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

2026/020) From the number $7^{2026}$ we delete its first digit, and then add the same digit to the remaining number. This process continues until the left number has ten digits. Show that the left number has two same digits.

We are deleting one digit and adding to the number. So value mod 9 shall not change . Given number mod 9 is not zero . So when we get a 10 digit number it shall not be divisible by 9. If all 10 digits are different then the number is divisible by 9. So all ten digits cannot be different. So upto 9 digits shall be different so at least one digit must repeat or two same digits will be there.