2026/043) Show that there exists an n digit number each digit being odd and the number is divisible by $5^n$

We shall prove the same  by construction. But before that set us try to understand the pattern

One digit number divisible by 5there is only one number 5 and the digit is odd 

2 digit number divisible by $5^2=25$ the numbers are  25,50,75 and 75 has both digits odd 

3 digit number divisible by $5^3=125$ the numbers are  125,250,375 and so on 375 has all there digits odd

4 digit number divisible by $5^4= 625$ I am not enumerating  and a number 9375

We shall use this as a basis for construction of number by induction we shall expand the number from n digits to n+1 digits by adding a a digit to the left.

Let there be an n digit number with all n digits odd and divisible by  $5^n$ and let it be $k*5^n$. Note that k has to be odd else digit in unit place shall be zero which i even.

Now we know   that $10^n$ is divisible by $5^n$

So adding $p *10^n$ we can convert the n digit number to n+1 digit number and this is divisible by $5^n$.

We have n+1 digit number $p * 10^n  + k * 5^n= (p *2 ^n + k) 5^n$

Now we require and do we have $p *2^n + k$ divisible by 5

That is $p * 2^n \equiv -k \pmod 5$

As 3 is multiplicative inverse of 2 we get

$p  \equiv -k * 3^n \pmod 5$

p cannot be zero as $gcd(3,5) = 1$

So p is 1 or 2 or 3 or 5

If p is 1 or 3 then we are done

If p is 2 or 4 add 5 to p to get p single digit and odd

 

 

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