Veronica wrote:How many integers from 0 to 50, inclusive, have a remainder of 1 when divided by 3?
a.15, b.16, c.17, d.18, e.19
For this problem, I chose B, is it correct?
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
Post one question per thread, if you please.
The fist member to the list is 1, next and onwards would obviously be in an arithmetic progression of 1 as its first term, a, and 3 as the common difference. d. To avoid lengthy steps of calculations, if we could eyeball the greatest such integer under 50, then isn't it 49?
Hence, in nth term = a + (n - 1) d formula, we have 49 = 1 + (n - 1) × 3 or n - 1 = 16 or [spoiler]
n = 17
C[/spoiler]
If a - b is even, then a and b are either both even or both odd, and this results in a + b as always even. If a/b is even, then a is always even, and if a is even, a + b is even, then b must also be even, Remember, an even number divided by another even number cannot generate an even quotient, if the dividend were not a multiple of 4, hence, we cannot try 2, 6, 10, 14, etc for a in our question. From now, what is sure about a is that it's an even multiple of 2*; but still, b could either be an even or an odd multiple of 2**.
Let's see through the choices, now
A. a/2 = 4/2 = 2 is not odd*, so move ahead
B. b/2 must be an odd integer only if b is an odd multiple of 2**, so move ahead
C. (a + b)/2 must be an odd integer only if b is an odd multiple of 2**, so move ahead
D. (a + 2)/2 must be an odd integer only if a is an even multiple of 2, and a is*, [spoiler]
so grab it
D[/spoiler]
