under the summation each term is divisible by n we get
$4^n = 9p + 3n +1 $ for sum mutilple of 9 with some p
Or $4^n -3n -1 = 9p$
Or $4^n -3n -1$ s divisible by 9
Hence proved
