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When you see a question like this related to factorials and highest powers, know that the expression containing the factorial always represents the dividend AND the number for which you are trying to find the highest power, always represents the divisor.

Therefore, the idea is to work on breaking down the divisor into its prime factors (if it is not a prime number already).

In this question, therefore, the huge expression, 30! * 31! * 32!*….. is the dividend and the number 111 is the divisor.

111 = 37 * 3.

Finding the highest power of 111 which divides the 30!*31!*32!*…. is equivalent to finding the highest power of 37 and 3 that is contained in the expression.

The bigger number of the two is 37. Clearly, the highest power of this number will be much lesser than the highest power of 3, because there are going to be far fewer number of 37s in the product than the number of 3s.

So, finding the highest power of 111 is essentially the same as finding the highest power of 37.

30! * 31! * 32! * 33! * 34! * 35! * 36! * 37!* 38!* 39! * 40! – in this expression, as we see, 37 comes up exactly 4 times i.e. in 37!, 38!, 39! and 40!
The highest power of 37, and hence 111, is 4.

The correct answer option is B.

Hope this helps!
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What is the sum of all roots of the equation \((x^2 - 4x -9)( 2x^2 + 8x - 11)=0\)?

A. 0
B. -4
C. 99
D. 4
E. 8

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Kinshook
What is the sum of all roots of the equation \((x^2 - 4x -9)( 2x^2 + 8x - 11)=0\)?

A. 0
B. -4
C. 99
D. 4
E. 8


Sum of all roots:
1) x^2 - 4x -9 =0
x1 + x2 = -(-4) = 4

2) 2x^2 + 8x - 11=0
x3 + x4 = -(8/2) = -4

x1+ x2 + x3 + x4 = 4 + (-4) = 0
Answer is (A)

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