ARML Sample Tryout Questions A - 1 Let N be a three-igit base ten integer whose mile igit is 0. N is a multiple of an N the quotient equals the sum of the squares of the igits of N. Compute N. A - If the thir an fourth terms in an arithmetic progression are increase by an 7 respectively, then the first four terms form a geometric progression. Fin all possible values of the fourth term of the arithmetic progression. C - 1 Let P be the set of consecutive integers > 90. (Specifically, P = {90, 91, 9, 9, 94, }). Determine the 5 r member of the set P that is not ivisible by,, 4 or 8. C - A square is inscribe in a 90 sector of a circle so that two of its vertices are on the circle an the other two vertices are on the pair of perpenicular raii. If a sie of the square is 6 meters long an the area of the region boune by the entire circle is kπ square meters, fin the value of k.
ARML Sample Tryout Questions D - 1 How many integers from 1 to 100 inclusive have no repeate prime factors? D - The equation ( x ) ( x ) ( x ) sin + cos 1 + tan = 0 has exactly 8 solutions in the range 0 < x < M, where x an M are given in egrees. Fin the smallest possible value of M. F - 1 4 5 If sin t = (0 < t < 90 ), compute the value of ( logsin t) + ( logcot t). 9 F - An n-gon has six times as many iagonals as sies. Let p enote the number of iagonals in a polygon with (n 1) sies. Let q enote the number of iagonals in a polygon with (n + 1) sies. Determine the number of ifferent prime factors of the prouct pq.
A 1 Let N be a three-igit base ten integer whose mile igit is 0. N is a multiple of N an the quotient equals the sum of the squares of the igits of N. Compute N. 100x Let the three-igit integer N be x0y. Then x + y = an + y = x + y. Thus, 100x + ( x) = 99x + = (9x + 1) = x + ( x) 9x + 1 = x + ( x) x 1x + 10 = (x 15)(x 8) = 0 x = 8 an y = N = 80 Alternative: Here is a list of the -igit integers ivisible by an the sum of the squares of their igits {09 (85), 08 (7), 407 (65), 506 (61), 605 (61), 704 (65), 80 (7), 90 (85)} Only in the highlighte pair is the first number a multiple of the secon. A If the thir an fourth terms in an arithmetic progression are increase by an 7 respectively, then the first four terms form a geometric progression. Fin all possible values of the fourth term of the arithmetic progression. Let the terms of the A.P. be a, a+, a+, a+ (the usual suspects) Then the terms of the G.P. are: t 1 = a, t = ar = a +, t = ar = a + + an t 4 = ar = a + + 7 t 4 ar ( 1) = +7 t ar ( 1) = 4 Thus, t a ( r 1) a ( r 1)( r + r + 1) 1 +7 = = = r + r+ = (Eqtn #1) t a( r 1) a( r 1) t a+ + r = r = (Eqtn #) t a+ t t r = t 1 t = 1 t = t t (a + ) = a(a + + ) = a (Eqtn #) Substituting for a in equation #, + + r + 4 + 4 ( + ) + = = = = (Eqtn #4) or = - + ( + ) + = - a = AP:, 0, -, -4 an the GP:, 0, 0, However, the latter is not a GP an = - is extraneous. + + + 7 Substituting for r in #1, + + 1 = ( + ) + ( + ) + = ( + 7) + 6 + 4 = + 7 = 4 an a = 8 a + = 0 [AP: 8, 1, 16, 0 (with = +4) an the GP: 8, 1, 18, 7 (with r = /) ] Detaile Solutions ARML Tryout Questions 008
A continue The above algebraic blizzar coul have been avoie if the original AP ha been represente as a, a, a, a + to simplify the representation of the r an 4 th terms of the GP. If the GP is a, a, a +, a + + 7, then equating the quotients of consecutive terms, i.e. he multiplier r, an cross multiplying, we have (a ) = (a )(a + ) an (a + ) = (a + + 7)(a ). + 4 = a an + 7 = a 4 Subtracting, = a 4 or a = + 4 Substituting, + 4 = ( + 4) 8 = ( 4)( + ) = 0 = - or 4 This prouces the same answers as above with a lot less stress. Don t always jump into a problem assuming that trying the usual suspects will shorten the trial. C 1 Let P be the set of consecutive integers > 90. Specifically, P = {90, 91, 9, 9, 94, }. Determine the 5 r member of the set P that is not ivisible by,, 4 or 8. Examining consecutive integers, every n integer is a multiple of, every r integer is a multiple of, etc. Since the LCM of,, 4 an 8 is 4, we examine blocks of consecutive integers. Every block of 4 integers contains exactly 8 integers that are not ivisible by any of these 4 integers. 1 4 5 6 7 8 9 10 1 1 14 15 16 17 18 19 0 1 4 5 6 7 8 9 0 1 4 5 6 Notice the repeating gaps of an 1 (or vice versa) epening on where you start. Since the quotient 5 8 prouces a quotient of 1 an a remainer of 5, we want the 5th integer in the n block of 4 integers, starting with 90. If 90 = (4) + 18 starts the first block, then 4(4) + 18 = 84 starts the n block. Since 91, 95, an 97 start the first block, the first integer is start +1 an the spacing is 1. 85 89 841 845 847 C A square is inscribe in a 90 sector of a circle so that two of its vertices are on the circle an the other two vertices are on the pair of perpenicular raii. If a sie of the square is 6 meters long an the area of the region boune by the entire circle is kπ square meters, fin the value of k. Let the coorinates of O be (0, 0). Then the coorinates of P are (6, )an the raius r of the circle is OP. The area of circle O is πr. Thus, k = 7 + 18 = 90 O 6 45 45 P
D 1 How many integers from 1 to 100 inclusive have no repeate prime factors? We must eliminate any multiples of perfect squares from 1 to 100 inclusive. Between 1 an 100 inclusive, the perfect squares are 4, 9, 5 an 49 an there are 5 multiples of 4, multiples of 9, 4 multiples of 5 an multiples of 49. The only integers in more than one multiple list will be multiples of 4 9 = 6 or 4 5 = 100. The integers occurring in more than one multiple list are 6, 7 an 100. Thus, we have 100 (5 + + 4 + ) + ( + 1) = 61 D The equation ( x ) ( x ) ( x ) sin + cos 1 + tan = 0 has exactly 8 solutions in the range 0 < x < M, where x an M are given in egrees. Fin the smallest possible value of M. Each term is nonnegative an, therefore, each term must be zero to prouce a sum of 0. Thus, the first term requires x to be coterminal with 60 or 10. The secon term requires x to be coterminal with 60 or 00. The thir term requires x to be coterminal with 60 or 40. The only common groun is 60 + 60n. If the first solution is 60 + 0 60, then the 8 th solution is 60 + 7 60 = 580. F 1 If sin t 4 5 (0 9 t 90 ) = < <, compute the value of ( log sin t) ( log cot t) +. Since sin t + cos t = 1 an t is in the first quarant, we have cost = 1 9. ( log sin t) ( log cot t) + = log (sin t cot t) = = log = - log cost ( ) F - An n-gon has six times as many iagonals as sies. Let p enote the number of iagonals in a polygon with (n 1) sies. Let q enote the number of iagonals in a polygon with (n + 1) sies. Determine the number of ifferent prime factors of the prouct pq. nn ( ) We require n such that 6n =. This quaratic equation has a solution of 15 (0 is extraneous). Thus, a 14-gon has 77 iagonals an a 16-gon has 104 iagonals an pq = 77(104) = 7 1 4