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