2024/032) What is the proof that 11 is the only prime number of the form $n^2 + 2$ where n is prime?

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

2024/031) Find triangular number which is one less than a multiple of 11.

 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

2024/030) Prove that for triangular numbers $t_{3k}+t_{4k+1}=t_{5k+1}$

We have by definition  triangular number $t_k = \frac{k(k+1)}{2}$

So   $t_{3k}+t_{4k+1}$

$= \frac{(3k(3k+1)}{2} + \frac{(4k+1)(4k+2)}{2}$

$= \frac{(9k^2+3k}{2} + \frac{16k^2 + 12 k + 2}{2}$

$= \frac{25k^2+15k + 2}{2}$

$= \frac{(5k+1)(5k + 2)}{2}= t_{5k+1}$

2024/029) Consider the equation $x^{2021} + x^{2020} + \cdots+x - 1=0$

 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)-

Descartes

2024/028)What is the next term in series 8, 15, 24, and 35 and find the general term

 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 

 

 

2024/027) A number n has sum of digits 100 while 44n has sum of digits 800 Find the sum of digits of 3n

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.

2024/026) What is the largest number of consecutive positive integers whose sum is exactly 2024?

 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

2024/025)What is the third number if the two numbers are 48 and 60? The L.C.M is 1680, and the GCD is 12 of all the numbers.

 LCM of 48 and 60 (that is 12 * 4 and 12 *5) = 240

Now LCM 1680 = 240 * 7

So 3rd number should have a factor 7 and it should be multiple of 12(else GCD cannot be be 12) and can have other that is 3rd factor that is a factor of 240/12 or 20 it could be 1,2,4,5,10,20. as GCD of 48 and 60 is 12 so taking a higher multiple shall not change GCD

So 3rd number could be 84/168/324/420/840/1680 that is any of these 6 numbers

2024/024) If $\cos\, A +\cos\, B + \cos\, C = 0$, prove that $\cos 3A + \cos 3B +\cos 3C = 12 \cos\, A \cos\, B \cos\, C$.

 We shall use 2 facts

$\cos 3x = 4 \cos^3 x - 3 \cos\, x\cdots(1)$

And 

if $a+b+c=0$ then $a^3+b^3+c^3 = 3abc\cdots(2)$

We are given

 $\cos\, A +\ cos\, B + \cos\, C = 0\cdots(3)$

Hence $\cos^3 A +\ cos^3 B + \cos^3 C = 3 \cos\, A \cos\, B \cos\, C\cdots(4)$

Now $\cos 3A + \cos 3B +\cos 3C$

$= 4 \cos^3 A - 3 \cos\, A$ + $4 \cos^3 B - 3 \cos\, B$ + $4 \cos^3 C - 3 \cos\, C$ using(1)

$=4(\cos^3 A +\ cos^3 B + \cos^3 C - 3 (\cos\, A + \cos\, B + \cos\, C)$

$=4 * 3 \cos\, A \cos\, B \cos\, C- 3 *0$  using (3) and (4)

$=12 \cos\, A \cos\, B \cos\, C$

Proved

2024/023) If a, b, c be in Arithmetic Progression, then the value of $(a+2b-c)(c+2b-a)(a+2b+c)$ is nabc. Find n

 We have a,b,c are in AP so 

$2b = a+ c$

Hence $a+2b -c = a + a+c -c = 2a\cdots(1)$

$c + 2b -a = c + a + c - a = 2c\cdots(2)$

$a + 2b + c= 2b + a +c = 2b+2b = 4b\cdots(3)$ 

Multiplying (1) (2) and (3) we get $(a+2b-c)(c+2b-a)(a+2b+c)= 16abc$

Hence n = 16

 

2024/022) N is a 50 digit no. all digit except the 26^th (from left) are 1.if n is divisible by 13,find the 26^th digit?

Let the $26^{th}$ digit be k

so $N = 1111\cdots( 50\, times) + (k-1) * 10^{24}$

or $N = \frac{10^{50}-1}{9} + (k-1) * 10^{24}$

or     $9N = 10^{50}-1 + 9 *  ((k-1) * 10^{24})$

because $GCD(9,13)=1$ so if n is divisible by 13 9N is divisible by 13.

as 13 ia prime we have

$10^{12} \equiv 1 \pmod {13}$

so  

$10{24} \equiv 1 \pmod {13}\cdots(1)$

 and

$10{48} \equiv 1 \pmod {13}$

hence

 $10{50} \equiv 100 \pmod {13}$

 or

 $10{50} \equiv 9 \pmod {13}$

$9N = 10^{50}-1 + 9 *  ((k-1) * 10^{24}) \pmod {13}$

$= 9-1  + 9 *  ((k-1) * 1) \pmod {13}$

 $= 9k -1 \pmod {13}$

this is zero and because k is a digit trying from 0 to 9 we get k=3 satisfies the condition

so the $26^{th}$ digit is 3