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If a and b are positive integers such that a – b and a/b are both even integers, which of the following must be an odd integer? A. a/2 B. b/2 C. (a+b)/2 D. (a + 2)/2 E. (b+2)/2

is the question correct?? none of the options seem to satisfy

here a>b since a/b is even integer

let a= 24 and b = 4.

A. a/2 = 12 => even B. b/2 = 2 => even C. (a+b)/2 = 14 => even D. (a+2)/2 = 13 => odd E. (b+2)/2 = 3 =>odd

now if a = 20 and b = 2 then a-b = 18 and a/b = 10 A. a/2 = 10 => even B. b/2 = 1 => odd C. (a+b)/2 = 11 => odd D. (a+2)/2 = 13 => odd E. (b+2)/2 = 2 => even

If a and b are positive integers such that a – b and a/b are both even integers, which of the following must be an odd integer? A. a/2 B. b/2 C. (a+b)/2 D. (a + 2)/2 E. (b+2)/2

\(a-b\) even --> either both even or both odd

\(\frac{a}{b}\) even --> either both even or \(a\) is even and \(b\) is odd.

As both statements are true --> \(a\) and \(b\) must be even.

As \(\frac{a}{b}\) is an even integer --> \(a\) must be multiple of 4.

Options A is always even. Options B can be even or odd. Options C can be even or odd. Options D: \(\frac{a+2}{2}=\frac{a}{2}+1\), as \(a\) is multiple of \(4\), \(\frac{a}{2}\) is even integer --> even+1=odd. Hence option D is always odd. Options E can be even, odd.

If a and b are positive integers such that a – b and a/b are both even integers, which of the following must be an odd integer? A. a/2 B. b/2 C. (a+b)/2 D. (a + 2)/2 E. (b+2)/2

is the question correct?? none of the options seem to satisfy

here a>b since a/b is even integer

let a= 24 and b = 4.

A. a/2 = 12 => even B. b/2 = 2 => even C. (a+b)/2 = 14 => even D. (a+2)/2 = 13 => odd E. (b+2)/2 = 3 =>odd

now if a = 20 and b = 2 then a-b = 18 and a/b = 10 A. a/2 = 10 => even B. b/2 = 1 => odd C. (a+b)/2 = 11 => odd D. (a+2)/2 = 13 => odd E. (b+2)/2 = 2 => even

what the heck.... ...D it will be...sorry for oversight from my end

Now if u take a as 4 and b as 2 then a-b is even a/b is even and a+b/2 is odd which is option C.

Please explain where i am going wrong?

First question: Why \(a\) has to be multiple of 4?

As we concluded \(a\) and \(b\) have to be even integers to meet both conditions in the stem: \(a=2m\) and \(b=2n\).

Then we have \(\frac{a}{b}\) is an even integer: \(\frac{a}{b}=2k\) --> \(a=2bk=4kn\) --> \(a\) is a multiple of \(4\).

Second question: Option C also COULD give an even result, so why not C?

The question asks: "which of the following MUST be an odd integer?"

If we take \(a=16\) and \(b=4\), then \(\frac{a+b}{2}=10=even\). So option C may or may not be odd and we are asked to determine which option is ALWAYS odd, hence C is out.

In fact: Options A is always even. Options B can be even or odd. Options C can be even or odd. Options D is always odd. Options E can be even or odd.

a - b and a/b are both even intergers ==> \(a - b = 2m_1\) \(\frac{a}{b} = 2m_2\) ==> \(a = (b)(2m_2)\) ==> a & b are both even, otherwise the above conditions won't be satisfied.

Now lets go with options one by one ... A) a/2 ==> can be even or odd ( a = 6 or a = 8)

B) b/2 ==> Can be either even or odd ( b = 6 or b = 8)

C) (a+b) /2 ==> Since, \(a = (b)(2m_2)\), \(\frac{a + b}{2}\) =\(\frac {(b)(2m_2) + b}{2}\) ==> \(\frac{a + b}{2}\) =\(\frac {(b)(2m_2 + 1)}{2}\) ==> Can be even or odd

D) (a+2)/2 ==> \(\frac{a + b}{2}\) =\(\frac{2bm_2 + 2}{2}\) ==> \(\frac{a + b}{2}\) =\(bm_2 + 1\) ==> Since, b is even, \(bm_2\) is even and \(bm_2 + 1\) is odd always

E) (b+2)/2 ==> \(\frac{b+2}{2} = 1 + \frac{b}{2}\) ==> Can be even or odd
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Cheers! Ravi

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a and b both have to be even to fulfill the condition. a could be = 4, 6, 8 ...., b = 2,4, .... so a - b = 2 (Even) a/b= 2k (Even) a/2=2k a = 4k so a must be multiple of 4 now try the options ans. D
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If a and b are positive integers such that a – b and a/b are both even integers, which of the following must be an odd integer? A. a/2 B. b/2 C. (a+b)/2 D. (a + 2)/2 E. (b+2)/2

is the question correct?? none of the options seem to satisfy

here a>b since a/b is even integer

let a= 24 and b = 4.

A. a/2 = 12 => even B. b/2 = 2 => even C. (a+b)/2 = 14 => even D. (a+2)/2 = 13 => odd E. (b+2)/2 = 3 =>odd

now if a = 20 and b = 2 then a-b = 18 and a/b = 10 A. a/2 = 10 => even B. b/2 = 1 => odd C. (a+b)/2 = 11 => odd D. (a+2)/2 = 13 => odd E. (b+2)/2 = 2 => even

small correction : (a+2)/2 = 11 as a= 20 ,so 22/2= 11 not 13

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