2026/042) 65 distinct natural numbers not exceeding 2016 are given. Prove that among these numbers we can find four a,b,c,d such that a+b-c-d is divisible by 2016.

Out of 65 numbers one can choose 2 numbers in ${65}\choose {2}$ 2080 ways.

We have 2080 pairs and when we divide by 2016 there can be 2016 remainders So there exists a, b and c,d such that dividing a + b by 2016 leaves the same remainder as c + d dividing by 2016.  

Or a + b -c - d is divisible by 2016 . This is based on pigeon hole principle

2026/041) For $x^2+x+5$ to be a factor of $x^4+px^2+q$ the values of p and q must be, respectively: (A) −2,5(B) 5,25(C) 10,20(D) 6,25(E) 14,25

 Because product is bi-quadratic and one factor is quadratic so other factor must be quadratic

Other factor is of the form $x^2+ax+b$

So we get 

 $(x^2+2x+5)(x^2+ax+b) = = x^4 + (2+a)x^3 + (b+2a+5)x^2 + (2b+5a)x + 5b$

Comparing with $x^4+px^2+ q$ we get a = - 2 (coefficient $x^3$) and $b = 5$ from coefficient of x

So q = 5b = 25

Comparing coefficient of x^2 we get p = 5 -4 + 5 = 6

So Ans is (D) 6,25

 

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.