Integer

[Veronica]

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

[narik11]

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

For 1st One B is right..

Numbers which leave 1 as reminder when divided by 3 are 1,4,7,10..
which can be written as 0*3+1,1*3+1,2*3+1,...

Therefore you have 16 multiples that give 16*3+1=(49)

For 2nd one..

a-b and a/b to be even .. both a &b have to be even
Plugin values for a &b. a=8 and b=4

Therefore Option d & e.

What's the OA

[sanju09]

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]

[diebeatsthegmat]

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

for question 1.
from 0 to 31 you will have 10 numbers which have a remainder of 1 when devided by 3
from 31 till 50 you will have only 5 : 34,37,40,43,47
tus the answer is B

for question 2
a-b is even integer thus a -even and b-even or a-odd and b-odd ( of course a>b)
a/b -even integer thus a-even and b-even or a even, b-odd( in some cases such as 6,3,18,3 and a>b)
thus we conclude take we should take a-even and b-even
a/2 is even because even/even is even
b/2 is not in integer eliminate it
a+b/2 is not an integer , eliminate it
a+2/2 =a/2+1 thus odd
b+2/2 is not an integer eliminate it
the correct answer is D

[Geva@EconomistGMAT]

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

For 1st One B is right..

Numbers which leave 1 as reminder when divided by 3 are 1,4,7,10..
which can be written as 0*3+1,1*3+1,2*3+1,...

Therefore you have 16 multiples that give 16*3+1=(49)

For 2nd one..

a-b and a/b to be even .. both a &b have to be even
Plugin values for a &b. a=8 and b=4

Therefore Option d & e.

What's the OA

Just to continue this solution method: plugging in a=8 and b=4 eliminates A, B and C, since they do not result in an odd integer for these numbers, and the answer we're looking for must be odd for any set of numbers that satisfies the question stem's requirements.
to decide between D and E, plug in a second set just for these two answers: b=6, a=12 will eliminate E (since 6+2=8 gives an even result when divided by 2), but not D (as 12+2/2 = 7, which is odd), leaving D as the single possible answer.

[sanju09]

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

for question 1.
from 0 to 31 you will have 10 numbers which have a remainder of 1 when devided by 3
from 31 till 50 you will have only 5 : 34,37,40,43,47
tus the answer is B

for question 2
a-b is even integer thus a -even and b-even or a-odd and b-odd ( of course a>b)
a/b -even integer thus a-even and b-even or a even, b-odd( in some cases such as 6,3,18,3 and a>b)
thus we conclude take we should take a-even and b-even
a/2 is even because even/even is even
b/2 is not in integer eliminate it
a+b/2 is not an integer , eliminate it
a+2/2 =a/2+1 thus odd
b+2/2 is not an integer eliminate it
the correct answer is D

for question 1.

How are there only 10 such integers from 0 through 31? How are there only 5 such integers after 31 through 50? How can 47 leave a remainder of 1 when divided by 3? Why not 49 is included in your hit list?

for question 2.

How is even/even always even? Why not b/2 and (a + b)/2 are integers? Why do you need to check E when D has already quenched your thirst?

[Veronica]

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

For 1st One B is right..

Numbers which leave 1 as reminder when divided by 3 are 1,4,7,10..
which can be written as 0*3+1,1*3+1,2*3+1,...

Therefore you have 16 multiples that give 16*3+1=(49)

For 2nd one..

a-b and a/b to be even .. both a &b have to be even
Plugin values for a &b. a=8 and b=4

Therefore Option d & e.

What's the OA

The answer keys are C and D

[Veronica]

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]

For the 1st question, can you explain why we could eyeball the value 49, why not 50??

[sanju09]

For the 1st question, can you explain why we could eyeball the value 49, why not 50??

The greatest multiple of 3 under 50 is 48, plus 1 is 49; 50 would leave a remainder of 2 if divided by 3, that's why not 50.