Math IOQM

Let abc be a three digit number with nonzero digits such that a^2 + b^2 = c^2. What is the largest possible prime factor of abc? Solution: Only single digit Pythagorean triplet is 3,4,5 so the number could be 345 or 435. 345 = 3 x 5 x 23 435 = 3 x 5 x 29 Answer: 29

7. On a clock, there are two instants between 12 noon and 1 PM, when the hour hand and the minute hand are at right angles. The difference in minutes between these two instants is written as a + b/c , where a, b, c are positive integers, with b < c and b/c in the reduced form. What is the value of a + b + c ? Solution: Assume that clock is made up of 360 degrees. Hour hand moves 360/12 = 30 degrees per hour => 30/60 degrees per minute => 0.5 degrees per minute. Minute hand moves 360 degrees per hour => 360/60 degrees per minute => 6 degrees per minute. Speed of minute hand relative to hour hand  = 6 - 0.5 = 5.5 degrees per minute. The angle between them would become 90 after 90/5.5 minutes. And it would become 270 after 270/5.5 minutes. Difference between those 2 instants  = (270-90)/5.5 = 180/5.5 = 32 + 8/11 a = 32, b = 8, c = 11 a + b + c = 51

14. Find the smallest positive integer n ≥ 10 n \ge 10 such that n + 6 n + 6 is a prime and 9 n + 7 9n + 7 is a perfect square.  Solution: p = n + 6 9n + 7 = k^2 Solving we get k^2 = 9p - 47 Given n >= 10 So starting value of p is 17(16 is not prime). Each time we increase p by 1, k^2 increases by 9. p = 17, k^2 = 106 (not valid) keep trying for p = 23,29,31,37,41,43,47,53,59 at p = 59, k^2 = 484 => k = 22 n = 53 answer.

Q. One day I went for a walk in the morning at x x minutes past 5 O’ clock, where x x is a two-digit number. When I returned, it was y y minutes past 6 O’ clock, and I noticed that (i) I walked exactly for x x minutes and (ii) y y was a 2-digit number obtained by reversing the digits of x x . How many minutes did I walk? [PRE-RMO–2019] This is quite simple actually. Let x = ab and y = ba => x = 10a + b, y = 10b + a Now x minutes past 5 = 300 + x since 5 hours have 300 minutes. So 360 + y - (300 + x) = x => 60 + y = 2x => 30 + y/2 = x => 30 + (10b+a)/2 = 10a + b => 30 + 5b + a/2 = 10a + b => a = 2/19(30 + 4b) => a = (60 + 8b)/19 Now a,b are between 0,9 So just keep putting values of 'b' to find an answer. On b = 2, a = 4 So x = 42 and y = 24

Prerequisites: Divisibility by 13. 11. Find the largest value of a b a^b such that the positive integers a , b > 1 a, b > 1 satisfy a b ⋅ b a + a b + b a = 5329 a^b \cdot b^a + a^b + b^a = 5329 Solution: Rewrite the given equation a b   b a    +    a b    +    b a    =    5329 a^b\,b^a \;+\; a^b \;+\; b^a \;=\; 5329 as ( a b + 1 ) ( b a + 1 )    =    5329 + 1    =    5330. (a^b+1)(b^a+1)\;=\;5329 + 1 \;=\;5330. Thus if we set x = a b x=a^b and y = b a , y=b^a, then ( x + 1 ) ( y + 1 ) = 5330. (x+1)(y+1)=5330. Next one factors 5330 5330 (which is 2 × 5 × 13 × 41 2 \times 5 \times 13 \times 41 ) and looks for factor‐pairs ( p , q ) (p,q) of 5330 5330 such that x = p − 1 x=p-1 and y = q − 1 y=q-1 can both be positive perfect powers a b a^b and b a b^a . Among the possible divisors, the only workable pairs turn out to be ( p , q )    =    ( 65 ,   82 ) and ( 82 ,   65 ) , (p,q) \;=\; (65,\,82)\quad\text{and}\quad(82,\,65), which give x = 64 ,    y = 81 or x = 8...

 Five persons are to be seated around a circular table they are wearing badges of consecutive numbers 1,2,3,4,5. In how many ways can they be seated such that no two consecutive numbered badge wearing people are seated next to each other ? Solution: Try to arrange it manually. Let's first put one of 2,3,4 because each of them has 2 adjacent numbers which we need to avoid whereas for 1 and 5 there is only 1 number to avoid. Let's start with 2.       2 Now adjacent numbers can be only 4,5             2      4           5 Now 3 can't go next to 4 so:             2      4           5         1    3 If you flip 4,5 initially             2      5           4         3    1 So effectively, ther...

In how many ways can a pair of parallel diagonals of a regular polygon of 10 sides be selected? Hint: First try for an octagon. Visual solution . Solution : If the decagon is P 1 ⋯ P 10 P_1 \cdots P_{10} then there are two types of sets of parallel diagonals. One is typified by P 3 P 10 ,   P 4 P 9 ,   P 5 P 8 P_3P_{10},\ P_4P_9,\ P_5P_8 (parallel to the side P 1 P 2 P_1P_2 ) and the other by P 1 P 3 ,   P 10 P 4 ,   P 9 P 5 P_1P_3,\ P_{10}P_4,\ P_9P_5 and P 8 P 6 P_8P_6 . There are five sets of each type. If you are counting unordered sets of parallel diagonals, then the answer is 5 ( 3 2 ) + 5 ( 4 2 ) = 45.