IOQM mock test 2 - pending from Q2 onwards

 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.
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:

$$\sum x = 1 = a, \quad \sum x^2 = 69 = b, \quad \sum x^3 = 271 = c, \quad \sum xy = d, \quad xyz = e$$ $$a^2 = b + 2d \Rightarrow d = -34$$ $$c - 3e = a(b - d) \Rightarrow 271 - 3e = (1)(69 - (-34)) = 103 \Rightarrow e = 56$$

Now form a cubic equation in $t$ with roots $x, y, z$:

$$t^3 - at^2 + dt - e = 0 \Rightarrow t^3 - t^2 - 34t - 56 = (t + 4)(t + 2)(t - 7) = 0$$ $$\Rightarrow t = -2, -4, 7$$

Therefore,

$$(x, y, z) = (7, -2, -4)$$

Then,

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

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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:
We will prove it like this:
1. p != q
2. p > q
3. q is odd
4. q <= 5
5. q = 3
6. p = 13.