ARCO GRE maths books
ARCO GRE maths books
The book includes 6 fulllength tests.
Most of the test dont have many abtract problems like EST sample tests and I think it is good to get used to with calculation
I'm working on it so if you are interested in, let's discuss
Most of the test dont have many abtract problems like EST sample tests and I think it is good to get used to with calculation
I'm working on it so if you are interested in, let's discuss
Last edited by Nameless on Fri Sep 05, 2008 8:01 pm, edited 1 time in total.
Okie, let warm up with the problem #49  Sample test 3
if n is a positive integer then which of following will be always a integer :
1. [ sqrt(2)+1]^2n + [sqrt(2)1] ^2n
2. [ sqrt(2)+1]^2n  [sqrt(2)1] ^2n
3. [ sqrt(2)+1]^(2n+1) + [sqrt(2)1] ^(2n+1)
4. [ sqrt(2)+1]^(2n+1)  [sqrt(2)1] ^(2n+1)
a. 1 & 2
b. 1 & 3
c. 1 & 4
d. 2 & 3
e . 3 & 4
if n is a positive integer then which of following will be always a integer :
1. [ sqrt(2)+1]^2n + [sqrt(2)1] ^2n
2. [ sqrt(2)+1]^2n  [sqrt(2)1] ^2n
3. [ sqrt(2)+1]^(2n+1) + [sqrt(2)1] ^(2n+1)
4. [ sqrt(2)+1]^(2n+1)  [sqrt(2)1] ^(2n+1)
a. 1 & 2
b. 1 & 3
c. 1 & 4
d. 2 & 3
e . 3 & 4
No body is interested in ?
Okie, here the solution of the above problem :
Answer is 1 & 3
We can you the identity (a+b)^n= a^n+.....+b^n but this method takes a lot of time so let try by another way :
Since the problem is true for all n, so let pick n=1
plug n=1 into 1, 2, 3, 4 the we can easily see that :
1 & 3 are always integer so Answer is 1 & 3
Okie, let's continue with broblem 26  Sample test 1
if n is a multiple of 4 then the sum :
S=1+2i+3i^2+....+(n+1)i^n , where i=sqrt(1), equals :
a. 1 i
b. 1/2(2n+1)
c. 1/2 (n+2ni)
d. 1/2(n+2+i)
e. 1/2(n^2+84ni)
Okie, here the solution of the above problem :
Answer is 1 & 3
We can you the identity (a+b)^n= a^n+.....+b^n but this method takes a lot of time so let try by another way :
Since the problem is true for all n, so let pick n=1
plug n=1 into 1, 2, 3, 4 the we can easily see that :
1 & 3 are always integer so Answer is 1 & 3
Okie, let's continue with broblem 26  Sample test 1
if n is a multiple of 4 then the sum :
S=1+2i+3i^2+....+(n+1)i^n , where i=sqrt(1), equals :
a. 1 i
b. 1/2(2n+1)
c. 1/2 (n+2ni)
d. 1/2(n+2+i)
e. 1/2(n^2+84ni)
Ok, i'll try..
the exponent of complex number has 4 periodic (1,1,i,i), eg i = i^5 = i^9, etc
so, if we change the series into S = 1 + (2i + 3i^2 + 4i^3 + ...)
and S' = 2i + 3i^2 + 4i^3 + 5i^4 + ...
so that S' can be divided by 4 series, into :
1. 5 + 9 + ...
2. 2i + 6i + 10i + ...
3. 3  7  ...
4. 4i  8i  ...
the sum for each one :
1. n/2(6+4n)
2. n/2(4in)
3. n/2(24n)
4. n/2(4i4in)
when we add all the sum, we have : n(22i)
since n in this is n for each series, then S' = (n(22i))/4
1 + 2i + 3i^2 + .. = 1 + (2i + 3i^2 + ..) = 1 + (n(22i))/4 = 1/2 (n+2ni)
(C)..
i hope it's correct
i dont know if there's any fastest solution
the exponent of complex number has 4 periodic (1,1,i,i), eg i = i^5 = i^9, etc
so, if we change the series into S = 1 + (2i + 3i^2 + 4i^3 + ...)
and S' = 2i + 3i^2 + 4i^3 + 5i^4 + ...
so that S' can be divided by 4 series, into :
1. 5 + 9 + ...
2. 2i + 6i + 10i + ...
3. 3  7  ...
4. 4i  8i  ...
the sum for each one :
1. n/2(6+4n)
2. n/2(4in)
3. n/2(24n)
4. n/2(4i4in)
when we add all the sum, we have : n(22i)
since n in this is n for each series, then S' = (n(22i))/4
1 + 2i + 3i^2 + .. = 1 + (2i + 3i^2 + ..) = 1 + (n(22i))/4 = 1/2 (n+2ni)
(C)..
i hope it's correct
i dont know if there's any fastest solution
Hi lovemath,
Yes, I'm going to take the test on November.
your answer is correct. I do have a question about group theory :
Let G be the group of symmetries of the regular pentagram ( the picture is :http://introvert.net/images/2005/02/basicpentagram), then G is isomorphic to :
1. S_5
2. A_5
3. a cylic group of order 5
4. a cylic group of order 10
5. dihedral group of order 10
I am going to show this question :
G is generated by the rotation 2*Pi/5 via its center and the reflexion though x axis so by the definition G is isomorphic to dihedral group pf order 10 .
The answer is 5)
Let try another problem :
the set of real numbers x for which the serie Sum(n=1 to infinitive ) (n! x^2n)/[n^n*(1+x^2n)] converges is :
a. {0}
b. (1,1)
c. [1,1]
d. [sqrt(e), sqrt(e)]
e. R
I don't know to insert the image so please show how to do it. Thanks
Yes, I'm going to take the test on November.
your answer is correct. I do have a question about group theory :
Let G be the group of symmetries of the regular pentagram ( the picture is :http://introvert.net/images/2005/02/basicpentagram), then G is isomorphic to :
1. S_5
2. A_5
3. a cylic group of order 5
4. a cylic group of order 10
5. dihedral group of order 10
I am going to show this question :
G is generated by the rotation 2*Pi/5 via its center and the reflexion though x axis so by the definition G is isomorphic to dihedral group pf order 10 .
The answer is 5)
Let try another problem :
the set of real numbers x for which the serie Sum(n=1 to infinitive ) (n! x^2n)/[n^n*(1+x^2n)] converges is :
a. {0}
b. (1,1)
c. [1,1]
d. [sqrt(e), sqrt(e)]
e. R
I don't know to insert the image so please show how to do it. Thanks
Thanks lovemath, your explanation is great.
Now Let talk about number theory problems . I am getting stuck with the following problems :
How many positive integers k does the ordinary decimal representation of the integer k! end in exactly 99 zeros?
the answer is 4 but I don't know why? if some one, please explain. Thanks in advance
Now Let talk about number theory problems . I am getting stuck with the following problems :
How many positive integers k does the ordinary decimal representation of the integer k! end in exactly 99 zeros?
the answer is 4 but I don't know why? if some one, please explain. Thanks in advance
hi, nameless..
I think the answer is five. let me try to guess..
* 99 zeros will happen exactly when there are 99 10's ==> 99 5's and 99 2's (we can ignore 10, coz k! is a multiple of primes)
* since number 5 is more than 2, then when 5's reach 99, the 2's will be more than 99 .. eg : 5! = 1.2.3.2.2.5 ==> there are 3 2's when 5's is 1.
* so we can focus only on the 5's.
* For example x! is the number when the first 99 5's happens. then before the 5's reach 100, there can only be number (x+1)!, (x+2)!, (x+3)!, (x+4)!.. that (x+5)! will make 5's reach 100..
* so, there should be only x,x+1,x+2,x+3,x+4 numbers when 5's reach 99
I think the answer is five. let me try to guess..
* 99 zeros will happen exactly when there are 99 10's ==> 99 5's and 99 2's (we can ignore 10, coz k! is a multiple of primes)
* since number 5 is more than 2, then when 5's reach 99, the 2's will be more than 99 .. eg : 5! = 1.2.3.2.2.5 ==> there are 3 2's when 5's is 1.
* so we can focus only on the 5's.
* For example x! is the number when the first 99 5's happens. then before the 5's reach 100, there can only be number (x+1)!, (x+2)!, (x+3)!, (x+4)!.. that (x+5)! will make 5's reach 100..
* so, there should be only x,x+1,x+2,x+3,x+4 numbers when 5's reach 99
Sorry the answer is 5  there is five numbers satisfying the requirement .* 99 zeros will happen exactly when there are 99 10's ==> 99 5's and 99 2's (we can ignore 10, coz k! is a multiple of primes)
* since number 5 is more than 2, then when 5's reach 99, the 2's will be more than 99 .. eg : 5! = 1.2.3.2.2.5 ==> there are 3 2's when 5's is 1.
* so we can focus only on the 5's.
* For example x! is the number when the first 99 5's happens. then before the 5's reach 100, there can only be number (x+1)!, (x+2)!, (x+3)!, (x+4)!.. that (x+5)! will make 5's reach 100..
* so, there should be only x,x+1,x+2,x+3,x+4 numbers when 5's reach 99
BUT I am confused because of your notation : what 10's, 5's , 2's mean?
B
In Zeitz, Art and Craft of Problem Solving
m=[n(n+3)][(n+1)(n+2)]+1
=(n^2+3n)(n^2+3n+2)+1
=(n^2+3n+2)^2  2(n^2+3n+2) +1
=(n^2+3n+21)^2
The combination on the right side of the first line is to get as much the same as possible in both resulting terms (n^2+3n)
By the way, I started browsing through a Problem solving book like this to get ideas for quick tricks to use on the exam. I don't know if it is worth the time involved, but it's fun too. I like the Zeitz book because it is not as heavyduty as some of the Putnam prep books, and I am looking for quick tricks, not something I would have to think about for an hour.
In Zeitz, Art and Craft of Problem Solving
m=[n(n+3)][(n+1)(n+2)]+1
=(n^2+3n)(n^2+3n+2)+1
=(n^2+3n+2)^2  2(n^2+3n+2) +1
=(n^2+3n+21)^2
The combination on the right side of the first line is to get as much the same as possible in both resulting terms (n^2+3n)
By the way, I started browsing through a Problem solving book like this to get ideas for quick tricks to use on the exam. I don't know if it is worth the time involved, but it's fun too. I like the Zeitz book because it is not as heavyduty as some of the Putnam prep books, and I am looking for quick tricks, not something I would have to think about for an hour.
yes, it takes alot of time to prove mathematically.
The best method is to pick n=1 then we have m= 25= 5^2
n=2 we have m=121=11^2
so we can get the answer must be b)
We can use the trick to solve the following complicated problem :
what the greatest integer that divides p^41 for every prime number greater than 5 ?
1. 12
2. 30
3. 48
4. 120
5. 240
It took me about an hour to solve this problem BUT if we choose p=7>5 then 7^41=24011=2400 so the answer must be 5) 240 > easy and save a lot of time
The best method is to pick n=1 then we have m= 25= 5^2
n=2 we have m=121=11^2
so we can get the answer must be b)
We can use the trick to solve the following complicated problem :
what the greatest integer that divides p^41 for every prime number greater than 5 ?
1. 12
2. 30
3. 48
4. 120
5. 240
It took me about an hour to solve this problem BUT if we choose p=7>5 then 7^41=24011=2400 so the answer must be 5) 240 > easy and save a lot of time
Continue discussing number theory :
Problems 14  GR0568
Find the units digit of 7^25 ?
this one is easy one, since we know 7^2=49
so 7^2== (1) (mod 10)
>7^4== 1 (mod 10)
sine 25=4.6 +1 so
7^25 ==(7^4).7==1. 7==7 (mod 10)
so the units digit is 7
Question : Find the the tenth digit of 7^25 ? Can we solve this problems?
( Ex : 720 the tenth digit is 2 )
Problems 14  GR0568
Find the units digit of 7^25 ?
this one is easy one, since we know 7^2=49
so 7^2== (1) (mod 10)
>7^4== 1 (mod 10)
sine 25=4.6 +1 so
7^25 ==(7^4).7==1. 7==7 (mod 10)
so the units digit is 7
Question : Find the the tenth digit of 7^25 ? Can we solve this problems?
( Ex : 720 the tenth digit is 2 )
yes, it takes alot of time to prove mathematically.
The best method is to pick n=1 then we have m= 25= 5^2
n=2 we have m=121=11^2
so we can get the answer must be b)
We can use the trick to solve the following complicated problem :
what the greatest integer that divides p^41 for every prime number greater than 5 ?
1. 12
2. 30
3. 48
4. 120
5. 240
It took me about an hour to solve this problem BUT if we choose p=7>5 then 7^41=24011=2400 so the answer must be 5) 240 > easy and save a lot of time yes, it takes alot of time to prove mathematically.
The best method is to pick n=1 then we have m= 25= 5^2
n=2 we have m=121=11^2
so we can get the answer must be b)
We can use the trick to solve the following complicated problem :
what the greatest integer that divides p^41 for every prime number greater than 5 ?
1. 12
2. 30
3. 48
4. 120
5. 240
It took me about an hour to solve this problem BUT if we choose p=7>5 then 7^41=24011=2400 so the answer must be 5) 240 > easy and save a lot of time
It is not clear to me that your explaination is coherent. Just because 240 is the greatest integer and it works for p =7 does not imply it would work for p = 11.
The best method is to pick n=1 then we have m= 25= 5^2
n=2 we have m=121=11^2
so we can get the answer must be b)
We can use the trick to solve the following complicated problem :
what the greatest integer that divides p^41 for every prime number greater than 5 ?
1. 12
2. 30
3. 48
4. 120
5. 240
It took me about an hour to solve this problem BUT if we choose p=7>5 then 7^41=24011=2400 so the answer must be 5) 240 > easy and save a lot of time yes, it takes alot of time to prove mathematically.
The best method is to pick n=1 then we have m= 25= 5^2
n=2 we have m=121=11^2
so we can get the answer must be b)
We can use the trick to solve the following complicated problem :
what the greatest integer that divides p^41 for every prime number greater than 5 ?
1. 12
2. 30
3. 48
4. 120
5. 240
It took me about an hour to solve this problem BUT if we choose p=7>5 then 7^41=24011=2400 so the answer must be 5) 240 > easy and save a lot of time
It is not clear to me that your explaination is coherent. Just because 240 is the greatest integer and it works for p =7 does not imply it would work for p = 11.
For this problem I used Fermat Little Theorem:
gcd(p,5)=1 so 5 divides p^41  eliminate a & c
p^41=(p^21)(p^2+1), gcd(p,3)=1, so 3 divides p^21
Then the question is how many powers of 2 divide p^41:
Here we should remember 8 divides p^21 (because 4 divides either p1 or p+1), and at least 2 divides p^2+1, giving us at least 16:
but 3*5*16=240 the largest answer given, so it is the answer.
gcd(p,5)=1 so 5 divides p^41  eliminate a & c
p^41=(p^21)(p^2+1), gcd(p,3)=1, so 3 divides p^21
Then the question is how many powers of 2 divide p^41:
Here we should remember 8 divides p^21 (because 4 divides either p1 or p+1), and at least 2 divides p^2+1, giving us at least 16:
but 3*5*16=240 the largest answer given, so it is the answer.
By that logic, if 2400 was one of the answer choices, you would choose that?It took me about an hour to solve this problem BUT if we choose p=7>5 then 7^41=24011=2400 so the answer must be 5) 240 > easy and save a lot of time
In the prior problem, try the number 1 was an easy computation and it eliminated all other choices. In this problem, try the number 7 is more difficult computation and didn't eliminate any other choices.
I have a question on first problem.
I think the answer is 1&4.
as you said when n=1, 1&4 are integer.
as you said when n=1, 1&4 are integer.
Last edited by ieoi on Fri Oct 31, 2008 10:07 pm, edited 1 time in total.
Thanks for posting these problems. I am curious how you got your answer for the convergence problem you posted:
the set of real numbers x for which the series Sum(n=1 to infinitive ) (n! x^2n)/[n^n*(1+x^2n)] converges is :
a. {0}
b. (1,1)
c. [1,1]
d. [sqrt(e), sqrt(e)]
e. R
when I did this problem, I used the ratio test:
lim (a_(n+1)/a_n)=x^2(1+x^2n)/(1+x^(2n+2))<1
So x^2(1+x^2n)<1+x^(2n+2)
x^2<1
which only holds for x in (1,1)..and checking the endpoints I find that they also converge to get c) as my answer.
I suspect I went wrong somewhere, I'm just not sure where.
the set of real numbers x for which the series Sum(n=1 to infinitive ) (n! x^2n)/[n^n*(1+x^2n)] converges is :
a. {0}
b. (1,1)
c. [1,1]
d. [sqrt(e), sqrt(e)]
e. R
when I did this problem, I used the ratio test:
lim (a_(n+1)/a_n)=x^2(1+x^2n)/(1+x^(2n+2))<1
So x^2(1+x^2n)<1+x^(2n+2)
x^2<1
which only holds for x in (1,1)..and checking the endpoints I find that they also converge to get c) as my answer.
I suspect I went wrong somewhere, I'm just not sure where.
 Charles.Rambo
 Posts: 9
 Joined: Wed May 22, 2013 7:33 pm
Re: ARCO GRE maths books
Hey guys, I solved all the problems on the GRE math form 68 practice test. You can view my solutions at http://rambotutoring.com/GREmathsubje ... utions.pdf.

 Posts: 2
 Joined: Wed Jul 31, 2013 1:49 pm
Re: ARCO GRE maths books
Does anyone have all of solutions for the GRE Subject Math exams?
Re: ARCO GRE maths books
To me is 1 and 4, yes
In this link there are the solutions for 2 exams.
sfmathgre.blogspot.com/2010/05/prob39limits.html
A lot of problems have been posted here. It's a matter of being patience and read the titles of the 26 pages. And there is a complete solution from Charles Rambo, whose lonk is a few messages above.
In this link there are the solutions for 2 exams.
sfmathgre.blogspot.com/2010/05/prob39limits.html
A lot of problems have been posted here. It's a matter of being patience and read the titles of the 26 pages. And there is a complete solution from Charles Rambo, whose lonk is a few messages above.