Math IOQM

 

 

1. Prove that if p/q is a rational number in its lowest form then it will terminate if and only if denominator = 2^m.5^n and m,n are whole numbers. And the number of decimals will be max(m,n). 2. Prove sqrt(2) is irrational. (Assume it isn't and then p/q and p,q are both even). 3. a < (a+b)/2 < b for all real a,b 4. Fact: A rational number in decimal form will either terminate or start repeating after some digits. It's not possible for a rational to never terminate and not cycle at the same time. 5. Fact: Pie = 4 * (1/1 - 1/3 + 1/5 - 1/7 + 1/9 ...) Basically alternating odd denominators.

Q1. A DNA chain is composed of basic building blocks in the form of four chemicals, known by the symbols A,C,T,G. Consider three-letter chains consisting of these symbols (with or without repetitions). How many such chains are there? S1. 4*4*4 = 64 Q2. Consider the set {a,b,c,d,e}. How many three-letter words can be made out of them, with or without meaning? How many of these will have at least one vowel in them? Answer these questions both for the case when repetitions of letters are allowed and for the case when repetitions of letters are not allowed. S2. 2.1 With repetition: 5*5*5 = 125 Without: 5*4*3 = 60 2.2 With repetition: Total words: 5^3 = 125 Without any vowels = 3^3 = 27 Answer = 125 - 27 = 98 Without: Total: 5*4*3 = 60 Without vowels = 3*2*1 Answer = 60 - 6 = 54 Q3. A permutation (a1,a2,a3,a4,a5) of (1,2,3,4,5) is called heavy-tailed if a1 + a2 < a4 + a5. What is the number of heavy-tailed permutations? S3. Total such: 5! = 120 Let's fix 1,2 to a1,a2 => 2!*3! = 12...

1. Find the total number of words formed by the letters a,a,b,b,c. S1. 5!/2!.2! 2. In how many ways can you arrange 2 beads of red colour, 2 beads of blue colour and 3 beads of black colour in a ring? Note that beads with the same colour are identical and arrangements that can be obtained by rotating (or) flipping the ring are identical. S2. First attempt(wrong): (7-1)! circular permutations Divide by 2!2!3! to account for repeated elements. Divide by 2 to account for flipping. Answer = 15 This is wrong, correct answer is 18. Why? Let's see using a smaller example. 2 blue, 3 black. (5-1)! circular permutations Divide by 2!3!. So far so good. Mistake is when we divide by 2 for flipping. Why do we even divide by 2 for flipping. Let's say we have 4 distinct beads. Here is one necklace: 1 --- 2 |       | 4 -- 3 and flipping it we get: 2 --- 1 |       | 3 -- 4 So 1-2-3-4 was clockwise and it became counter clockwise. Flipping gives a distinct perm...

1. 5 chairs, 4 people? How many ways can they sit? 5C4*4! = 120 2. 10 mangoes, 12 apples, 16 guava, 15 bananas. How many ways to eat 3 fruits? Case 1: Every single fruit is distinct. We have 53 fruits in total. So 53C3. That is if the order of eating doesn't matter. If the order of eating matters, then: 53.52.51 Case 2: Fruits of one type are all same.  Case 2.1 - Order of eating doesn't matter. Then this is simply stars and bars. (n+k-1)C(n) where 'n' is the number of items and 'k' is the number of buckets. Here,  n = 3 k = 4 So: All 3 fruits are same: 4C1 = 4 2 Same 1 distinct: 4C2 * 2C1 = 12 All 3 distinct: 4C3 = 4 Total: 20 Case 2.2 - Order of eating does matter. 4 * 4 * 4 = 64 Case 3: One fruit of each type, i.e. I eat 1 mango, 1 apple, 1 banana for e.g. Case 3.1 Fruits of same type are also distinct We have to consider 4 different cases(4C3): Mango, banana, apple = 10*15*12 Mango, banana, guava = 10*15*16 Mango, apple, guava = 10 * 12 * 16 Apple, banana, g...

  Problem 1. Let (a) be a positive real number. Prove that one of the positive real numbers [ a,;2a,;3a,...;(n-1)a ] is at most (1/n) distant from a positive integer. S1. Let's test for n = 2 The list is simply: 'a'. And any real positive number is at most 1/2 away from a positive integer. So it works for n = 2. Let 1 <= k <= n-1. We want to show that at least one k.a is at most 1/n away from a positive integer. Let {k.a} be the fractional part of k.a. Distance of {k.a} from a positive integer is min( {ka}, 1 - {ka}). To understand, see the number line: 0______{ka}_______1 Its distance from 0 is {ka} and 1 - {ka} from 1. Let's prove by contradiction. Let's say for each 'k', 1 <= k <= n-1, {k.a}is more than 1/n away from the nearest integer. Like this: 0___1/n___{ka}___1 - 1/n____1 So {ka} is always between 1/n and (n-1)/n. Distance between 1/n and (n-1)/n is (n-2)/n. So there are (n-2) gaps each of length 1/n. Using pigeonhole argument, fraction...

1) P(x) = x^4- 4x^3 + 6x^2 - 4x - 2005. Find the integer part of the product of all non-real roots of (P). S1) P(x) = 0 => (x-1)^4 = 2006 => (x-1) = K.(2006)^(1/4) where K = -+1, -+i So non real roots are x = 1 +-i.(2006)^(1/4) Multiply: 1 + 2006^(1/2) 45^2 = 2025 45^2 - 44^2 = 89 So 2006^(1/2) = 44.something Answer: 45 2) First we will prove this: In the incircle trisects the median then the sides are in the ratio 5:10:13 where 10 maps to the base where median is landing. Once we do that, since BC = 20, other sides will become 10 and 26. Then apply Heron's formula. s = 28 s - a = 28 - 20 = 8 s - b = 28 - 10 = 18 s - c = 28 - 26 = 2 [ABC] = sqrt(28.18.8.2) = sqrt(64.7.2.9) = 24.sqrt(14) m + n =38 Ok coming to the proof of the first part. Using power of the point theorem: If DF is the tangent from D, then DF^2 = DY.DX = d/3 * 2d/3 = 2d^2/9 Similarly Tangent from A = AE then AE^2 = AX.AY = DF^2 => AE^2 = DF^2 But AE = s - a from incircle properties. And DF = (s-c) - a/2 = a+...

  1) Find the largest natural number 'm' such that:  (m^2) is the difference of two consecutive perfect cubes. AND  (2m + 79) is a perfect square. S1) m^2 = (n+1)^3 - n^3 = 3n^2 + 3n + 1 Since m,n are both natural numbers, RHS will grow much faster than LHS and at some point it will irreversibly exceed it. m^2 - 1 = 3n(n+1) So RHS is a multiple of 6 => m^2 = 1 mod 6 m^2 = 25,49 satisfy this but 24 = 3n(n+1) => 8 = n(n+1) no go. 48 = 3n(n+1) => 16 = n(n+1) no go. what next? So let's try some other approach. Once you complete the square in 'n', you will get: 3n(n+1) + 1 = 3[(n+1/2)^2 - 1/4] + 1 = m^2 4m^2 = 4 + 3[ (2n+1)^2- 1] = 1 + 3.(2n+1)^2 4m^2 - 1 = 3(2n+1)^2 (2m-1)(2m+1) = 3(2n+1)^2 LHS has 2 consecutive odd numbers which are co prime. Why? gcd(2m-1,2m+1) = gcd(2,2m-1) = 1 So RHS should split cleanly into 2 factors which don't share any prime factors. Case 1: 2m-1 = a^2 2m + 1 = 3b^2 Case 2: 2m + 1 = a^2 2m - 1 = 3b^2 Case 2: 2 = a^2 - 3b^2 => a...

Q1) Let A be a set of m positive integers, where m >= 1. Show that there is a non-empty subset (B) of (A) such that sum of all elements in B is divisible by (m). S1) Let's create prefix sums like this: S1 = a1 S2 = a1 + a2 ... Sm = a1 + a2 .. am If any one of S1,S2...Sm has a 0 remainder we are done. If not, then at least 2 of them have same remainder. Then: Sk - Sj = 0 mod m If k > j Then Sk - Sj = ak + a_{k-1} ... a_{j-1} So these are the required elements. Q2)  Let (A) be a subset of ({1,2,...,2n}) such that (|A|=n+1). Show that there exist (a, b in A) such that gcd(a,b)=1. Part 2: Can we also say the same thing when |A| = n? S2) There are at least 2 consecutive numbers in A. Proof by PHP: {1,2} {3,4} ... {2n-1,2n} Each number belongs to one bucket out of these 'n' buckets. If there are n+1 numbers, at least 2 of them will fall into same bucket. Proof by contradiction(not very solid): Let's say A has only non-consecutive numbers. Then it can have only n number...

 Q 1) Find the number of ways in which three numbers in A.P. can be selected from (1, 2, 3, ..., n). (For both (n) odd and even.) S1) We have to select 2 numbers of same parity and the AM will lie within. Case 1: n is even. Select 2 odd numbers: (n/2)C2 Select 2 even numbers: same Case 2: n is odd. Select 2 odd numbers: (n+1)/2C2 Select 2 even numbers: (n-1)/2C2 simplify to get: (n-1)^2/4

Q1. Does there exist a natural number “(n)” such that (n^2 + n + 1) is divisible by (1955)? If “Yes”, provide an example and if “No”, provide a valid proof. S1. n^2 + n + 1 = 0 mod 5 (if it's div by 1955, it's also div by 5) n^2 + n = 4 mod 5 Let's check all possible remainders of n with 5. n = 0 mod 5 No. n = 1 no. n = 2 4 + 2 = 1 mod 5 n = 3 9 + 3 = 2 mod 5 n = 4 16 + 4 = 0 mod 5 So nothing works. "No". Q3. Using the properties of congruences, prove that 89 | 2^44 - 1 and 97 | 2^48 - 1 S3. 2^44 - 1= 0 mod 89 (2^11 - 1)(2^11 + 1)(2^22 + 1) = 0 mod 89 Since 89 is a prime, one of these factors has to be a multiple of 89. Let's test one by one. 2^11 = 1 mod 89. 2^5.2^6.3 = 3 mod 89 96.64 = 3 mod 89 = 7.64 = 448 = 3 mod 89 H.P. Part 2: 2^48 - 1 = 0 mod 97 = (2^6 - 1)(2^6 + 1)(2^12 + 1)(2^24 + 1) First 2 factors are small. Next: 2^12 + 1 = 0 => 2^12 = -1 mod 97 = 96 mod 97 = 32.128 => 128 = 3 mod 97 which is wrong Next: 2^24 = -1 = 96 mod 97 2^19 = 3 mod 97 ...

Q1)   A square of dimension (n-1) * (n-1) is divided into (n-1)^2 unit squares in the chess board manner. Now, each of these (n^2) vertices of these squares are to be painted black or blue such that each unit square has exactly two black vertices. In how many ways can this be done? S1. Look at an individual square. There are 2 ways for it to yield 2 black vertices. One horizontal edge blue(or black), other black(or blue). Or 1 color each on top edge, and similarly on bottom edge. Once you expand the pattern, there will be only 2 ways to get it done. Each row alternates or each column alternates or both alternate. Case 1: Each row alternates. So each row can start with one of 2 colors and then the pattern is fixed. So 2^n ways since there are 'n' rows. Case 2: Each column alternates. 2^n again. Case 3: Both alternate. Exactly 2 ways like a chessboard. Total: 2^n.2 - 2 = 2^(n+1) - 2. Q2. S2. As the diagram shows a 2x2 bounding box can hold 2 different squares. That rule actually ...

 If an odd prime number leaves remainder of 1 when divided by 4 then it can be expressed as sum of two integer squares and those 2 integers will be unique for this prime. Examples:  17 = 16 + 1 29 = 25 + 5 41 = 25 + 16 1013 = 22^2 + 23^2

Q1.  Find the number of triangles whose angular points are at the angular points of a given polygon of (n) sides, but none of its sides are the sides of the polygon. S1. Simple application of  Kaplansky's first lemma . Formula derived is: (n-k-1)C(k-1) + (n-k)C(k) Here k = 3 (n-4)C(2) + (n-3)C(3) = (n-4)(n-5)/2 + (n-3)(n-4)(n-5)/6 = (n-4)(n-5)/2[1 + (n-3)/3] = n(n-4)(n-5)/6

1. Is 2026 a semiprime(product of only 2 primes)? Yes 2. Is 2026 a deficient number(sum of its proper divisors is less than the number itself)? Yes 3. Is it a happy number? If you keep squaring the digits and adding and then repeating - does it reduce to 1? Yes 4. Is it sum of 2 squares? 45^2 + 1^2 5. Is 1013 sum of 2 squares? = 22^2 + 23^2 ( Fermat theorem on sum of 2 squares )

Q1. (n) is a positive integer. (A) is a set such that A={1,2,...,n}. Let (t_n) denote the number of subsets of (A) such that the arithmetic mean (AM) of the elements is an integer. Prove that (t_n) and (n) are both odd or both even. S1. 1. Let us ignore empty subset since A.M. is not defined for that. 2. Subsets of size 1: {1}, {2} ... {n}. There are 'n' such subsets. And each of them has their A.M. as integer. 3. Now let's consider the subsets with size >= 2 which have an integer A.M. Let G be the set of all such subsets. Let's consider a function 'f' defined for subsets of size >= 2. f(S) = S - {k} where k is the A.M. of S and k is present in S. f(S) = S + {k} where k is the A.M. of S and k is not present in S. For e.g. S = {1,3,8}, k = (1+3+8)/3 = 4 4 is not there in S. f(S) = {1,3,4,8} S = {1,2,3}, k = (1+2+3)/2 = 3 3 is there in S. f(S) = {1,2} Note 1: S and f(S) have the same A.M. Why? Let S = {a1,a2...,ak} AM = sigma(a_i)/k If you remove AM, the ne...

Q1. Find integer solutions for x^2 + y^2 = 2025 S1. Mod 9 of a square gives 0,1,4,7 Here RHS is 0. For LHS to be 0, both have to give mod 0. => x = 3a, y = 3b => a^2 + b^2 = 225 Again mod 9 is 0 a = 3c, b = 3d c^2 + d^2 = 25 (c,d) = (0,+-5) (-+5,0) (-+3,-+4) (-+4,-+3) Total solutions = 2 + 2 + 4 + 4 = 12 To find actual solutions multiply c,d with 9.

Theorem: If a1x1 + a2x2 + .... an.xn = K then this Diophantine equation has a solution if d | K where d = gcd(a1,a2...an). We can note the similarities with Bezout's lemma here. Practice: Which of these Diophantine equations have integer solutions? 1. 21x + 15y = 17 2. 7x + 17y = 27 Theorem: If ax + by = c and d | c, where d = gcd(a,b), then this Diophantine equation has infinitely many solutions, and the solutions are of the form x = x0 + (b/d).k, y = y0 - (a/d).k, where k is integer and (x_0, y_0) are particular solutions of (ax + by = c); and x0, y0 are integers.

Q1. Solve for x: 32x = 79 mod 1225 S1. 32x = 79 mod 25 => 7x = 4 mod 25, since 7*3 = 21 = -4 m 25 => 7*-3 = 4 m 25 => x = 22 m 25 32x = 79 mod 49 => 32x = 30 m 49 => 16x = 15 m 49, since 16.4 = 64 = 15 m 49 => x = 4 m 49 Use CRT now: x = p.25 + q.49 m 25 22 = q.-1 m 25 => q = -22 = 3 mod 25 m 49 4 = p.25 mod 49 => p = 8 mod 49, since 25.8 = 200 = 4 mod 49 x = 8.25 + 3.49 mod 1225 = 347 mod 1225

Q1. (combinatorial geometry problem) There is a 21-sided regular polygon with vertices (A1, A2 .... A21). Triangles are formed by joining these vertices. a) How many of these triangles are acute-angled ? b) How many of these triangles are right-angled ? c) How many of these triangles are obtuse-angled ? d) How many of these triangles are equilateral ? e) How many of these triangles are isosceles ? S1. Regular polygon can be inscribed in a circle. Each arc being of equal length. For e.g. equilateral triangle will create 3 arcs of 120 degrees each. So 21-gon will create 21 equal-sized arcs. Total possible triangles = 21C3 = 21.20.19/6 = 19.70 = 1330 b) 0 right angle triangles. Why? By Thales' theorem, diameter inscribes right angle on circumference. And a diameter divides circle into 2 equal parts. In an odd sided regular polygon, we cannot have equal number of arcs on both sides of the diameter. c) By inscribed angle theorem, for an obtuse angle on circumference, the intercept...

Q1. Two friends, Marco and Ian, are talking about their ages. Ian says, "My age is a zero of a polynomial with integer coefficients." Having seen the polynomial p(x) Ian was talking about, Marco exclaims, "You mean, you are seven years old? Oops, sorry I miscalculated! p(7) = 77 and not zero." "Yes, I am older than that," Ian's agreeing reply. Then Marco mentioned a certain number, but realizes after a while that he was wrong again because the value of the polynomial at that number is 85. Ian sighs, "I am even older than that number." Determine Ian's age. S1. Hint: (a-b) divides (a^n - b^n) for all integer 'n'. Let the root be 'a'. p(a) = 0 p(7) = 77 p(b) = 85 and b > 7. a > b > 7. For a polynomial with integer coefficients and x,y as integers: x-y | p(x) - p(y) So a-7 | p(a) - p(7) a-7 | -77 So a-7 can be 1,7,11,77 => a can be {8,14,18,84} b - 7 | 8 => b-7 can be 1,2,4,8 => b can be {8,9,11,15} a-b| 85 ...

Q1. Prove that if the two angle bisectors of a triangle are equal, then the triangle is isosceles. ( Steiner-Lehmus Theorem ) S1. So, let's see what are we trying to prove. BE = CF and angle B = angle C We know that length of angle bisector in triangle is given by the below formula. BE = 2ac.cos(B/2)/(a+c) Similarly, CF = 2ab.cos(C/2)/(a+b) make the equal: c.cos(B/2)/(a+c) = b.cos(C/2)/(a+b) cos(B/2)/cos(C/2) = ab + bc/ac + bc We will prove by contradiction. WLOG, Angle B > Angle C => b > c also => cos(B/2) < cos(C/2) (since B/2 and C/2 are both < 90). So LHS < 1 But RHS > 1 why? Because  ab + bc > ac + bc (since b > c) And if Angle C < Angle B, again LHS > 1 and RHS < 1. So only option is that both are equal. H.P. A slippery slope: If you go a different way, you will miss the proof! It's quite tricky. c.cos(B/2)/(a+c) = b.cos(C/2)/(a+b) from here: b/c = sinB/sinC => cos(B/2)/(a+c) = sinB/sinC * cos(C/2)/(a+b) sinB = 2sin(B/2)cos(B/2) ...

  [ABC] = a.c.sin(B)/2 [ABC] = [ABE] + [EBC] [ABE] = x.c.sin(B/2)/2 [EBC] = a.x.sin(B/2)/2 x = ac.sin(B)/sin(B/2).(a+c) x = ac/(a+c) * 2cos(B/2) BE = 2ac.cos(B/2)/(a+c)

Q1. Determine all solutions in the positive integers of 18x+5y=48 S1. y = (48-18x)/5 48-18x = 0 mod 5 3 = 3x mod 5 x = 1 mod 5 x = 1 => y = 6 x >= 6 not possible else y will become negative. what if I had done x = (48 - 5y)/18 in that case 48 = 5y mod 18 => 12 = 5y mod 18 Now, the thing is that y mod 18 can range from 0 to 17, so my trial and error will take much longer. So for these kind of equations it's always better to go for a smaller modulus, as we did in the first approach. Q2. 5x+3y=52. S2. x = (52 - 3y)/5 => 2 = 3y mod 5 => y = 4 mod 5 y = 4,9,14 => x = 8, 5, 2 Q3. Find all residues r with 0≤r≤7 such that x^2 ≡r(mod8) has a solution. S3. 0,1,4 Q5. A single bench section at a school event can hold either 7 adults or 11 children. When N bench sections are connected end to end, an equal number of adults and children together will occupy all the bench space. What is the least possible positive integer value of N? S5. a + c = N 7a = 11c 7,11 are primes => s...

Q1. S1. QR || AC by MPT. => line BHE is perpendicular to QR. In triangle AHB, using MPT, QP || BH. => QR is perpendicular to QP => angle PQR = 90 H.P. Q2. ABCD is a quadrilateral in which AB = AD . The bisectors of ∠BAC and ∠CAD intersect the sides BC and CD at E and F , respectively. Prove that (EF ||BD). S2. Using Angle bisector theorem: Triangle ABC: AB/BE = AC/EC Triangle ADC: AD/DF = AC/FC => AB/AC = BE/EC = AD/AC = DF/FC (since AB = AD) Now in triangle BDC: EF divides the sides BC and DC in same ratio => EF || BD H.P.

  Q1) Let (ABCD) be a rectangle such that (BC = 3AB). (P) and (Q) are points on the side (BC) such that BP = PQ = QC. Show that \angle DBC + \angle DPC = \angle DQC. S1. Q2. In a quadrilateral ABCD, given that angle A + angle D = 90. Pr. th. AC^2 + BD^2 = AD^2 + BC^2. S2. BC^2 = OB^2 + OC^2 AD^2 = OA^2 + OD^2 BC^2 + AD^2 = OB^2 + OC^2 + OA^2 + OD^2 Look at RHS: OA^2 + OC^2 = AC^2 OB^2 + OD^2 = BD^2 H.P. Q3.  In ( \triangle ABC ), (BM) and (CN) are perpendiculars from (B) and (C) respectively on a line passing through (A). If (L) is the midpoint of (BC), prove that [ ML = NL. ] S3. Method 1: Let LP be perpendicular to the same line passing through A. LP || BM || CN According to Intercept theorem, if 3 parallel lines cut 2 equal segments from a traversal, they will do the same with any other traversal. So MP = PN Now we will show triangle LMP congruent to LNP. LP is common side. Angle P is 90 in both. MP = PN. => LM = LN. Method 2: co-ordinate geometry Let the line thro...

Q1. The side AB of a parallelogram ABCD is produced both ways to F and G, so that AF = AD and BG = BC. Prove that FD and GC produced intersect at right angle. S1. There are 2 ways to draw the diagram here. Both give us the desired proof. Q2. In triangle AQB, points P and D lie on sides AB and AQ, respectively, such that [APQ]=[ABD]. Through D, draw the line DR∥AB, meeting BQ at R. Through B draw a line || to AQ which meets DR at C. Pr. th. RC=AP. S2. ABCD is a ||gram. AB = CD____[1] AD = BC Given [APQ] = [ABD] => Sin(A).AP.AQ = Sin(A).AB.AD => AP/AB = AD/AQ => Triangles APD and ABQ are similar by SAS. => PD || BR => BPDR is a ||gram => PD = BR and BP = DR___[2] From [1]: AP + PB = DR + RC Using [2]: AP = RC H.P. Q3. ABCD is a parallelogram. Through C, a straight line RQ is drawn outside the parallelogram, and AP, BQ, DR are drawn perpendicular to RQ. Show that DR + BQ = AP. S3.  

Q1. Let n be the least positive integer greater than 1000 for which gcd(63,n+120) = 21 and gcd(n+63,120) = 60. What is the sum of digits of n? S1. First wrong attempt: n = 21p - 120 = 60q - 63 => q = (7p - 19)/20 p = 17 works between p = 1 to 20. So p = 17 mod 20. p = 17,37,57,77,97... 21p - 120 >= 1000 => p>=880/21 => p = 57 => q = 19 n = 21*57-120 = 1077 sum_digits(n) = 15 Why is this wrong? gcd(63, 1077 + 120) = 63 not 21. gcd(63, n + 120) = 21 means that gcd(63/21,(n+120)/21) = 1 gcd(3,p) = 1 => p can't be a multiple of 3 but we chose 57. Similarly: gcd(n+63,120) = 60 => gcd((n+63)/60,2) = 1 => gcd(q,2) = 1 => q is odd. So coming back to: q = (7p - 19)/20 for p = 77, q = (77*7 - 19)/20 = 26 but q has to be odd. for p = 97, q = (97*7 - 19)/20 = (630 + 49 - 19)/20 = 660/20 = 33. It works. So, n = 60*33 - 63 = 1980 - 63 = 1917 sum_digits(n) = 18 Q2. Prove that 10^(3n+1) cannot be represented as a sum of 2 cubes(integers). S2. Typically when dealing wi...

Q1. S1. First do angle chasing and get all angles. Now apply Sine rule in AMC. AC/sin(150) = MC/sin(7) sin(150) = sin(180-150) = sin(30) = 1/2 => AC = MC/2sin(7)___[1] In BMC: BC/sin(M) = MC/sin(97-M) BC = AC and sin(97-M) = sin(90 + 7 - M) = cos(7-M) = cos(M-7) => AC = MC.sin(M)/cos(M-7)  ____[2] Using [1] and [2]: 1/2sin(7) = sin(M)/cos(M-7) => 2sin(7)sin(M) = cos(M-7) = cosMcos7 + sinMsin7 sin7sinM = cos7cosM tan7 = cotM = tan(90-M) => M = 83 = answer Q2. S2. Isosceles triangle, altitude bisects the base. x^2 = h^2 + 225 y^2 = h^2 + 36 (x+y)(x-y) = 189 = 3^3*7 189 = 189 * 1 = 63 * 3 = 21 * 9 = 7 * 27 case 1: x+y = 189, x-y=1 => 2x = 190 => x = 95, y=94 => AD = CD = 95, perimeter = 190 + 30 = 220 case 2: x+y = 63, x-y=3 => 2x = 66 => x = 33, y=30 => AD = CD = 33, perimeter = 66 + 30 = 96 case 3: x+y = 21, x-y=9 => 2x = 30 => x = 15, y=6 => AD = CD = 15, invalid since AC = 30 case 4: x+y = 27, x-y=7 => 2x = 34 => x = 17, y=10 => AD =...

Q1. Is there a solution to 5x = 3 mod 11? Yes. Why? Since 5,11 are co-prime we can always find 'x' s.t. 5.x = r mod 11 for all 'r' from 0 to 10. Q2. Simple way to solve x = 1 mod 3 x = 2 mod 5 x = 3k + 1 3k + 1 = 2 mod 5 3k = 1 mod 5 => k = 2 mod 5 => k = 5t + 2 => x = 3(5t + 2) + 1 = 15t + 7 = 7 mod 15 Similarly x = 0 mod 8 x = 59 mod 125 125p + 59 = 0 mod 8 5p + 3 = 0 mod 8 5p = 5 mod 8 p = 1 mod 8 So x = 125*1 + 59 = 184

Let phi(n) = totient function, i.e. number of integers upto 'n' which are co-prime to it. Then: a^(phi(n)) = 1 mod 'n' if 'a' and 'n' are co-prime. You can see the similarity with Fermat's little theorem: a^(p-1) = 1 mod 'p' where 'p' is a prime and 'a','p' are co-prime. As we know all integers upto 'p' are co-prime to it.

 newton's method

Q1. Does there exist an integer x satisfying the following conditions? 10x = 1 mod 21 5x = 2 mod 6 4x = 1 mod 7 S1. No. Why? [1] =>  10x = 1 mod 21 => 10x = 1 mod 3 and 10x = 1 mod 7 => x = 1 mod 3 and 3x = 1 mod 7 => 36x = 12 mod 7 => x = 5 mod 7 So we have x = 1 mod 3 and x = 5 mod 7 [2] =>  5x = 2 mod 6 => 5x = 6k + 2 => 5x = 2 mod 3 => 2x = 2 mod 3 => 4x = 4 mod 3 => x = 1 mod 3 And 5x = 0 mod 2 => x = 0 mod 2 So we have x = 1 mod 3 x = 5 mod 7 x = 0 mod 2 [3] => 4x = 1 mod 7 => 8x = 2 mod 7 => x = 2 mod 7 Which is a contradiction to existing solutions. Q2. For how many n between 1 and 2520 do we have n^3 = 1 mod 2520? S2. 2520 = 2^3 * 3^2 * 5 * 7 So n^3 = 1 mod 8 => n = 1 mod 8 n^3 = 1 mod 9 => n = 1,4,7 mod 9 n^3 = 1 mod 5 => n = 1 mod 5 n^3 = 1 mod 7 => n = 1,2,4 mod 7 For each of those options there is a unique solution (using CRT) between 1 and 2520. So total 9 valid solutions. Q3. Find positive integers a,b,c ...