OG 10th Q48

[fr743]

If n is a positive integer, then n(n+1)(n+2) is

 

a) even only when n is even

b) even only when n is odd

c) odd whenever n is odd

d) divisible by 3 only when n is odd

e) divisible by 4 whenever n is even

 

Please, explain me how to solve such problems. Thanks.

[gmathintsdotcom]

n(n+1)(n+2) is a fancy way of saying the product of three consecutive integers

 

if n is odd, n+1 is even, and n+2 is odd

n(n+1)(n+2) = odd x even x odd = even x odd = even

 

if n is even, n+1 is odd, and n+2 is even

n(n+1)(n+2) = even x odd x even = even

 

(if you don't understand this, you should review number properties of odd's and evens)

 

from above, n is even regardless if n is even or odd; thus A and B are false we can eliminate them.

 

when n is odd, n(n+1)(n+2) is even so C is false.

 

Now we have choices D and E left.

 

choice D:

any three consecutive positive integers will have one of the numbers divisible by 3.

 

proof:

when a positive integer is divided by 3, it has remainders of 0 (i.e. divisible by 3), 1, or 2

if n has a remainder of 0, it is divisible by 3 and therefore n(n+1)(n+2) is divisible by 3

if n has a remainder of 1, n+1 has a remainder of 2, and n+2 has a remainder of 0. n+2 is divisible by 3 and therefore n(n+1)(n+2) is divisible by 3

if n has a remainder of 2, n+1 has a remainder of 0, and n(n+1)(n+2) is divisible by 3

 

Eliminate D.

 

We could pick by process of elimination but for the sake of completion:

 

If n is even, n can be expressed as 2k, for k an integer

if n = 2k

then n + 2 is also even and n + 2= 2m for m an integer

 

n(n+1)(n+2) = 2k(n)(2m) = 4knm

here knm is an integer, so 4knm and consequently n(n+1)(n+2) is divisible by 4.

 

enjoy!

 

p.s. for more OG alternate explanations, click here

[gmathintsdotcom]

another way you can do this problem is to try and find counterexamples:

 

n(n+1)(n+2)

 

a) even only when n is even

if n=3 3 x 4 x 5 = 60 n is odd and the product is even FALSE.

 

b) even only when n is odd

if n=2 2 x 3 x 4 is clearly even FALSE

 

c) odd whenever n is odd

from a, when n is odd the product is even FALSE

 

d) divisible by 3 only when n is odd

if n=6, 6 x 7 x 8 is divisible by 6 and so also must be divisible by 3 FALSE

 

e) divisible by 4 whenever n is even

Process of elimination, only choice left.

[fr743]

Thank you very much.

 

Gmathintsdotcom, i have some more question if you dont mind. I dont want to open a new thread. So I think i could post them right here. Could I use your further help? :)

 

OG 10th Q80

Jack is now 14 years older than Bill. If in 10 years Jack will be twice as old as Bill, how old will Jack be in 5 years?

 

a 9

b 19

c 21

d 23

e 33

 

Q92

 

Which of the following CANNOT be the greatest common divisor of two positive integers x and y?

 

a 1

b x

c y

d x-y

e x+y

[gmathintsdotcom]

sure.

 

OG 10th Q80

Jack is now 14 years older than Bill. If in 10 years Jack will be twice as old as Bill, how old will Jack be in 5 years?

 

a 9

b 19

c 21

d 23

e 33

 

j = jack's age now

b = bill's age now

j = 14 + b (i.e. jack is 14 years older than bill)

 

(j+10)=2(b+10)

j+10 = 2b + 20

j = 2b + 10

 

substitute

14 + b = 2b + 10

b = 4

 

bill is now 4

jack is now 18

 

in 5 years, jack will be 23. The answer is D.

 

 

 

Q92 (This is question 98 in the OG Quant Review 2nd Edition - Green Book)

 

Which of the following CANNOT be the greatest common divisor of two positive integers x and y?

 

a 1

b x

c y

d x-y

e x+y

 

a) Can 1 we be the greatest common divisor of two positive integers x and y?

Yes. Suppose x and y are prime. (e.g. the gcd of 2 and 3 is 1)

 

b) yes, suppose x = y or x = 2 and y = 4

 

c) yes, suppose x = y or y = 2 and x = 4

 

d) yes, suppose x = 4 and y = 2, the GCD is 2 and x-y = 2

 

All of these can be the GCD so we could stop here.

 

e) A number cannot be divisible by a larger number. Could 5 be divisible by 17. No. 5 cannot be divisible by any number larger than 5.

 

because x and y are positive, x + y > x and x + y > y

 

y / (x+y) is

 

x / (x+y) is

 

The answer is E.

[fr743]

gmathintsdotcom thank you!

 

I really need some tutor like you :)

 

Still have a lot of questions. If you dont mind AGAIN I would post my questions right here. :) Would be very happy and pleasant to learn something from you.

 

P.S. By the way, I also do need some advanced book for math theory. All books that I bought are full bull *****. Barrons, McGraw, even OG. All I got in that books are basics which dont help at all. Could you suggest me something really effective to read?

[fr743]

If x=a/3+b/3^2+c/3^3, where a, b, and c are each equal to 0 or 1, then x could be each of the following EXCEPT:

 

A. 1/27

B. 1/9

C. 4/27

D. 2/9

E. 4/9

[gmathintsdotcom]

sure.

 

If x=a/3+b/3^2+c/3^3, where a, b, and c are each equal to 0 or 1, then x could be each of the following EXCEPT:

 

A. 1/27

B. 1/9

C. 4/27

D. 2/9

E. 4/9

 

Rewrite the equation

x = a/3 + b/9 + c/27

 

if it helps, further we could rewrite it as

x = (1/3)a + (1/9)b + (1/27)c

x = (9/27)a + (3/27)b + (1/27)c

 

a, b, c can be either 0 or 1

 

starting with choice A:

1/27 possible? Yes, let c=1 and a=0, b=0

 

choice B

1/9 or 3/27 possible? Yes, let b=1 and a=0, c=0

 

choice C

4/27 possible? Yes, let b=1 and c=1 and a=0

 

 

choice D

2/9 or 6/27 possible? No possible way.

 

choice E

4/9 or 12/27 possible? Yes, let a=1 and b=1 and c=0

 

Choice D is not possible.

to get 6/27, clearly a cannot be 1.

if b=1 and c=0, we end up with 3/27

if b=1 and c=1, we end up with 4/27

if b=0 and c=1, we end up with 1/27

 

As far as books for math theory, I would say Manhattan GMAT books are pretty good and all you will need (for quant at least). I have written about my gmat experience here

 

While I have not used them personally, Veritas Prep and EZ Solutions books do get good reviews on Amazon so they may be worth checking out.

 

best.

[Abhishek009]

If n is a positive integer, then n(n+1)(n+2) is

 

a) even only when n is even

b) even only when n is odd

c) odd whenever n is odd

d) divisible by 3 only when n is odd

e) divisible by 4 whenever n is even

 

Please, explain me how to solve such problems. Thanks.

 

Plug in some values for n

 

Let n be 1 then n(n+1)(n+2) is = 1*2*3 =>6

 

Let n be 2 then n(n+1)(n+2) is = 2*3*4 =>24

 

Let n be 3 then n(n+1)(n+2) is = 3*4*5 => 60

 

Let n be 4 then n(n+1)(n+2) is = 4*5*6 => 120

 

Let n be 5 then n(n+1)(n+2) is = 5*6*7 => 210

 

Now go thru the options.

 

a) even only when n is even

 

We can find that it is not true , irrespective of the value of n ( be it odd or even) the product is always even

 

 

b) even only when n is odd

 

We can find that it is not true , irrespective of the value of n ( be it odd or even) the product is always even

 

c) odd whenever n is odd

 

False see the values of n = 1 , 3 , 5 ; U will find the value is always even

 

d) divisible by 3 only when n is odd

 

All the numbers are divisible by 3 , irrespective of the values of n ( Be it odd or even)

 

e) divisible by 4 whenever n is even

 

True , all of them are divisible by 4 except n =5 , so we can claim all even values of n are divisible by 5

[rnbp9]

gmathintsdotcom thank you!

 

I really need some tutor like you :)

 

Still have a lot of questions. If you dont mind AGAIN I would post my questions right here. :) Would be very happy and pleasant to learn something from you.

 

P.S. By the way, I also do need some advanced book for math theory. All books that I bought are full bull *****. Barrons, McGraw, even OG. All I got in that books are basics which dont help at all. Could you suggest me something really effective to read?

 

Kaplan is usually a very good source for tough questions. I also heard that their CD that comes with several books is pretty helpful!

[fr743]

Cool! Thanks :)

[fr743]

If a three-digit number is selected at random from the integers 100 ti 999, inclusive, what is the probability that the first digit and the last digit of the integer will both be exactly two less thand the middle digit?

 

A. 1:900

B. 7:900 correct answer

C. 9:1000

D.1:100

E.7:100

 

We got 131.242.353.464.575.686.797. 7 numbers matching the requirements. So, the answer must be B or E. But why 900 (B)?

[fr743]

Which of the following inequalities is equivalent to -4

 

A. |x-1|

B. |x+2|

C. |x+3|

D. |x-2|

E. None of the above

 

Here is (previous owner of my book wrote that near the question) also something concerning the question. This must be solution or explanation. But still couldn't understand.

[fr743]

p q r s

p 0 144 171 186

q 144 0 162 X

r 171 162 o Y

s 186 X Y 0

 

The table above shows the one-way driving distance, in miles, between four cities: P, Q, R and S. For example, the distance between P and Q is 144 miles. If the round trip between S and Q is 16 miles further than the round trip between S and R, and the round trip between S and R is 24 miles less than the round trip between S and P, what is the value of X?

 

A. 174

B. 182

C. 186

D. 348

E. 364

[fr743]

In a sample of associtates at a law firm, 30 percent are second-year associates, and 60 percent are not first-year associates. What percentage of the associates at the law firm have been there for more than two years?

 

A. 10

B. 20

C. 30

D. 40

E. 50

[dharan]

I didn't get your explanation for e). How all of them are divisible by 4 except when n =5, what about when n = 1.

divisible by 4 whenever n is odd also possible like when n = 3.

[dharan]

Rewrite the eqn 27 x = 9a + 2B + c

 

sub abc = 000,001,010,011,100,101,110,111,

First three subs we can eliminate A, B and C

We can also eliminate E and only remaining is D.

[dharan]

ewrite the eqn 27 x = 9a + 2B + c

 

sub abc = 000,001,010,011,100,101,110,111,

First three subs we can eliminate A, B and C

We can also eliminate E and only remaining is D.

[800]

Which of the following inequalities is equivalent to -4

 

A. |x-1|

B. |x+2|

C. |x+3|

D. |x-2|

E. None of the above

 

Here is (previous owner of my book wrote that near the question) also something concerning the question. This must be solution or explanation. But still couldn't understand.

 

For answering an inequality problem first get the FEELING of question. Yes feeling.

In this problem the question stem says the range of x . the range is - 4 to 8. Now observe the options. A and C will not give us the range as x is without co-efficient. We should not weste valuable time for calculation of A & C. NOW, we have to do the solid calculation for Option B and option D. We have to consider BOTH negative and positive value for each option.

For option B. if l x+2 l is positive, x + 2

For option D. if l x -2 l is positive, x - 2 -4. Have we got another end value of the range. YES. BINGO !!!

 

It is D.

[800]

If a three-digit number is selected at random from the integers 100 ti 999, inclusive, what is the probability that the first digit and the last digit of the integer will both be exactly two less thand the middle digit?

 

A. 1:900

B. 7:900 correct answer

C. 9:1000

D.1:100

E.7:100

 

We got 131.242.353.464.575.686.797. 7 numbers matching the requirements. So, the answer must be B or E. But why 900 (B)?

 

Because here we have 900 three digit number (s). HTH