If 22^n is a divisor of 97!+98! ,what is the maximum possible

In questions related to divisibility, breaking down the dividend will give a lot of information about what the divisor needs to be if it HAS to divide the dividend fully. Therefore, let’s break the dividend down, first.

In 97! + 98!, we can take 97! as common. When we do this, we have 97! + 98! = 97! (1 + 98) = 97! * 99. This helps us understand that \(22^n\) should be able to divide 97! * 99 fully.

Let’s now look at 22. 22 = 2*11, therefore \(22^n\) = \(2^n\) * \(11^n\).
So, essentially, we are trying to find the highest power of 11 in the dividend since the number of 11s would be far fewer compared to the number of 2s.
Finding out the highest power of any prime number in a given factorial can be done by successive division. In our case, it can be done as shown below in the diagram.



So, the highest power of 11 in 97! is 8. But, 99 also has a 11 in it. Therefore, the highest power of 11 in the numerator (or the dividend) is 9. So, there need to be 9 11s in the denominator as well. As such, the highest power of 22 that can divide 97! + 98! is 9.
The correct answer option is B.

Hope that helps!

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