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.

2026/019) Determine all positive integers that are equal to 300 times the sum of their digits.

 

The number must end with 2 zeroes . Excluding the 2 zeroes the sum of digits multiplied by 3 is the number. It is not one digit number as multiplied by 3 of the digit changed number

Maximum sum of digits of a 3 digits number is 27. Multiplied by 3 gives 81 which is a 2 digits number . So number of digits cannot be 3 or more. So it is a 2 digit number if it exists. 

Let it be $10a +b$. 

We have 

$10a +b =3(a+b)$ 

Or $7a =2b$ 

As a and b are single digit number so a is 2 and b is 7 and number is 27. We have removed 2 zeroes and so number is 2700 .