We have a prime number is 2 or 3 of of the form $6n\pm 1$ for $n \gt 0$ let us compute $n^2+2$
$2^2 + 2 = 6 = 2 * 3$ not a prime
$3^2+ 2 = 11$ is a prime
$(6n\pm 1)^2+ 2 = (36n^2 \pm 12 n + 1) + 2 = (36n^2 \pm 12 n + 3) = 3(12n^2 \pm 4 n + 1)$ which is not a prime
hence 11 is the only prime
We have $n^{th}$ triangular number $t_n= \sum_{k=1}^n k = \frac{n(n+1)}{2}$
So we must have $\frac{n(n+1)}{2} \equiv -1 \pmod {11}$
Or $n(n +1) \equiv -2 \pmod {11}$
Or $n^2 + n + 2 \equiv 0 \pmod {11}$
Or $4 n^2 + 4n + 8 \equiv 0 \pmod {11}$ (the purpose of doing this is to covert to perfect square as evident from next line)
Or $(2n+1)^2 + 7 \equiv 0 \pmod {11}$
Or $(2n+1)^2 = \equiv -7 \pmod {11}$
Or $(2n+1)^2 = \equiv 4 \pmod {11}$
Let us find the square mod 11 for n = 0 to 5 we get (0,0),(1,1),(2,4),(3,9),(4,5),(5,3)$
So the numbers are 2 and 9 ( that is 11 -2)
$2n + 1 \equiv 2 \pmod {11}$ gives n = 6 and $2n + 1 \equiv 9 \pmod {11}$ gives n = 4
So we have the triangular numbers are $t_{11k+4}$ and $t_{11k+6}$ for any non negative k
We have by definition triangular number $t_k = \frac{k(k+1)}{2}$
So $t_{3k}+t_{4k+1}$
$= \frac{(3k(3k+1)}{2} + \frac{(4k+1)(4k+2)}{2}$
$= \frac{(9k^2+3k}{2} + \frac{16k^2 + 12 k + 2}{2}$
$= \frac{25k^2+15k + 2}{2}$
$= \frac{(5k+1)(5k + 2)}{2}= t_{5k+1} $
Then
a) exactly one real root is -ve
b) all real roots are positive
c) exactly one real root is positive
d) no real root is positive.
Solution
we have
$x^{2021} + x^{2020} + \cdots+x - 1=0$
or $x^{2021} + x^{2020} + \cdots+x + 1=2$
Note that 1 is not a root of this equation.
Multiplying both sides by $x-1$ we get
$x^{2022} - 1 = 2(x-1)$
or $f(x) = x^{2022} - 2x + 1 = 0$
Note that as we have multiplied by x-1 so x = 1 so there is at least one positive root that is 1
As there is change of sign two times as per Descarte rule there are two or zero positive roots root but as it has at least one positive root so there are two positive roots so original equation has one positive root.
Now $f(-x) = x^{2022} + 2x + 1 = 0$
As there is no change of sign so there is no -ve root
So the equation has has one root and it is positive.
So answer is (c)-
We have the terms 8,15,24,35
Let us compute 1st order difference 7,9,11 which is an AP
Next difference is 13 and so next term is 35 + 13 = 48
So the nth term of the given sequence is quadratic say $an^2+bn + c$
Putting n =1 ,n= 2 and n =3 we get following
$a+b + c = 8\cdots(1)$
$4a+2b + c = 15\cdots(2)$
$9a+3b + c = 24\cdots(3)$
Subtracting (1) from (2) we get
$3a+b = 7\cdots(4)$
Subtracting (2) from (3) we get
$5a+b = 9\cdots(5)$
Subtracting (4) from (3) we get $2a=2$ or $a=1$
Putting $a=1$ in (4) we get $b=4$ and putting in (1) we get $c=3$
So $n^{th}$ term = $n^2+4n + 3$
we get the same results by putting n=1,2,3,4 so on
The number can contain only the digits 1,2 besides 0. 1 * 44 = 44 and
there is no overflow( if in the number the sum is one then it becomes 8,
and if it is 2 the the sum of digits is 16) so the sum of digits is 8
times. if any digit is 3 to 9 then sum of digits less than 8 times so
this shall give a lesser sum
(just an observation and not required for the result) Again 1 may be preceded/succeeded by 0 or 1 as 11 * 44 = 484 . but 2
has to be preceded/succeeded by 0 as 21 * 44 = 924 and 12 * 44 = 538
and the sum of digits become less
So the number shall have 0 1 and 2 meeting above conditions so that sum
of digits 100 and when we multiply by 3 (that is 3n) the digits shall be
0,3,6 and there is no overflow and sum of digits 300.
Let the number of numbers be n and it starts at a
So we have the sum $= an + frac{n(n-1)}{2} = 2024$
Or $2an + n(n-1) = 4048$ and $a\ge 1$
Or $n(2a+n-1)= 4048 = 16 * 253 = 2^4 * 23 *11$
Now out of n and 2a+n-1 one is even and one is odd so n = 16 ( 2a +n -1 = 253) or 1 ( 2a +n -1 = 4048) or 11 ( 2a +n -1 = 368) or 23 ( 2a +n -1 = 176)
Clearly n = 23 is the largest giving 2a + 22 = 176 and a = 77
So largest number of consecutive positive integers is 23 and it starts with 77
