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The book includes 6 full-length 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

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

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+2-ni)
d. 1/2(n+2+i)
e. 1/2(n^2+8-4ni)

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(-2-4n)
4. n/2(-4i-4in)

when we add all the sum, we have : n(2-2i)
since n in this is n for each series, then S' = (n(2-2i))/4

1 + 2i + 3i^2 + .. = 1 + (2i + 3i^2 + ..) = 1 + (n(2-2i))/4 = 1/2 (n+2-ni)

(C)..

i hope it's correct

i dont know if there's any fastest solution

Are there any? i'd love to get another question..

And, are u gonna take the november test?
I am planning to do that (just hoping i'll finish reading all the material first (not a math major, and there are a loooot of readings.. ))

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/basic-pentagram), 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

Oh, I was so silly:(
this problem is easy one , just use the Weierstrass's condition then we can say that the serie converges for all x so the answer is 5) R

about the pentagram: it's 5 rotational symmetry and 5 reflection symmetry, so it belongs to D10 symmetry group..




to post images, u use the button Img above, but in this special case, the url doesn't end with image extension, so i thought i add png at the end to see if it works, and it does..

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

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

Sorry the answer is 5 - there is five numbers satisfying the requirement .
BUT I am confused because of your notation : what 10's, 5's , 2's mean?

sorry, i mean, in 5! = 1*2*3*2*2*5 , there are 1 5's means number 5 only occurs 1 time, 99 5's means number 5 occurs 99 times.. etc..
or it can be said : there are 1 factor of number 5, 99 factors of 5 in k!, etc..

Thanks lovemath

Continue discussing number theory :

Problem : let m= n(n+1)(n+2)(n+3)+1 then m is :
a. divisible by 5
b. a perfect square
c. divisible by 4
d. a prime number
e. a perfect number

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+2-1)^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 heavy-duty 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^4-1 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 7^4-1=2401-1=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 )

ok.. i'll try

7^4 = 2401 = 1 mod 100
so 7^25 = 7*7^(4*6) = 7*(1)^6 mod 100
= 7 mod 100..

so, the answer is 0..

Lovemath,

awesome

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^4-1 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 7^4-1=2401-1=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^4-1 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 7^4-1=2401-1=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^4-1 - eliminate a & c
p^4-1=(p^2-1)(p^2+1), gcd(p,3)=1, so 3 divides p^2-1

Then the question is how many powers of 2 divide p^4-1:

Here we should remember 8 divides p^2-1 (because 4 divides either p-1 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.

Since p^4-1 works for all prime number p so it must work when p=7, p=11,...etc . You can check it by yourself

By that logic, if 2400 was one of the answer choices, you would choose that?

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 think the answer is 1&4.
as you said when n=1, 1&4 are integer.

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.