If n is an integer such that n

Question

If n is an integer such that n < 61, and  (n - 1)(n - 3)(n - 5)...(n - 53)(n - 55)(n - 57) > 0 , how many distinct values are possible for n ?

A. 15
B. 17
C. 29
D. 30
E. 31

chetan2u · Answer

(1) When n<58 Whenever n is odd, that is 1,3,5….,57, the Left hand side is 0, so answer is NO for 0>0. So we are looking at only even values<57 Even values: There are 28 even values : 2 to 58. To make (n-1)*(n-3)…(n-57) positive, we have to have even number of terms negative. There are 29 terms in the expression. When n =2, we have n-1 positive while all other 29-1 or 28 are negative terms. Similarly, n as 6,10,14…..54 will give even negative terms. Thus 14: (2,6,10,…..54) (2) When n>57 Also all values greater than 57 will give all terms in the expression as positive. Hence, 58, 59, 60 Total 14+3 or 17. B

BrushMyQuant · Answer

Given that n is an integer such that n < 61, and \((n - 1)(n - 3)(n - 5)...(n - 53)(n - 55)(n - 57) > 0\) and we need to find how many distinct values are possible for n?

(n - 1)(n - 3)(n - 5)...(n - 53)(n - 55)(n - 57) > 0
=> to find the terms we can just find how many numbers are in the series 1, 3, 5, 7,... 57

Now, this is an Arithmetic series with first term 1 and last term 57 and common difference = 2
=> Number of terms, n = \(\frac{Last Term - First Term }{ Common Difference}\) + 1 = \(\frac{57 - 1}{2}\) + 1 = \(\frac{56}{2}\) + 1 = 29 terms
( Watch this video to learn about the Basics of Arithmetic Series )

Now, product of 29 terms has to be greater than 0
=> We need to have

Now, if we take any value of n≤0 then all terms will become negative
Ex: n = 0
=> (n - 1)(n - 3)(n - 5)...(n - 53)(n - 55)(n - 57) = (0 - 1)(0 - 3)(0 - 5)...(0 - 53)(0 - 55)(0 - 57) = odd number of negative terms => < 0
=> NOT POSSIBLE

Now, n cannot be equal to 1, 3, 5, ..., 55, 57 as this will make the entire expression 0
=> NOT POSSIBLE

Now, If we take Even values of n which are not a multiple of 4 then we get odd number of negative numbers and even number of positive numbers, which will not give us < 0 value for entire expression
Ex: n = 2
=> (n - 1)(n - 3)(n - 5)...(n - 53)(n - 55)(n - 57) = (2 - 1)(2 - 3)(2 - 5)...(2 - 53)(2 - 55)(2 - 57) = + * - * -.... * - => < 0
=> NOT POSSIBLE

Now, If we take Even values of n which are a multiple of 4 then we get even number of negative numbers and odd number of positive numbers, which will give us > 0 value for entire expression
Ex: n = 4
=> (n - 1)(n - 3)(n - 5)...(n - 53)(n - 55)(n - 57) = (4 - 1)(4 - 3)(4 - 5)...(4 - 53)(4 - 55)(4 - 57) = - * - * +.... * + => > 0
=> POSSIBLE
=> n can be 4 , 8, 12, ..., 56 => \(\frac{56-4}{4}\) + 1 = \(\frac{52}{4}\) + 1 = 13 + 1 = 14 values

n = 58, 59, 60 will give us +ve value for all expressions
=> POSSIBLE

=> Total values of n possible are 14 + 3 = 17

So, Answer will be B
Hope it helps!

Watch the following video to learn the Basics of Inequalities

bumpbot · Answer

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If n is an integer such that n < 61, and (n - 1)(n - 3)(n - 5)...(n -

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