2026/040) The number $2^{29}$ has exactly 9 distinct digits. Find the missing digit.

Let us work mod 9.

We have $2^3 \equiv -1 \pmod 9$

Hence $2^{27} \equiv (-1)^9  \equiv -1 \pmod 9$ 

Hence $2^{29} \equiv (-1) * 4  \equiv -4 \pmod 9$

If we have all the digits(once) that is 10 digits  then sum of digit is 45 so it is divisible by 9 or 0 mod 9 

So removing 4 we shall have -4 mod 9. 

So  missing digit is 4

2026/039) The product (8)(888…8), where the second factor has k digits, is an integer whose digits have a sum of 1000. What is k? (A) 901(B) 911(C) 919(D) 991(E) 999

Solution 

The above value = 8 * (k 8s) = 8 * 8 * (k ones) ths we find by taking 8 out 

$= 64 * \frac{(10^k-1)}{9}$ as $(n) ones * 9 = \frac{10^n-1}{9}$

 $= (7 *9 +1) * \frac{(10^k-1)}{9}$  as denominator is 9 we put 64 as multiple of 9 and plus 1

 $ =7 * 9  * \frac{10^k-1}{9}+ \frac{10^k-1}{9}$ expanding

 $ =7 * (10^k-1)+ \frac{10^k-1}{9}$

  $ =7 * 10^k-7 + \frac{10^k-1}{9}$ 

  $ =7 * 10^k + \frac{10^k-1}{9}-7 $

 The 1st term gives 7 followed by k zeroes the 2nd term gives k ones and sum total shall be 7 followed by k ones. . when we subtract 4 we get   7 followed by k-2  zeroes followed by 04.

This gives sum of digits = 7 + k -2 + 4 = 1000

or  = 991 

 so Answer is (D)

2026/038) Find all $n \in N$ so that 7 divides $5^n+1$-.

 Basically we need to find n such that $5^n = -1 \pmod 7$

Now we have as 7 is a prime number as per Fermat's Little Theorem $5^6 \equiv 1 \pmod 7$

So $5^{6k} \equiv 1 \pmod 7$

Now as $5^6 \equiv 1 \pmod 7$ so we need to check for power of 5 to a factor of 6 that is 1 or 2 or 3

$5^1,5^2$ do not satisfy and $5^3 \equiv -1 \pmod 7$ satisfies.

so $n \in 6k+3 $ for all $k \in \mathbb{N}$ 

 

 

 

 

2026/037) GCD of 2472,1284 and a third number n is 12.If their LCM is $2^3* 3^2*5*103 * 107$.

 Because this is problem of CGD and LCM it makes sense to find prime factors of all numbers. 

Because GCD is 12 my approach is to mention it as product of 12 and other prime factors. let n = 12k

2472 = 12 * 206 = 12 * 2 * 103

1284 = 12 * 107

n =  = 12 *k

LCM = 12 * 2 * 3 * 5 * 103 * 107

Let us see that is  

After the 12 there is additional 2 and that comes from 2472 ( so there can be 0 or 1 2 in k)

There is an additional 3 and it has to come from k as it does not come from other numbers

There is an additional 5 and it has to come from k as it does not come from other numbers

There is a 103 in 2472 not in 1284 putting 0 or 1 103 shall not change GCD or LCM

There is a 107 in 1284 not in 2742 putting 0 or 1 107 shall not change GCD or LCM

So $n = 12k = 12 * 3 *5 *2^a * 103^b * 107^c = 180 * 2^a *103^b * 107^c$ where each of a,b,c is 0 or 1

2026/036) Show that a positive integer m is a sum of two triangular numbers if and only if 4m+1 is a sum of two squares.

m is a sum of 2 triangular numbers 

Let the 2 triangular numbers be $t_n$ and $t_p$

We have 

$t_n = \frac{n(n+1)}{2}$

$t_p = \frac{p(p+1)}{2}$

So we have

$m = t_n +t_p =\frac{n(n+1)}{2} + \frac{p(p+1)}{2}$

Or $4m = 2n(n+1) + 2p(p+1)$

Or $4m + 1= 2n(n+1) + 2p(p+1) + 1$

 Or $4m+1 = 2n^2+2p^2 + 2n + 2p + 1$

using the fact that $2(a^2+b^2) = (a+b)^2 + (a-b)^2$ one can expand the RHS and check 

 we get $4m+1 = (p+n)^2 + (p-n)^2  + 2n + 2p + 1$

Now $4m + 1 = 2t(t+1) + 2k(k+1) + 1 = 2t^2 +2t + 2k^2 + 2k + 1$

$= (t-k)^2 + (t+k)^2 + 2(t+k) +1$

$= (t-k)^2 + (t+k+1)^2$

is sum of 2 squares.  As each step is reversible we can start from bottom and go backwards to prove the other part.

 

 

2026/035) Prove that circle l(0,2) with equation $x^2+y^2=4$ contains infinite points with rational coordinates.

 Solution

Solution to this is $x = 2 \sin 2t$ and $y = 2\cos 2t$ (deliberately chosen angle in form of 2t to avoid fraction angle)

We can represent $\sin 2t$ and $\cos 2t$ expressible in form $\tan t$ as

$\sin 2t = \frac{(2 \tan\, t)}{(1+ tan^2 t)}$

$\cos 2t = \frac{(1-tan ^2t)}{(1+ tan^2 t)}$

So we have 

$x = \frac{(4 \tan\, t)}{(1+ tan^2 t)}$

$y = \frac{2(1-tan ^2t)}{(1+ tan^2 t)}$

If $\tan\, t$ is rational then both x and y are rational and point is rational co-ordinate

We can any rational value of $\tan\, t$ to get rational co-ordinate of a point

Hence it has infinite points with rational co-ordinates

2026/034) Prove that the number of integral solutions of the equation $x^3+y^4=z^{31}$ is infinite

Note: This is one method of solution. Other method exist 

Because the left had side has 2 terms and right had side has one and  

$2 * 2^p = 2^{p+1}$

If we can make 

  $x^3= y^4\cdots(1)$ 

Same as some power of 2 we have a solution

Because   $x^3$ is a power of 2 so  x has to be some power of 2 say 

$x=2^m\cdots(2)$ 

And similarly y has to be some power of two say 

 $x=2^n\cdots(3)$

From (1),(2), and (3) we have

$(2^m)^3 = (2^n)^4$

So 3m = 4n

As m and n are integers we have m must be divisible by 4 and n by 3 so 3m and 4n which are same by 12

So we have

$3m = 4n = 12k$

So

$m = 4k\cdots(4)$

$n= 3k\cdots(5)$

And from (3) and (4)

$x = 2^{4k}\cdots(6)$

$y =  2^{3k}\cdots(7)$

Putting in the given equation we get

$(2^{4k})^3 + (2^{3k})^4 = z^{31}$ 

Or $2^{12k} + 2^{12k} = z^{31}$

Or $2^{12k+1} = z^{31}$

Because LHS is a power of 2 so RHS is also a power of 2 so z has to be a power of 2 say 

$z= 2^t\cdots(8)$

Thus we get  $2^{12k+1} = 2^{31t}$

Or  $12k+1 = 31t\cdots(9)$

We can solve it using Extended Euclidean Algorithm to solve the same.

However as 12 and 31 are small numbers we can use the following approach as well

As $12 | 31t-1$ so $12 | 7t-1$ as 31 is 7 mod 12

By putting values of t from 0 to 11 we get (we need not put all values but upto the solution) and kowing that t is odd as t even shall make the number odd and not divisible by 12 we get t = 7.

Putting $t=7$ in (3) to get $k=18$

So one soultion is k =18, t = 7

As $12k+1 = 31t$ adding 12 * 31 a on both sides shall not change the values

So 12(k+31a) + 31(t +12a)

As one set of solution is (18,7) so parametric solution is  t = 7+12a, k= 18 + 31a

From (6) and (7) and (8) 

We get $x=2^{4(18+31a)}$,  $y=2^{3(18+31a)}$,  $z=2^{7+12a}$ 

This is a parametric solution and by varying a any whole number we can get any number of solution

Hence infinite number of solutions 

 

 

 

2026/033) Find the number of ordered pairs (a,b) of positive integers that are solutions of the following equation: $a^2+b^2=ab(a+b)$.

We have 

$a^2(b-1) + b^2(a-1) =0$

As both terms are non negative and sum is zero both are zero or $a=b=1$ giving one ordered pair 

2026/032) The graph of $x^2−4y^2=0$ is: (A) a parabola (B) an ellipse (C) a pair of straight lines (D) A point

We have $x^2–4y^2=0$

Or $(x-2y) (x+2y) = 0$

Hence $x-2y = 0$ (this is a straight line)   or $x + 2y = 0$ (this is another straight line)

Hence ans is (C) A pair of straight lines

2026/031) What is the greatest integer less than or equal to $(2+√3)^2$

 

Using $(a + b) ^2 + ( a -b)^2 = 2(a^2 + b^2)$

We get $(2 + √3)^2 + (2 -√3)^2 = 2(4+3) = 14$

As $0 < 2 -√3 < 1$ so $0 < ( 2 - √3)^2 < 1$

hence

$13 < (2+√3)^2 < 14$

Hence integral part is 13 or greatest integer is 13

2026/030) How can we solve this equation $a^-ab+b^2=1 such that a and be are positive integers?

Multiply by 4 on both sides to get

$4a^2 -4ab + 4b^2 = 4$

Or $4a^2 - 4ab + b^2 + 3b^2 = 4$

Or $(2a -b)^2 + 3b^2 = 4$ 

Looking at integer solutions $2a - b = 1, b = 1$ as b above 1 becomes larger

Giving $a = 1, b = 1$

2026/029) What is the smallest perfect square larger than 1 with a perfect square number of positive integer factors?

 $2^2$ has 3 factors and $3^2$ has 3 factors product 36 has 9 factors . Let us check if smaller square has 16 has 5 and 25 has 3 . So 36 is the smallest perfect square number having perfect square number of  positive factors

2026/028) Let n be a positive integer such that $12n^2+12n+11$ is a 4-digit number with all 4 digits equal. Determine the value of n.

Let it be 1111x . Add 1 in both sides to get

$12n^2 + 12n + 12 = 1111x + 1$

Work mod 12 to get 

$7x + 1 = 0  \pmod {12}$

So x is odd

Trying x odd values that is $1,3,5,9,11$ we get x is 5 .

So $12n^2 + 12n + 12 = 5556$

Or $n^2 + n + 1 = 463$

Or $n = 21$

2026/027) Let n be a positive integer such that $12n^2+12n+11$ is a 4-digit number with all 4 digits equal. Determine the value of n.

All 4 digits are same

Let it be 1111x . Add 1 in both sides to get

$12n^2 + 12n + 12 = 1111x + 1$

Work mod 12 to get 

$7x \equiv = 0 \pmod {12}$

So x is odd and 

Trying x 1 3 5 9 11 we get x is 5 .

So $12n^2 + 12n + 12 = 5556$

Or $n^2 + n + 1 = 463$

Or $n = 21$

 

2026/026) The number of solution-pairs in the positive integers of the equation 3x+5y=501 is:(A) 33(B) 34(C) 35(D) 100(E) none

We have 501 = 495 (multiple of 15) + 6

6 = 3 * 2

So 501 = 2 * 3 + 99 * 5

As 3 * 5 = 5 * 3

We have 501 = 3 (2 + 5t) + 5 ( 99 - 3t)

If t is zero or positive coefficient of 3 is positive and t -ve coefficient of 3 is -ve . We need to ensure 99 - 3t positive so t is less than 33 so 33 values 0 to 32

Hence answer (A) 33

2026/025) Show that the sum of three consecutive perfect cubes can always be written as the difference between two perfect squares.

We have sum of 1st n cubes ($\frac{(n(n+1)}{2})^2$

Hence $\sum_{k=1}^{k=n} k^3$ =  $(\frac{(n(n+1)}{2})^2$

Also  sun of 1st n+p cubes is

 $\sum_{k=1}^{k=n+p} k^3 $ =  $(\frac{((n+p)(n+p+1)}{2})^2$

Hence sum of p cubes from $(n+1)^3$ to $(n+p)^3$ is

  $\sum_{k=n+1}^{k=n+p} k^3 $ =  $(\frac{((n+p)(n+p+1)}{2})^2$ - $(\frac{(n(n+1)}{2})^2$

p = 3 is a special case of the problem 

2026/024) Which of the following numbers is a perfect square? A) $\frac{14!15!}{2}$, B) $\frac{15!16!}{2}$,C) $\frac{16!17!}{2}$,D) $\frac{17!18!}{2}$

 

We are having numbers in the form $\frac{n!(n+1)!}{2}$ .

This is $\frac{(n!)^2 * (n+1)}{2}$

It  is a perfect square if $\frac{(n+1)}{2}$ is a perfect square $\frac{18}{2}$ is a perfect square so (D) is the answer

2026/023) Let P be the product of any three consecutive positive odd integers. The largest integer dividing all such P is: (A) 15(B) 6(C) 5(D) 3(E) 1

 Because this is 3 consecutive odd numbers one of them is divisible by 3. None may be divisible by 5 as in case of 17,19,21. If the form are 5n + 2, 5n+ 4, 5n+ 4 where n is odd.  None is divisible by 2 as numbers are odd . So answer is 3 that is (D) 

2026/022) Let n be certain positive integers divisible by 17 & n+1 divisible by 13. How to determine such integers n?

 Let n be divisible by 17 . So n is of the form 17k . 17k +1 is divisible by 13 so 4k +1 is divisible by 13 . So k is 3 is a solution or any multiple of 13 plus 3. Say $k=13 t +3$ or $n =221t +51$

2026/021) Let a,b,c be integers satisfying $ab+bc+ca=1$. Prove that $(1+a^2)(1+b^2)(1+c^2)$ is a perfect square.

We have putting $ab +bc +ca$ for 1 in $1+a^2$

$1+a^2 = ab + bc +ca + a^2 = (a+b)(a+c)$

Similarly $1+b^2 = (b+c)(b+a)$

And $1+ c^2 = (c+b)(c+a)$

So $(1+a^2)(1+b^2)(1+c^2) = ((a+b)(b+c)(c+a))^2$ a perfect square

2026/020) From the number $7^{2026}$ we delete its first digit, and then add the same digit to the remaining number. This process continues until the left number has ten digits. Show that the left number has two same digits.

We are deleting one digit and adding to the number. So value mod 9 shall not change . Given number mod 9 is not zero . So when we get a 10 digit number it shall not be divisible by 9. If all 10 digits are different then the number is divisible by 9. So all ten digits cannot be different. So upto 9 digits shall be different so at least one digit must repeat or two same digits will be there.

2026/019) Determine all positive integers that are equal to 300 times the sum of their digits.

 

The number must end with 2 zeroes . Excluding the 2 zeroes the sum of digits multiplied by 3 is the number. It is not one digit number as multiplied by 3 of the digit changed number

Maximum sum of digits of a 3 digits number is 27. Multiplied by 3 gives 81 which is a 2 digits number . So number of digits cannot be 3 or more. So it is a 2 digit number if it exists. 

Let it be $10a +b$. 

We have 

$10a +b =3(a+b)$ 

Or $7a =2b$ 

As a and b are single digit number so a is 2 and b is 7 and number is 27. We have removed 2 zeroes and so number is 2700 .

 

 

2026/018) Let $lcm(a,b)$ denote the least common multiple of a and b. Find the sum of all positive integers x such that $x \le 100$ and $lcm(16,x)=16x.$

 x must be co-prime to 16 . That means x is odd and every odd number satisfies the criteria . So result  is sum of all odd numbers less than 100. That is 2500

2026/017) Let m and n be positive integers such that 5 divides $2^n+3^m$. Prove that 5 divides $2^m+3^n$

 

Because 5 is a small number we can work as below each exponent

Working in mod 5 we have

$,2^1 = 2,2^2= 4, 2^3 = 8 = 3,2^4=1$

$3^1=3, 3^2 = 9 = 4 , 3^3 = 27 = 2,3^4=1$

so $2^4 + 3^2$ = 0 mod 5 also $3^4 +2^2 = 0$ mod 5

$2^1+ 3^1 = 0$ mod 5 ( n=m)

$2^3 + 3^3 = 0$ mod 5(n=-m)

We have checked for all combinations that it is true

2026/016) Find all solutions of the linear congruence $3x−7y \equiv 11 \pmod {13}$

We have as 13 is a prime number we can choose any value of a x and the choose y in terms of x

We have $7y \equiv  3x -11  \pmod  {13}$

Now 7 needs to be multiplied by its inverse to get coefficient of y as 1 so multiplying by 2 (inverse of 7) we get

$y \equiv  6x - 22 \pmod  {13}$

Or   $y \equiv 6x-9  \pmod {13} $

For x = 0 to 12 mod 13 we get corresponding  value of y 

2026/015) Show that difference between squares of 2 conscutive triangular numbers is a perfect cube

We have $T_n = frac{n(n+1)}{2}$ 

To avoid fraction we have $2T_n = n(n+1)$

Now we have 

$ (2T_{n+1})^2 -   (2T_{n})^2 = ((n+1)(n+2))^2 -((n+1)(n+2))^2 $

Or $4(T_{n+1}^2-T_n^2) = (n+1)^2((n+2)^2 - n^2) = (n+1)^2 * 4 (n+1))= 4(n+1)^3$

Or   $T_{n+1}^2 -   T_{n}^2 = (n+1)^3$

 

2026/014) Find all integer N such that $\lfloor \sqrt{n} \rfloor$ divides n

 Any number from $k^2$ to $k(k +2)$ shall have square root between $k$ and $k+1$ as $k(k+2) +1$ shall have square root $k + 1$

So the floor function of any number $k^2$ to $k(k+2)$ shall be $k$. For $k$ to divide the number with in the range it has to be of the form $k^2$ or$ k(k+1)$ or $k(k+2)$

 So for any $k $ $k^2 , k(k+1),k(k+2)$ satisfy the criteria

2016/013) Show that the tens digit of $3^n$ is always even

We know working in modulo 20 as 3 is co-prime to 20 

$3^0  \equiv 1 \pmod {20}$

$3^1  \equiv 3 \pmod {20}$

$3^2  \equiv 9 \pmod {20}$

$3^3  \equiv 7 \pmod {20}$

$3^4  \equiv 1 \pmod {20}$

It repeats from this point,

As $3^n \pmod {20}$  is single digit so tens digit is even.

2026/012) The sum of two numbers is 667, the ratio of their LCM to GCF is 120. What are the two numbers?

 

Let $GCD(m,n) = k$ then we have

$m = ka$

$n = kb$

For some integers a and b where GCD(a,b) = 1

LCM = kab and GCD = k

So $ab = 120$

And $m +n = k(a+b)$

$m +n = 667 = 29 * 23$

So k can be $1,23,29, 667$

If $k =$1$  $a +b = 667$ and $ab = 120$ this is not possible

If $k = 667$ $a + b = 1$ so this is not possible

If $k = 23$ $a +b = 29, ab = 120$ so $a = 24,b =5$ or vice versa giving $(552,115)$  or $(115,552)$

if $k = 29$ $a +b = 23, ab = 120$ so $a = 15, b =8$ or vice versa giving$ (345,232)$ or $(232,345)$

So solution set is $\{(552,115), (115,552), (345,232), (232,325)\}$

2026/011) Solve in integer $x^2+4x +2 \equiv 0 \pmod 7$

 To complete square add 2 on both sides to get 

$x^2+4x +4 \equiv 2 \pmod 7$

or $(x+2)^2  \equiv 2 \pmod 7\cdots(1)$

now working mod 7

 $(0^2  \equiv 0 \pmod 7\cdots(2)$

 $(1^2  \equiv 1 \pmod 7\cdots(3)$

 $(2^2  \equiv 4 \pmod 7\cdots(4)$

 $(3^2  \equiv 2 \pmod 7\cdots(5)$ 

 from  (1) and (5) we have

 $(x+2)  \equiv 3 \pmod 7\cdots(6)$ or  $(x+2)  \equiv 3 \pmod 7\cdots(7)$ 

or $x\equiv 1 \pmod 7$ or    $x\equiv 2 \pmod 7$

 

 

 

2026/010) Is it true that $a^2+ab+b^2$ is divisible by 7 for infinitely many coprime pairs (a,b)

As a = 0 gives b= 0 so a = 7k and b= 7m satisfy the given equation but they ate not co-prime  

Because of symmetry let is assume $a=mb \pmod 7$ 

Putting in the given equation we get $m^2+m+1 \equiv 0\pmod 7$

As $m=1$  is not a solution to it multiplying by (m-1) on both sides   we get 

$m^3 \equiv 1 \pmod 7$ 

So $ m \equiv 1  \pmod 7$ or $ m \equiv 2  \pmod 7$ or $ m \equiv 4  \pmod 7$ but as $ m \equiv 1  \pmod 7$ is not a root of original eqution so solution is $ m \equiv 2  \pmod 7$ or $ m \equiv 4  \pmod 7$

So we can chose $(7p+m, 7q+2m)$ or $(7p+m, 7q+2m)$  p and q to be chosen such that the pairs form a co-prime

One set the infinite co-primes $(7p+1,7p+2)$ for any p 

 

 

 

 

2026/009) If $abc=1$ and $a+b+c=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$ how can I show that a or b or c equals 1

We have $abc=1\cdots(1)$

 So $\frac{1}{a} = \frac{abc}{a} = bc$

Similarly 

$\frac{1}{b} = ca$

And 

$\frac{1}{c} =   ab$

Hence

$\frac{1}{a}+\frac{1}{b}+\frac{1}{c} = bc+ca+ab\cdots(2)$

From above and

$a+b+c= \frac{1}{a}+\frac{1}{b}+\frac{1}{c}$

We have  

Hence $a+b+c = ab+bc+ca\cdots(3)$ 

a,b c are roots of equation

$P(x) = x^3-(a+b+c)x^2 +  (ab+bc+ca)x - abc=0\cdots(4)$

Let $a + b+c = m\cdots(5)$

So we get  from (4), (3), (5) 

$P(x) = x^3-mx^2+mx -1=0$ 

1 is a root of above equation as P(1) is zero

As a , b,c are roots one of $a,b,c$ is 1

Proved  

2025/008) Is it true that every odd number divides some number of the form $2^n−1$

 

Let the odd number be k. now let us take the number sequence $2^2–1, 2^3–1, 2^4–1$ so on k values that is upto $2^k-1$

Let us calculate all k remainder values when the above expressions are divided by k (as the remainder can be from zero to k-1) then we have on difference is zero and we are done.

If all k values are not different then we have say for 2 values of m and n the remainder must be same. without loss of generality let us assume $m > n$

So $(2^m-1) - (2^n-1) = 2^(m-n) -1 * 2^n$ is divisible by k. As k is odd $gcd(2^n,k) = 1$ so k must divide $2^(m-n) -1$

Note: The above is for the persons who are not familiar with number theory  or want to know from $1^{st}$ principle. 

n divides $2^{\phi(n)} -1$ as n is co-prime to 2 as per  Eulers theorem in number theory.

2025/007) Can two different primes $p\ne q$ satisfy $p^q−q^p=1$

Because difference is one one of the term is even and other is odd . So either p or q (being prime) is 2.

Let p = 2

so $2^q-q^2$, $q =3$ gives $-1$,$ q = 5$ gives $7$ and this keeps going increasing so no solution

Now let us check with $q = 2$ 

So $ p^2 -2^p$  $p =3$ satisfies$ p = 5$ gives $-7$ and larger p gives smaller value decreasing so no more solution

So only solution $p =3, q = 2$

2026/006) Prove that for real $x_1,x_2,x_3$ $x_1^2 + x_2^2+ x_3^2 \ge \frac{4}{3}(x_1x_2 + x_2x_3)$

We have

 $x_1^2 + x_2^2+ x_3^2-\frac{4}{3}(x_1x_2 + x_2x_3)$

$=(x_1^2- \frac{4}{3}x_1x_2) + ( x_3^2-  \frac{4}{3} x_2x_3)+x_2^2$  Combining $x_1x_2$ with $x_1^2$ term and   $x_3x_2$ with $x_2^2$ term 

$=(x_1^2- 2 *\frac{2}{3}x_1x_2 + (\frac{2}{3}x_2)^2) + ( x_3^2-  2* ( \frac{2}{3} x_2x_3+  (\frac{2}{3}x_2)^2)+(x_2^2-  (\frac{2}{3}x_2)^2) - (\frac{2}{3}x_2)^2)$  completing the square and subtracting the same 

$=(x_1 - \frac{2}{3}x_2)^2 + ( x_3- \frac{2}{3}x_2)^2+\frac{1}{9} x_2^2$

This is positive or zero

So 

  $x_1^2 + x_2^2+ x_3^2-\frac{4}{3}(x_1x_2 + x_2x_3)>=  0$

Hence  $x_1^2 + x_2^2+ x_3^2 \ge \frac{4}{3}(x_1x_2 + x_2x_3)$ 

2026/005) How do I the smallest four-digit integer N, such that $\sqrt{3\sqrt{N}}$ is an integer?

We have one 3 under the square root so we should have another 3 that should be part of $\sqrt{N}$ So it should be $N= (3x)^2$ and x has to be a square so smallest 4 digit number giving $N= 9x^4$ satisfying that it is smallest 4 digit number    when $x=4$ giving  $N= 9 * 256$ or 2304 

2026/004) If roots of the equation $a(b−c)x^2+b(c−a)x+c(a−b)=0$ are equal, then prove that 'a', 'b', 'c' are in harmonic proportion.

 Because roots are equal so the descrminamt is zero

or $b^2(c-a)^2-4ac(b-c)(a-b) = 0$

becuase we need to show that $a,b,c$ are in HP so $\frac{1}{a} = p , \frac{1}{b} = q, \frac{1}{c} = r$ are in AP

now given relation is

$\frac{1}{q^2} (\frac{1}{r} - \frac{1}{p})^2- 4(\frac{1}{p})(\frac{1}{q})(\frac{1}{q}- \frac{1}{r}) (\frac{1}{p}- \frac{1}{q})=0 $ 

or $\frac{(p-r)^2}{p^2q^2r^2}$$ - 4(\frac{(r-q)(q-p)}{q^2r^2p^2})=0 $

or $(r-p)^2 -4 (r-q)(q-p) = 0$

Let $x= r-q$ and $y = q-p$

so we get $r - p = x + y$

or $(x+y)^2 - 4xy= 0$

or $(x-y)^2 = 0$

or $x = y$ or $q-p = r-q$ so   $\frac{1}{a} = p , \frac{1}{b} = q, \frac{1}{c} = r$ are in AP and hence $a,b,c$ are in HP 

 

 

2026/003) Solve in positive integers $a^2=b^4 + 16b + 1$

 b has to even because if b is odd then we have RHS is of the form 4k +2 and it cannot be a perfects squares

now next square after $b^4$ is $(b^2+1)^2$ and the given expression can be a square(not necessary it will be) if greater than or equal to $(b^2+1)^2$

$b^4 + 16b + 1 >= (b^2+1)^2$

or $b^4 + 16b + 1 >= b^4 + 2b^2 +1 $

or $16b >= 2b^2$ or $ b <= 8$

So we need to try even values of b less or equal to 8 giving $b= 2 =>  a^2 = 49 =>a = 7$ 

$b =4 => a^2 = 256 + 64 +1 = 321$ no solution

$b = 6=> a^2 = 1395$ no solution

$b = 8 => a = b^2+1 = 65$

Solution $(7,2)$ and $(65,8)$

2026/002) Let a,b,c be integers. If $4a+5b−3c$ is divisible by $19$, prove that $ 6a−2b+5c$ is also divisible by $19$

 As 4 and 6 have LCM 12 so we should multiply by 3 and proceed

$4a + 5b -3c$ is divisible by 19 so multiply by 3 to get $12a + 15b -9c$ is divisible by 19. adding $19c-19b$ we get

$12a - 4b +10c$ is divisible by 19

or $2*(6a-2b + 5c)$ is divisible by 19 and as 2 is co-prime to 19 hence $6a-2b+5c$  is divisible by 19

2026/001) How do I find the smallest positive integer n, such that each digit of n is either 0 or 7, and n is divisible by 15?

It is divisible by 15. So it is divisible by both 3 and 5

Because it is  divisible by 5 so unit digit is 0 or 5, As we have to chose from 0 and 7 so it is zero. Now we cannot have a zero in other position as it will increase the value of the number as it is additional digit and does not serve any other purpose.Sso other digits shall be 7 and we require 3 7s to be divisible by 3 or the number as 7770( 15 * 518) .