IOQM mock test 2 - Pending(11,15,16,18,20,22,24,28,29)

 Q1(geometry): AB is a chord in a circle with center O and radius 52 cm. The point M divides the chord AB such that AM = 63 cm and MB = 33 cm. Find the length OM in cm.
Solution:

Let's say OD is the perpendicular on AB.

OD^2 = 52^2 - 48^2 = 400
OM^2 = 15^2 + OD^2 = 625
=> OM = 25

Q2(combinatorics,set-theory): 

If the number of triplets of set {A, B, C} such that

(a) A, B, C ∈ {1, 2, 3, 4, ..., 8}
(b) |A ∩ B| = |B ∩ C| = |C ∩ A| = 2
(c) |A| = |B| = |C| = 4

is λ, then find the sum of digits of λ. (where |S| denotes the number of elements in set S)

Solution:
A can be constructed in 8C4 = 70 ways.
B has to pick 2 numbers from A and 2 from A'(A's complement): 4C2*4C2
Let's say A is {1,2,3,4} B is {3,4,5,6} now.
C has to pick 4 digits in 3 different ways:
1. It picks 3,4 from A (the same digits which B picked). Now it has to pick 7,8 from A' otherwise (b) given in question won't hold. So C = {3,4,7,8}. 
2. It picks 1,2 from A (the digits which B didn't pick). Now it has to pick 5,6 from B. C = {1,2,5,6}
3. It picks one digit from A ∩ B  = {3,4} and one from {1,2}. There are 4 such pairs. Then it picks one digit from {5,6} and one from {7,8}. Again 4 ways. So total 16 ways.
Total ways to make C = 18

Finally: 70*36*18 = 45360.
Sum of digits = 18

Q3(modulo arithmetic). The number 1003009027081243 has one or more prime factors greater than 10^5 let x be the sum of all such primes. What is the sum of digits of x.
Solution:
1,003,009,027,081,243 
= 1.10^15 + 3.10^12 + 3^2.10^9 + 3^3.10^6 + 3^4.10^3 + 3^5.10^0
It's a G.P. with n = 6, a = 1.10^15, r = 3/1000
Sum  = a.(1-r^n)/(1-r) = 1000^6 - 3^6/1000-3
Denominator is 997.
Numerator is x^6 - y^6 = (x^3 + y^3)(x^3 - y^3) = (x+y)(x^2 - xy + y^2)(x-y)(x^2 +xy + y^2)
 = 1003.997.(1000,000 - 3,000 + 9)(1000,000 + 3,000 + 9)
= 1003.997.997,009.1003,009
So after removing 997 it is:
= 1003.997,009.1003,009
1003,009 = 7*143287
997,009 = 13*76693
So only factor more than 100,000 is 143,287.
Answer: 25 

Q4. (combinatorics) There are five cities A, B, C, D, E on a certain island. Each city is connected to every other city by road. In how many ways can a person starting from city A come back to A after visiting some cities without visiting a city more than once and without taking the same road more than once? (The order in which he visits the cities also matters: e.g., the routes A→B→C→A and A→C→B→ A are different).
Solution:
Routes like A->B->A are not allowed as the road between A,B will be traveled twice.
So we have to choose at least 2 cities.
Routes like A-B-C-A => Choose second city in 4 and third city in 3 ways => 12
Routes like A-B-C-D-A => 4*3*2 = 24
Routes like A-B-C-D-E-A => 4*3*2*1 = 24
Total: 24+24+12 = 60

Q5. (algebra)

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Let $a$ and $b$ be real numbers such that

$$\frac{a}{a^2 - 5} = \frac{b}{5 - b^2} = \frac{ab}{a^2b^2 - 5}$$

where $a + b \ne 0$. Find

$$\frac{a^4 + b^4}{313} = ?$$

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Solution:
Use first 2 fractions to get ab = 5.
Then use first and last fractions to get a = 5,-1 and similarly b = 1,-5.
Since ab = 5 so only possible solutions for a,b are 5,1 and -1,-5.
=> a^4 + b^4 = 626
Answer: 2

Q6. Let x be the product of all possible values of abc; if abc +1 = ab + bc + ca where a, b, c, are positive integers. Find sum of digits of x. (Diophantine)
Solution:

w.l.o.g.    a ≥ b ≥ c > 0 
let b = B + 1 ,    c = C + 1 ,    a = A + 1 
⇒   A = (B + C + 1) / (BC - 1)
Now the denominator here grows faster than numerator.
So after few values the solution won't be feasible.
BC - 1 >= 1 so let's try B=2,C=1 => A = 4
Next B=2,C=2 => A = 5/3
B=3,C=2 => A = 6/5
B=3,C=3 => A = 7/8 Now nothing left to try.
So only combination we found is A,B,C = 4,2,1
⇒   a, b, c = 2, 3, 5 
abc = 30 = x

Q7(algebra). 

Let P(x) be the unique polynomial of minimal degree with the following properties:

• P(x) has leading coefficient 1,

• 1 is a root of P(x) − 1,

• 2 is a root of P(x − 2),

• 3 is a root of P(3x),

• 4 is a root of 4P(x)

The roots of P(x) are integers, with one exception. The root that is not an integer can be written in the form of $\frac{m}{n}$, where m and n are relatively prime positive integers.
What is m + n?

Correct Answer
47

Solution
The first condition states that when x = 1, P(x) − 1 = 0 so we get P(1) − 1 = 0 so P(1) = 1.
Next, when x = 2, P(x − 2) = 0, so P(0) = 0.
For the 3rd condition x = 3, P(3x) = 0, so P(9) = 0.
When x = 4, 4P(x) = 0, 4P(4) = 0, P(4) = 0.
So we have x = 0, x = 9, x = 4 as the roots of P(x). These are all integer roots, and since we are minimizing the degree, we only need 1 more root of $x = \frac{m}{n}$. So we have:

$$P(x) = (x−0)(x−9)(x−4)\left(x−\frac{m}{n}\right)$$

We know we can use the first condition
P(1) = 1, to solve for $x−\frac{m}{n}$:

$$P(1) = 1 = (1)(1 − 9)(1 − 4)\left(1 − \frac{m}{n}\right) = (−8)(−3)\left(1 − \frac{m}{n}\right)$$ $$24\left(1 − \frac{m}{n}\right) = 1$$ $$1 − \frac{m}{n} = \frac{1}{24}$$ $$− \frac{m}{n} = \frac{1}{24} − 1 = −\frac{23}{24}$$ $$\frac{m}{n} = \frac{23}{24}$$

m and n are relative primes so m = 23, n = 24.
m + n = 23 + 24 = 47

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Q8(Combinatorics) A student council must select a two-person welcoming committee and a three-person planning committee from among its members. There are exactly 10 ways to select a two-person team for the welcoming committee. It is possible for students to serve on both committees. In how many different ways can a three-person planning committee be selected?

Solution:
nC2 = 10 => n = 5 => nC3 = 10

Q9 (Geometry) 

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Given that $ABC$ is an equilateral triangle of side 1, $\triangle BDC$ is isosceles with $DB = DC$ and $\angle BDC = 120^\circ$. If points M and N are on AB and AC respectively such that $\angle MDN = 60^\circ$, find the perimeter of $\triangle AMN$.

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Solution:
There are 2 cases possible here.
D is inside the triangle or outside.
Case 1: Let's consider the inside case.
DB = DC and triangle BDC is isosceles => Angles DCB = DBC = 30. => D is the incenter as DB and DC are angle bisectors.
In an equilateral triangle incenter and centroid are same.
Now if I extend BD then it meets AC at P => DP = 1/3(Median) = 1/3(sqrt(3)/2) = 1/(2.sqrt(3))
Similarly if I extend CD then it meets AB at Q => DQ = 1/3(Median) = 1/3(sqrt(3)/2) = 1/(2.sqrt(3))

Angle MDN is 60.
I can choose M and N in such a way that AM = AN so that everything is symmetrical.
(It can be shown that even if M and N are chosen in a different way, the result would be same).
Now, due to symmetry angle NDP  = angle MDQ = 30
because angle PDQ = 120 and PDQ = MDN + NDP + MDQ = 60 + 2NDP => NDP = MDQ = 30.

In triangle NDP, DN.cos30 = DP => DN = (2/sqrt(3)).(1/2.sqrt(3)) = 1/3.
DM = DN = 1/3
Since DMN is isosceles with angle at D being 60, it's equilateral => MN = 1/3
AN = AP - NP = 1/2 - cos60/3 = 1/2 - 1/6 = 1/3
=> AM = 1/3
Perimeter of AMN = 1/3 + 1/3 + 1/3 = 1.

Case 2: Let's consider the outside case.
Now BD = CD  = 1/sqrt(3)
Angle DBC = DCB = 30
=> Angle DBA = DCA = 90
Again to keep things simple we can choose AM = AN and DM = DN.
Let AM = x => BM = 1-x = CN
In triangle MBD, DM^2 = BD^2 + BM^2 = 1/3 + (1-x)^2
In triangle MDN by cosine rule,
MN^2 = DM^2 + DN^2 - cos60.2.DM.DN = DM^2 = 1/3 + (1-x)^2

In triangle AMN by cosine rule:
MN^2 = AM^2 + AN^2 - cos60.2.AM.AN = AM^2 = x^2

Equating both:
1/3 + 1 - 2x = 0 => x = 2/3
AM = AN = MN = 2/3
So perimeter = 2.

Another solution:

Q10(Combinatorics, co-ordinate geometry). 

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There are $n^2$ triangles of positive area that have one vertex $A(0,0)$ and the other two vertices whose coordinates are drawn independently with replacement from the set $\{0,1,2,3,4\}$ e.g. (1,2), (0,1), (2,2) etc. Find the value of $n$.

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Solution:
24C2 - 3*4C2 - 2*2C2 = 256
=> n = 16
Explanation:
There are 5*5 = 25 possible points.
(0,0) is already gone so we need to choose 2 from 24: 24C2.
Now we subtract those cases when all 3 points are collinear.
There are 5 lines possible using the points in the grid: 
x-axis, y-axis, y = x, y = 2x, y = x/2
First 3 lines have 4 points available each: 3*4C2
Last 2 lines have 2 points available each: 2*2C2.

More details:
Let's say we choose (x1,y1) and (x2,y2).
If they are collinear with (0,0) then y2/y1 = x2/x1.
y-axis is the special case where x1 = 0.
Otherwise ratio of y's and x's should be same.

Q11(algebra). 

Prerequisite: Common algebraic identities, rational-root test.
For 

$x, y, z \in \mathbb{R}$ such that;
$x > y > z$ and if

$$x + y + z = 1$$ $$x^2 + y^2 + z^2 = 69$$ $$x^3 + y^3 + z^3 = 271$$

Find $2x + 3y + z$.

Correct Answer: 4

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Solution:

1. Let

$$S_1 = x+y+z = 1,\qquad S_2 = xy+yz+zx,\qquad S_3 = xyz.$$

2. From

$$x^{2}+y^{2}+z^{2}=69,$$

use $x^{2}+y^{2}+z^{2}=S_1^{2}-2S_2$:

$$69 = 1^{2} - 2S_2 \;\;\Longrightarrow\;\; S_2 = -34.$$

3. From

$$x^{3}+y^{3}+z^{3}=271,$$

use $x^{3}+y^{3}+z^{3}=S_1^{3}-3S_1S_2+3S_3$:

$$271 = 1^{3} - 3(1)(-34) + 3S_3 = 1 + 102 + 3S_3 = 103 + 3S_3 \;\;\Longrightarrow\;\; S_3 = 56.$$

4. The numbers $x,y,z$ are the roots of

$$t^{3}-S_1 t^{2}+S_2 t-S_3 = t^{3}-t^{2}-34t-56=0.$$

By the rational-root test, $t=7$ is a root. Factoring:

$$(t-7)(t^{2}+6t+8)=0\quad\Longrightarrow\quad t\in\{7,-2,-4\}.$$

With $x>y>z$, we have $(x,y,z)=(7,-2,-4)$.

5. Finally,

$$2x+3y+z = 2(7)+3(-2)+(-4)=14-6-4 = 4.$$

So $2x+3y+z = \boxed{4}$.

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Shortcut:

Once you have arrived at xyz = 56 you could guess that factors are 7,-4,-2 because x + y + z = 1
Then you cross verify with the given equations.

Q12. Find the number of ordered triples (a, b, c) of positive integers such that abc = 108.
Solution:
108 = 2^2*3^3
Three buckets, k = 3
Apply stars and bars twice with n = 2,3
(2+3-1)C(3-1) = 4C2
(3+3-1)C(3-1) = 5C2
6*10 = 60

Q13. A postman has to deliver five letters to five different houses. Mischievously, he posts one letter through each door without looking to see if it is the correct address. In how many different way could he do this so that exactly two of the five houses receive the correct letter?
Solution:
5C2*D3 = 10*2 = 20

Q14. There are n - 1 red balls, n green balls, and n + 1 blue balls in a bag. The number of ways of choosing two ball from the bag that have different colors is 362. What is the value of n?
Solution:
1R1G: n.(n-1)
1R1B: n.(n+1)
1G1B: (n-1).(n+1)
Total: n^2 - n + n^2 + n + n^2  - 1 = 3.n^2 - 1 = 362 => n^2 = 121 => n = 11

Q15. ABCD is a trapezium with AB || CD and AB < DC. AC and BD intersect at E, EF || AB, intersecting BC at F. Given that AB =20, CD=80, BC=100, then EF is:
Solution:
Whenever you see 3 parallel lines, try to find similar triangles. Especially one triangle inside a bigger triangle. The only killer move here is to spot BF + CF = BC.


Triangles BEF and BDC are similar.
Triangles CEF and CAB are similar.
EF/CD = BF/BC => BF = 5.EF/4
EF/AB = CF/CB => CF = 5EF
BF + CF = BC = 100 = 25EF/4 => EF = 16.

Question 16 
Given that

$$\frac{1}{2!17!} + \frac{1}{3!16!} + \frac{1}{4!15!} + \frac{1}{5!14!} + \frac{1}{6!13!} + \frac{1}{7!12!} + \frac{1}{8!11!} + \frac{1}{9!10!} = \frac{N}{19!}$$

find the greatest integer that is less than $\frac{N}{1000}$.

Correct Answer: 13

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Solution
Multiplying both sides by 19! yields:

$$\frac{1}{2!17!} + \frac{1}{3!16!} + \frac{1}{4!15!} + \frac{1}{5!14!} + \frac{1}{6!13!} + \frac{1}{7!12!} + \frac{1}{8!11!} + \frac{1}{9!10!} = \frac{N}{19!}$$ $$\Rightarrow \binom{19}{2} + \binom{19}{3} + \binom{19}{4} + \binom{19}{5} + \binom{19}{6} + \binom{19}{7} + \binom{19}{8} + \binom{19}{9} = 19N$$

Recall the combinatorial identity:

$$2^{19} = \sum_{n=0}^{19} \binom{19}{n}$$

Since $\binom{19}{n} = \binom{19}{19-n}$, it follows that

$$\sum_{n=0}^{9} \binom{19}{n} = \frac{2^{19}}{2} = 2^{18}$$

Thus,

$$19N = 2^{18} - \binom{19}{0} - \binom{19}{1} = 2^{18} - 1 - 19$$ $$= (2^9)^2 - 20 = (512)^2 - 20 = 262144 - 20 = 262124$$

So,

$$N = \frac{262124}{19} = 13796 \quad \text{and} \quad \left\lfloor \frac{N}{1000} \right\rfloor = \left\lfloor \frac{13796}{1000} \right\rfloor = 13$$

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Q17.  "How many ordered pairs of integers (m, n) where 

$0 < m < n < 2008$ satisfy the equation $2008^2 + m^2 = 2007^2 + n^2$?"
Solution:
n^ - m^2 = (2008+2007).1 = 4015 = (n+m)(n-m)
4015 = 5.11.73
Total divisors = 2*2*2 = 8
(n+m)(n-m) = 4015*1 = 803*5 = 365*11 = 73*55
Other factors are invalid since n+m > n-m.
We get 4 pairs for (n,m) = (2008,2007), (404,399), (64,9), (188,177)
First one is invalid so 3 answers in total.

Q18. Triangle ABC is the right angled triangle with the vertex C at the right angle. Let P be the point of reflection of C about AB. It is known that P and two midpoints of two sides of ABC lie on a line. Find the ∠CBA of the triangle ABC (Consider AC > BC).

Solution:

Since we have to find reflection along AB, let it be x-axis.
No lengths are given so we can assume AB = 1 WLOG.
B is (0,0), A = (1,0) C = (x1,y1) P = (x1,-y1)
To find angle ABC we need to find slope of line BC = y1/x1.

BC and AB are perpendicular =>
y1/x1 * y1/(x1-1) = -1
=> y1^2 = x1 - x1^2    _________[1]

Since AC > BC => midpoints of AB and AC are colinear with P.
So (x1, -y1), (1/2,0), ((x1+1)/2,y1/2) are colinear.________[2]

Equating the slopes from [2]:
-y1/(x1 - 1/2) = (y1/2)/(x1/2)
=> 2/(1-2x1) = 1/x1 => 2x1 = 1-2x1 => x1 = 1/4
Put x1 in [1]:
y1^2 = 1/4 - 1/16 = 3/16 => |y1| = sqrt(3)/4
=> |y1/x1| = sqrt(3) = slope of BC.
Now slope of BC is negative so it's -sqrt(3)  = tan(120)  => Angle ABC = 180 - 120 = 60
Answer: 60

Once you equate the slopes in [2] and use [1] you will get y1/x1 = -sqrt(3) = slope of BC.
=> Angle ABC = 60 degrees.

Q19. Let p(x) = x³ + ax² + bx + c be a polynomial where a, b, c are distinct non-zero integers. Suppose p(a) = a³ and p(b) = b³. Find the sum of digits of p(13).
Solution:
Since p(a) = a^3 and p(b) = b^3 => ax^2 + bx + c = 0 is a quadratic with roots a,b.
Sum of roots = -b/a = a + b and product ab = c/a => c = b.a^2
a+b = -b/a
=> a^2 + ab + b = 0
=> b = -a^2/(a+1)
b is an integer so a^2 has to be divisible by a+1.
From modulo arithmetic, we know that a^2/(a+1) = 1 mod (a+1).
But here it should be 0 for b to be an integer.
=> (a+1) = -+(1)  is the only way to resolve this contradiction.
a+1 = 1 => a = 0 but question says a can't be zero.
=> a = -2
=> b = 4
=> c = 16

p(13) = 13^3 -2.13^2 + 4.13 + 16 = 11*169 + 68 = 1690 + 169 + 68 = 1927
Sum of digits = 19

Q20.  p^3 - q^7 = p - q. Find p + q if p,q are prime. 
Solution:

Step 1: p > q, p > q^2, p > q^(7/3)
Step 2: q^(7/3) < p <= (q^2 + q + 1)
Step 3a: p = q^2 + q + 1
OR
Step 3b: Check for q's values 2,3,5,7. At 7 it won't hold.
So check at 2,3,5.
q = 3 will hold and give p = 13.

You can roughly see that p > q since p^3 is balancing out q^7 when p's and q's are moved to either side.
Let's take it further.
p^3 > p^3 - p = q^7 - q = q(q^6 - 1) >= 2(q^6 - 1) > q^6
p^3 > q^6 => p > q^2 > q
But p^3 = q^7 + p - q => p^3 > q^7 => p > q^(7/3)

Also:
p(p^2 - 1) = q(q^6 - 1) = q.(q^3 - 1).(q^3 + 1) = q.(q - 1)(q^2 + q + 1).(q + 1)(q^2 - q + 1)
p has to divide something from the RHS. It can't divide q,(q-1),(q+1),(q^2 - q + 1) as they are all smaller than q^2 and hence smaller than p.
Only factor divisible by p is (q^2 + q + 1).
We can prove p = q^2 + q + 1 OR take the route below.
3a. prove p = q^2 + q + 1
q^2 + q + 1 = kp for k >= 1
q^2 + q + 1 < 2q^2 for any prime starting from 2.
=> q^2 + q + 1 < 2p
So only k which fits here is 1
=> q^2 + q + 1 = p
Now if you put this value in
p(p^2 - 1) = q(q^6 - 1) = q.(q^3 - 1).(q^3 + 1) = q.(q - 1)(q^2 + q + 1).(q + 1)(q^2 - q + 1)
and factorize/cut common factors, you will get:
(q^2 + 1)(3-q) = 0 => q = 3

3b.
=> q^(7/3) < p <= (q^2 + q + 1)
Let's check for some primes here:
q = 2 => (128)^(1/3) <= 7 which does hold.
q = 3 => 3^(7/3) = 3^2.3^(1/3) = 9.3^(1/3) <= 13
To verify whether this holds, we need an approximate value of 3's cube root.
3's square root is 1.732 so cube root will be less than that.
1.4^2 is 1.96 and 1.96 *1.4 < 2*1.5 = 3
1.5^2 is 2.25 and 2.25 * 1.5 > 2*1.5 = 3
So 3's cube root is 1.45 approximately.
9*1.45 ~ 13 something. So both sides are quite close.

q = 5 => 5^(7/3) = 25*5^(1/3) <= (25 + 5 + 1 = 31)
Again a bit close.

q = 7 => 49*7^(1/3) <= (57)
Now this is hard to hold since 7's cube root will be very close to 2.
So let's test the original equation at q = 2,3,5 and p > q^2.
p^3 - p = q^7 - q
q = 2 =>  p^3 - p = 126, at p = 5 LHS = 120 and p = 7 crosses.
q = 3 (p > 9) => p.(p+1).(p-1) = p^3 - p = q.(q^3 - 1)(q^3 + 1) = 3.26.28 = 12.13.14 => p = 13
Answer: q = 3, p = 14


Q21. 

Let $r_1, r_2, r_3, r_4$ be the four roots of the polynomial
$x^4 - 4x^3 + 8x^2 - 7x + 3.$
If the value of

$$\frac{r_1^2}{r_2^2 + r_3^2 + r_4^2} + \frac{r_2^2}{r_1^2 + r_3^2 + r_4^2} + \frac{r_3^2}{r_1^2 + r_2^2 + r_4^2} + \frac{r_4^2}{r_1^2 + r_2^2 + r_3^2}$$

is $k$, find $|k|$.

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Solution:

Add 1 to each fraction to get

$$\frac{r_1^2 + r_2^2 + r_3^2 + r_4^2}{r_2^2 + r_3^2 + r_4^2} + \frac{r_1^2 + r_2^2 + r_3^2 + r_4^2}{r_1^2 + r_3^2 + r_4^2} + \frac{r_1^2 + r_2^2 + r_3^2 + r_4^2}{r_1^2 + r_2^2 + r_4^2} + \frac{r_1^2 + r_2^2 + r_3^2 + r_4^2}{r_1^2 + r_2^2 + r_3^2}$$

This seems like a difficult problem until one realizes that

$$r_1^2 + r_2^2 + r_3^2 + r_4^2 = (r_1 + r_2 + r_3 + r_4)^2 - 2(r_1r_2 + r_1r_3 + r_1r_4 + r_2r_3 + r_2r_4 + r_3r_4) = 4^2 - 2 \cdot 8 = 0$$

Thus, our current expression is 0. Noting that we added 4, the original value had to be -4.

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Q22. 
Find the number of ordered pairs (x,y) where 1/x + 1/y = 1/2024 and x,y are positive integers.

=> 2024.(x + y) = xy
=> xy - 2024.x - 2024.y = 0
Add 2024^2 to both sides:
=> xy - 2024.x - 2024.y + 2024^2 = 2024^2
=> (x - 2024).(y - 2024) = 2024^2
Let x - 2024 = p, y - 2024 = q
=> Number of divisors of 2024^2 will give us the unique (d,d') pairs and hence unique (x,y) pairs.
2024 = 2^3.11.23
2024^2 = 2^6.11^2.23^2
Number of divisors = (6+1).(3).(3) = 63

Q23.

Let $p(x) = x^2 - 5x + a$ and $q(x) = x^2 - 3x + b$, where $a$ and $b$ are positive integers. Suppose $\text{hcf}(p(x), q(x)) = x - 1$ and $k(x) = \text{lcm}(p(x), q(x))$. If the coefficient of the highest degree term of $k(x)$ is 1, the sum of the roots of $(x - 1) + k(x)$ is:

Solution:
p(x) has one root = 1, other root = 5 - 1 = 4
q(x) has one root = 1, other root = 3 - 1 = 2
p(x) = (x-1)(x-4)
q(x) = (x-1)(x-2)
LCM = (x-1)(x-2)(x-4)
k(x) + (x-1) = (x-1){(x-2)(x-4) + 1} = (x-1).(x-3)^2
roots  = 1,3,3
Sum = 7
Answer: 7

Q24. 

Given that P is an inner point of the equilateral triangle ABC, such that
$PA = 2, \quad PB = 2\sqrt{3}, \quad PC = 4$.
Then square of the side of $\triangle ABC$ is.

Solution:

Rotate $\triangle BPA$ around B in anti-clockwise direction by $60^\circ$, then $A \rightarrow C$
Let the image of P be M under the rotation.
Then $BM = BP$, $\angle MBP = 60^\circ$,
So $\triangle MBP$

MP = $2\sqrt{3}$. From MC = PA = 2 and
$MP^2 + MC^2 = 12 + 4 = 4^2 = PC^2$
so $\angle PMC = 90^\circ$, $\angle BMC = 150^\circ$.
Further, PC = 2MC implies $\angle MPC = 30^\circ$,
so $\angle BPC = 90^\circ$, and
$BC^2 = PB^2 + PC^2 = 12 + 16 = 28$, thus BC = 28.

Q25. 

Point P is inside ABC. Line segments APD, BPE, and CPF are drawn with D on BC, E on CA, and F on AB (see the figure below). Given that AP = 6, BP = 9, PD = 6, PE = 3, and CF = 20, find half of area of △ABC.

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Solution:
Let's compute CP and PF using mass points.
AP:PD = 1:1
BP:PE = 3:1 => mB = 1, mE = 3
Let mass of P = mP = 4 = mB + mE
mA = mD = mP/2 = 2
mC = mD - mB = 2 - 1 = 1
mF = mA + mB = 3
mC:mF = 1:3 = FP:PC => FP = CF/4 = 20/4 = 5 => PC = 20 - 5 = 15

In triangle PCB,
D is the midpoint of BC since mC = mB = 1.
By Apollonius theorem, PC^2 + PB^2 = 2(PD^2 + BD^2)
=> (15^2 + 9^2)/2 - 6^2 = BD^2
=> 306/2 - 36 = 153 - 36 = 117 = BD^2
In triangle PBD = BD^2 = 117 = PB^2 + PD^2 = 81 + 36 => PBD is right angle at P.
=> [PBD] = 9*6/2 = 27
[PCB] = 2*[PBD] = 54.

Now:
[ABC]/[PBC] = [AD]/[PD] = 2
=> [ABC]/2 = [PBC] = 54 = Answer.

Q26. Let the rational number p/q be closest to but not equal to 22/7 among all rational numbers with denominator <100. What is the value of p - 3q?
Solution:
| (p/q) - (22/7) | should be smallest.
=> |  (7p - 22q)/7q  | should be smallest.
For this to be smallest we can try the numerator = 1 and q = 99.
Why q  = 99? Because the questions says in p/q the denominator < 100.

Let's try 7p - 22q = 1
=> p = (22*99 + 1)/7 = 2179/7.
But p should be an integer and 2179 is not divisible by 7.

So let's try 22q - 7p = 1
=> p = (22*99 - 1)/7 = 2177/7 = 311.
So this one works.

=> p/q = 311/99
p - 3q = 14 is the answer.

What would have happened if weren't able to find an integer p after assuming q  = 99 and numerator of the difference = 1?
Then we have 2 options. Either we decrease the denominator, i.e. q = 98 or increase the numerator, i.e. |7p - 22q| = 2. Which one would we choose? q = 98. Why? Because making the numerator 2 would double the difference. While decreasing q will increase it by a small margin.

Q27. 

A four term sequence is formed by adding each term of a four-term arithmetic sequence of positive integers to the corresponding term of four-term geometric sequence of positive integers. The first three terms of the resulting four term sequence are 57, 60 and 91. If the fourth term of this sequence is $k$, find

$$\frac{2024}{k + 47}.$$

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Solution:
a + p = 57
a + d + pr = 60
a + 2d + pr^2 = 91

Subtract [1] from [2] and [2] from [3]:
d + p(r-1) = 3
d + p.r.(r-1) = 31

Subtract again:
p.(r-1)^2 = 28
One solution is p  = 7, r = 3, another is p = 28, r = 2.
We will try out the first one and if there is any discrepancy, we will try out the second one.
What kind of discrepancy?
Questions says that 4 terms of both the A.P. and G.P. have positive integers. If that is not violated, we are good.
p = 7, r = 3 => d = -11, a = 50.
A.P.: 50, 39, 28, 17
G.P.: 7,21,63,189
4th term: 189 + 17 = 206
2024/(206 + 47) = 2024/253 = 8
Answer: 8

Q28.

In $\triangle ABC$, $\angle C = 90^\circ$. $\angle BAC < 45^\circ$ and $AB = 7$. P is point on AB such that
$\angle APC = 2\angle ACP$ and $CP = 1$. If $\frac{AP}{BP} = 6 + \sqrt{a}$, then $a =$

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Solution:

Step 1: Let O be the mid-point of AB, which is also the circumcentre of $\triangle ABC$. Extend CP to meet the circumcircle at D.

Step 2: Note that P is between O and B (if P is between O and A then $\angle ACP$ is less than $45^\circ$ while $\angle APC$ is obtuse, contradicting $\angle APC = 2\angle ACP$).

Let's explore this point in some more detail.
Consider the triangle △OAC.

OA = OC => ∠OCA = ∠OAC.
Given that ∠BAC (=∠OAC = ∠PAC) < 45 => ∠OCA < 45.
If P is between O and A, then ∠PCA < ∠OCA => ∠PCA < 45.
In triangle △PAC, ∠PCA + ∠PAC < 90 => ∠APC > 90
But ∠APC = 2∠ACP which is a contradiction => P is between O and B.

Step 3.

Let $\angle ACP = \theta$. Then $\angle DPB = \angle ACP = \theta$ and $\angle AOD = 2\theta$ (angle made by AD at the centre). It follows that $DO = DP$, both being 3.5 (radius of the circle).

Why is ∠DPB=∠ACP=2θ?
Because they are vertical opposite angles.

And, why is DO = DP?
Because △ODP is isosceles. Same external angles => same internal angles.

Step 4.

Using the power chord theorem $PA \times PB = PC \times PD$,
we have $PA \times (7 - PA) = 1 \times 3.5$

Solving, we get

$$PA = \frac{7 \pm \sqrt{35}}{2}$$

The positive square root is taken as P is between O and B(=> PA > 7/2). It follows that

$$PB = \frac{7 - \sqrt{35}}{2}$$

and so the answer is

$$\frac{7 + \sqrt{35}}{7 - \sqrt{35}} = 6 + \sqrt{35}$$

Q29:

Let $a = \sqrt{3} + \sqrt{5} + \sqrt{7},\ b = \sqrt{3} - \sqrt{5} + \sqrt{7},\ c = \sqrt{3} + \sqrt{5} - \sqrt{7}$.
Evaluate

$$\frac{a^4}{(a-b)(a-c)}+\frac{b^4}{(b-c)(b-a)}+\frac{c^4}{(c-a)(c-b)}$$

Solution:

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1. Put the three fractions over a common denominator

$$S=\frac{a^{4}}{(a-b)(a-c)}\;+\;\frac{b^{4}}{(b-c)(b-a)} \;+\;\frac{c^{4}}{(c-a)(c-b)} .$$

The common denominator is

$$D\;=\;(a-b)(b-c)(c-a).$$

Writing each term with this denominator gives

$$S=\frac{N}{D}, \qquad\text{with}\qquad N=a^{4}(b-c)-b^{4}(a-c)+c^{4}(a-b).\tag{1}$$

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2. Show that $(a-b)\,(b-c)\,(c-a)$ divides $N$

Evaluate $N$ when two variables coincide:

• If $a=b$: $N=b^{4}(b-c)-b^{4}(b-c)+c^{4}(b-b)=0.$

• If $b=c$: $N=a^{4}(c-c)-c^{4}(a-c)+c^{4}(a-c)=0.$

• If $c=a$: $N=a^{4}(b-a)-b^{4}(a-a)+a^{4}(a-b)=0.$

Hence $a-b,\;b-c,\;c-a$ are three linear factors of $N$, so we can write

$$N=(a-b)(b-c)(c-a)\,Q(a,b,c),\tag{2}$$

for some polynomial $Q$.

Because $N$ is degree $5$ and the product of the three linear factors is
degree $3$, $Q$ must be a degree-2 polynomial.

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3. Write $Q$ as a linear combination of the two basic symmetric quadratics

$N$ changes sign when two variables are swapped (it is alternating).
The factor $(a-b)(b-c)(c-a)$ is also alternating, so their quotient $Q$ is
fully symmetric.

Every symmetric quadratic in three variables is a combination of

$$A=a^{2}+b^{2}+c^{2},\qquad B=ab+bc+ca;$$

hence

$$Q=\alpha\,A+\beta\,B,\qquad\text{for some constants }\alpha,\beta.\tag{3}$$

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4. Find $\alpha$ and $\beta$

Pick a simple substitution that keeps the three linear factors distinct;
$c=0$ is convenient.

Compute $N$ and the triple product for $c=0$

$$\begin{aligned} N(a,b,0)&=a^{4}b-b^{4}a =ab(a^{3}-b^{3}) =ab(a-b)(a^{2}+ab+b^{2}),\\[4pt] (a-b)(b-c)(c-a)\Big|_{c=0}&=(a-b)(b)(-a)=-ab(a-b). \end{aligned}$$

The value of $Q$ for $c=0$

$$Q(a,b,0)=\frac{N}{(a-b)(b-c)(c-a)}=-\bigl(a^{2}+ab+b^{2}\bigr).$$

But with $c=0$

$$A=a^{2}+b^{2},\qquad B=ab,$$

so

$$\alpha\,A+\beta\,B=-\,(A+B)\;\Longrightarrow\; \boxed{\;\alpha=-1,\quad\beta=-1\;}.$$

Thus

$$Q=-(a^{2}+b^{2}+c^{2}+ab+bc+ca).$$

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5. The simplified form of $S$ and the final numerical answer

Putting (2) and the value of $Q$ into (1),

$$N=-(a-b)(b-c)(c-a)\bigl(a^{2}+b^{2}+c^{2}+ab+bc+ca\bigr).$$

But $D=(a-b)(b-c)(c-a)$, so

$$S=\frac{N}{D}=a^{2}+b^{2}+c^{2}+ab+bc+ca =\tfrac12\Bigl[(a+b)^{2}+(b+c)^{2}+(c+a)^{2}\Bigr].\tag{4}$$

Plugging the given numbers

$$a=\sqrt3+\sqrt5+\sqrt7,\; b=\sqrt3-\sqrt5+\sqrt7,\; c=\sqrt3+\sqrt5-\sqrt7.$$ $$\begin{aligned} (a+b)^{2}&=(2\sqrt3+2\sqrt7)^{2}=40+8\sqrt{21},\\ (b+c)^{2}&=(2\sqrt3)^{2}=12,\\ (c+a)^{2}&=(2\sqrt3+2\sqrt5)^{2}=32+8\sqrt{15}. \end{aligned}$$

Insert these into (4):

$$\boxed{\,S =\frac12\Bigl[(40+8\sqrt{21})+12+(32+8\sqrt{15})\Bigr] =42+4\sqrt{15}+4\sqrt{21}\;\approx 75.82.}$$

(The earlier answer “30” came from mistakenly cancelling the
$\sqrt3$ and $\sqrt7$ terms in $(a+b)$ and $(c+a)$; the calculation above shows they do not cancel.)

Q30:

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There are real numbers $x, y$ and $z$ such that the value of

$$x + y + z - \left( \frac{x^2}{5} + \frac{y^2}{6} + \frac{z^2}{7} \right)$$

reaches its maximum of $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers.
Find $m + n + x + y + z$.

Solution:
Let's complete the square in x:
x - x^2/5 = -(x^2/5 - x) = -1/5(x^2 - 5x) = -1/5(x^2 - 2.x.5/2 + 25/4 - 25/4) = -1/5{(x-5/2)^2 - 25/4}
 = 5/4 - (1/5).(x-5/2)^2
Similarly for y:
6/4 - (1/6).(y-6/2)^2

Similarly for z:
7/4 - (1/7).(z-7/2)^2

In total:
= 5/4 - (1/5).(x-5/2)^2 + 6/4 - (1/6).(x-6/2)^2 + 7/4 - (1/7).(z-7/2)^2
= 18/4 - {sum of squares}
For this to be max, x = 5/2, y = 3, z = 7/2
m/n = 9/2
Final answer: 9 + 2 + 5/2 + 3 + 7/2 = 20 is the answer.