SOLUTION FOR 10/03/2025 MATH CHALLENGE// BY @AKWASH360

Hello everyone, welcome to another episode of the math challenge where I will be submitting my solution for 10/03/2025 math challenge. Even though it’s quite challenging but I’m glad I could find my way around it and I hope that this will be of help to many students out there.

Image is mine
So, the first thing I did to solve this problem was the obtain the prime factors of each of the numbers provided, which is also known as the divisors of that number. I started by obtaining the divisors for 6400 as seen in the above image of which I got 2 raises to the power of 8 and also 5 raise to the power of 2.

Image is mine
Furthermore, I proceeded to obtain the divisors or the prime factors of 9600 of which I have it to be 2 raises to the power of 7, 3 raise to the power of 1 and 5 raise to the power of 2 as seen in the above image.

Image is mine
After which I went further to obtain the common factors of 6400 and 9600 to be 2 raises to the power of 7 and 2 raises to the power of 2. Which I further obtain the total divisor by applying the formula (a + 1) (b + 1), of which I have 24 to be the positive divisors in common for 6400 and 9600 as we are asked to find.

Below is the complete process for the solution;

Image is mine

Thanks for going through my post and I hope you've learned from it.

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