Thank you all for your attempts. Here is the solution for this question.

Given: \(p\) is the smallest perfect cube greater than 1 such that difference between the tens digit of \(p^2\) and the units digit of \(p^2\) is 2.

We are also given that \(Z = 1*2*3*…*p\)

The question asks us to find the total number of factors of \(Z\).

Approach: To find the total number of factors of \(Z\), we need to know the power of every prime factor of \(Z\). Let’s say Z is of the form \(Z = a^m * b^n * …\). where \(a, b\),… are primes, then the total number of factors of \(Z\) will be \((m+1)*(n+1)…\)

Therefore to be able to express \(Z\) in its prime factorized form, we need the value of \(p\).

We will try to find the value of \(p\) based on the constraints imposed on it:

a. \(p\) is a perfect cube greater than 1

b. difference between the tens digit of \(p^2\) and the units digit of \(p^2\) is 2

Working Out: \(p\) is a perfect cube greater than 1.

Possible values of \(p\) are 8, 27, 64…

When \(p = 8\), \(p^2 = 64\).

Notice that this satisfies the second condition too (difference between the tens digit of \(p^2\) and the units digit of \(p^2\) is 2). So we need not check other values for \(p\).

Therefore \(p = 8\)

This essentially means \(Z = 1*2*3*4*5*6*7*8 = 2 * 3 * 2^2 * 5 * 2*3 * 7 * 2^3 = 2^7 * 3^2 * 5 * 7\)

Therefore total number of factors of \(Z\) = \((7+1)*(2+1)*(1+1)*(1+1) = 96\)

Correct Answer:

Option DHere is another question that tests your conceptual understanding of primes and factors.

x-y-are-integers-find-the-number-of-even-factors-of-4x-197375.htmlHope this helps.

Regards,

Krishna

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