2026/044) Each digit in the n-digit number N is 1. What is the smallest value of n for which N is divisible by 333,333?

We know 333333 = 3 * 111111

Now the number which is 111111 or $\frac{10^6-1}{9}$ 

Any number which is having  1's is $\frac{10^n-1}{9}$ 

 $\frac{10^6-1}{9}$ shall divide  $\frac{10^n-1}{9}$ only when n is multiple of 6

Or n = 6k for some integer k

Number is divisible by 111111 and 9 as GCD(111111, 9) = 333333 

This is so because 111111 is divisible by 3 and 9 = 3* 3

So 6k should be divisible by 9 and smallest k = 3 or 6k = 18. 

N = 6k = 18  

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 $k \in \mathbb{N}$