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