Odd divided by even is never an integer.

Let x = 3
(A) \(\frac{(x + 1)}{2} = \frac{(3 + 1)}{2} = \frac{4}{2} = 2\)
(B)\(\frac{x}{7} = \frac{3}{7}\) ---------- Not integer
(C) \(\frac{{x^2}}{3}\) = \(\frac{3^2}{3} = \frac{9}{3} = 3\)
(D) \(\frac{x^3}{3} = \frac{3^3}{3} = \frac{27}{3} = 9\)
(E) \(\frac{x}{24} = \frac{3}{24} = \frac{1}{8}\) ------------ Not integer

B, C and D eliminated.

Let x = 21
(B) \(\frac{x}{7} = \frac{21}{7} = 3\)
(E) \(\frac{x}{24} = \frac{21}{24} = \frac{7}{8}\) --------- Not integer. Answer E...

+ E

If x is not divisible by 2 then it won't be divisible by 24 either

Good observation , for the number to be divisible by 24 we must have the numerator as a multiple of both 2 and 3 , which contradicts the statement -



Thus, answer must be (E)
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We are looking for an odd multiple of 3.

Evaluating option 3, just take the example of to be 9

9/24 is not an integer. Hence, E
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IMO, E

In this question you are looking for X divided by some even number (or X+-1 divided by 3)

B, C, D do not have either of these

A) (X+1)/2 -> if our x is divisible by 3 and not 2, then it is odd so (X+-1) is always divisible by 2

E) X/24 -> 24 is even so this is not possible, and thus is our answer

E
A) If X is not divisible by 2, it means that it is odd, hence (X+1)/2 is always an integer
B) X/7 could be true if the X is 21, 63, 105...
C) X^2/3 is always an integer 3^2/3=3 9^2/3=27 etc
D) The same logic as in C. 3^3/3=9
E) Will never be an integer as X doesn't contain any even factor(not divisible by 2), while 24 has 3 2s

for this to hold true,
x is an odd multiple of 3

statement 1: x + 1 is even hence divisible by 3 - can be an integer
statement 2: if x is 21, x/7 is an integer
Statement 3: x^2/3 always an integer as x is a multiple of 3, so x^2 is a multiple of 3
Statement 4: same as above, x is a multiple of 3, x^3 is a multiple of 3 hence x^3/3 is an integer

Statement 5- correct answer- not an integer as 24 has two even integers and hence x is odd - CAN NEVER BE AN INTEGER

x is divisible by 3 and not by 2. So x is odd multiple of 3
so, lets get down to choices to check which one is an integer and which is not.

(A) x+1 is even and hence (x+1)/2 = I
(B) x/7 may be an integer or may be not. We can't say anything based on the data available with us.
(C) x^2 must be a multiple of 3. So, x^2/3 = I
(D) x^3 must be a multiple of 3. So, x^3/3 = I
(E) 24 = 2*2*2*3 since x is odd multiple of 3. x is not divisible by 24. Hence x/24 is not integer.

Answer E.

Because x is divisible by 3 and not 2, x must be an odd multiple of 3, such as 3 or 9.

Since 24 is an even number, x/24 will NEVER be an integer.

Answer: E
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x / 3 ! by 2

x = 3 , 9, 15 , 21

x/24, will never give an integer for the present conditions, since it is 12*2 (this goes against the question)

E
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what if we take X as 240 for option E.?

[quote="bharat8484"]what if we take X as 240 for option E.?[/quote]


X can not be 240 as we have to find the value of x which is NOT divisible by 2 .

240 is divisible by 2 , hence x can not take this value . This is the given condtion in question.


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