Q1).

a(a+!)=b(b+1)(b+2)(b+3)

 as this =n will be equal when both sides will  be 0

when a=0 ,b can be 0,-1,-2,-3

when a=-1,b can be 0,-1,-2,-3

so total of 8 pairs

Q2)only 3n and 3n+2(3n-1) will be divisble by 3

3n===3,6........2007 ---------> 669

3n-1==2,5......2006 ---------> 669

total=669+669=1338

According to me answers are
1)a
2)b
3)b
4)c
5)a
q1)1. Find all ordered pair of integers (a,b) such that a(a + 1) = b(b + 1)(b + 2)(b + 3)
Ans1)It is given that a,b are any integers
So this condition only fulfill when any out of b,b+1,b+2,b+3 are 0.So there are 4 conditions when b(b+1)(b+2)(b+3)=0
Similarly for a,a+1 there are 2 conditions for which a(a+1)=0
Hence a)4*2=8(answer)

q2)Let a1 = 1 and an be the number formed by putting the digits of n at the end of an−1. For k = 1 to 2008, compute the number of ak that are divisible by 3.
Ans2)We can do this question using your intellegence
we know that 1,12,123 out of these three conditions two numbers are divisible by 3(because there sum is divisible by 3) and therefore for every three number 2 are divisible by 3 and also keep one thing in mind that first number after each 3rd number is not divisible by 3
now divide 2008/3=669 and remainder=1
so total number must be=669*2
Hence b)1338
q3) If a + b + c = 3, a2 + b2 + c2 = 9, a3 + b3 + c3 = 24 , Then find a4 + b4 + c4
Ans3)Now it looks difficult so go by option but before that find out some simple conditions
(a2+b2+c2)2=a4+b4+c4+2(a2b2+b2c2+a2c2)
81-2(a2b2+b2c2+a2c2)=a4+b4+c4
We know that 2*any number must be even
So 81-2*even number=odd number
Hence we leave with three answers 27,69,none of these
most of the chances that none of these is not the answer
So we leave with 27,69 and also it is not possible difference between a4+b4+c4 and a3+b3+c3 is only 3
So correct answer is b)69
q4)The positive integers a and b are such that the numbers 15a+16b and 16a− 15b are both squares of positive integers. Find the least possible value that can be taken by the minimum of these two squares.
Ans4)15a+16b=x2.........1
16a-15b=y2.........2
and we have to find out minimum value of y2(16a-15b is less than that of 15a+16b because a and b are positive integer)
Now multiply equation 1 by 15 and 2 by 16 and add them
then 225a+256a=15x2+16y2.........3
now multiply equation 1 by 16 and 2 by 16 and subtract them
then 225b+256b=16x2+15y2..........4
now multiply 3 by 16 and 4 by 15 and then subtract them
then 256y2-225y2=(225+256)(16a-15b)
31y2=481(16a-15b)
y2=481(16a-15b)/31
we know 481 is not divisible by 31
so 16a-15b=31*481 for minimum value of y2
Hence c)y=481
q5)N is a natural number . How many values of N exist, such that N2 + 24N + 21 has exactly three factors?
Ans5)We know that every number other than 1 atleast 2 factor i.e(1,itself number)
exactly three factors is possible for square of prime number i.e prime number2
Hence N2+24N+21=prime number2
(N+21)(N+1)+2N=prime number2
We know than N is natural number if prime number2 less number 21 there is no value for which we get natural number and also if prime number2 greater than 21 it is not possible to get any natural value
Hence a)0

Q5: We are to find the value of N such that X=N²+24N+21 has exactly 3 factors. This means X should be perfect sq of prime nos only because nos other than prime nos can be resolved into more factors.

We know that Perfect squares end with digits 0,1,4,5,6,9 out of which squares of prime factors end with only 1,5,9.

Put any no ending in 1     Last digit=6

                                 2                      3

                                 3                      2...and so on. we will find that only nos ending in 6 and 0 would give digit ending in 1...so all other options gets eliminated.

 Put N=10 X= 361= 19².

Likewise check for N=16, 20, 26,30...etc

I havent checked for larger nos...but at least 1 sol for N=10 is possible

ans = option b.

@ admin...kindly post the answers at least.

:)

Q2. a₁=1 not-divisible, a₂=12 divisible, a₃= 123 divisible, a₄= 1234 not-divisible, a₅ =12345 divisible, a₆= 123456 divisble, a₇ = 1234567 not-divisible, a₈ = 12345678 divisible, a₉= 123456789 divisble, a₁₀ = 12345678910 not-divisible.

 so similarly we can see out of every  3 a_k 2 are divisible.

 so among 2007 we have 2/3x(2007) = 1338 divisible by 3

and a₂₀₀₈ being the 1st one after divisble is again not divisble.

so overall 1338 a_k is divisble....

Q3 is pretty simple no much logic is required

(a+b+c)(a+b+c)= a² + b² + c²+2(ab+bc+ca)

so we get ab+bc+ca=0 from the values given

and by using the very famous eq.

(a+b+c)(a² + b² + c²-ab-bc-ca)=a³ + b³ + c³+3abc

we get abc = -1

now just    square of (a² + b² + c²) i.e., =  a⁴ + b⁴ + c⁴ + 2z

                square of  (ab+bc+ca)     = z+ 2abc(a+b+c)     

from this u can find z and from the first eq u can find the ans

solution of question one ...

a(a+1)=b(b+1)(b+2)(b+3)...

write this in qaudratic equation form...

determinant would be

D=1-4b(b+1)(b+2)(b+3) ...

for (a,b) to be interger i.e. real....

D>=0...for 0< b<-3 it is not possible....

so b=0,-1,-2 and -3....

corrosponding to each this values of b we will get two values of a ....hence 2x4=8 answer...

====================================================

solution of 5^th quetion..

if N²+24N+21(=p say)...has three factor ..then it can be factorize as it has atleast one factor other than 1 and itself.

now if it can be factorize... N²+24N+21=0..must have real solution..asN is natural number ...p=0 must have positive integer solution...

determinant of this equaton....D=24²-4x21=123...this not perfect square so eqution dosenot have any integer solution....

 so answer 0...