Algeba Toolkit Rules of Thumb. Make sue that you can pove all fomulas you use. This is even bette than memoizing the fomulas. Although it is best to memoize, as well. Stive fo elegant, economical methods. Look fo symmety. Look fo invaiants. Sometimes symmety and economy conflict. Fo example, it is not always best to squae both sides of equations like p A + p B = C. Sometimes it makes bette sense to change it to p A = C p B, and then squae. (Example: 2006B/5). Zeo is you best fiend. One is you second-best fiend. Neve multiply out, unless you have to. Always look fo factoizations, instead. To know the zeos of a polynomial is to know the polynomial. Look fo telescoping tems. Don t woy about being cleve. Dumb methods wok, too. Low-tech is bette than high-tech. But if things stat getting too dity and messy, step back and ask youself if thee is a bette way to poceed. 2 Aithmetic. Know all squaes up to d p 203e 2 Know all pefect powes < 203 Know all factoials up to 0! Know the fist 9 o 0 ows of Pascal s Tiangle Facto the cuent yea! Know you pimes, at least unde 00, ideally up to 200 o so. Be able to mentally squae numbes and multiply numbes, using the factoization x 2 y 2 =(x y)(x + y). Fo example, to compute 73 2, use 73 2 3 2 = 70 76 = 4900 + 420 = 5320. 3 Factoing. 00 = 7 3 (x + y) 2 = x 2 + 2xy + y 2. (x y) 2 = x 2 2xy + y 2. (x + y) 3 = x 3 + 3x 2 y + 3xy 2 + y 3 = x 3 + y 3 + 3xy(x + y). (x y) 3 = x 3 3x 2 y + 3xy 2 y 3 = x 3 y 3 3xy(x y). x 2 y 2 =(x y)(x + y). x n y n =(x y)(x n + x n 2 y + x n 3 y 2 + + y n ) fo all n. x n + y n =(x + y)(x n x n 2 y + x n 3 y 2 + y n ) fo all odd n (the tems of the second facto altenate in sign). a 3 + b 3 + c 3 3abc =(a + b + c)(a 2 + b 2 + c 2 ab bc ac)
4 Polynomials. FTA, conjugate complex solutions Completing the squae Facto/Remainde theoem Relationship between oots and coefficients 5 Sequences and Seies. Aithmetic seies: The best sum fomula is S = n( f ist + last)/2, since it tends to get you thinking about aveages which gets you thinking about symmety. Know the fomulas fo sums of squaes and cubes. Be able to deive fomulas fo sums of highe powes if needed. TELESCOPING is the mothe of all sequence/summation methods. Know the classic telescopes, fo example n(n + ) = and (n + )! n! = n n!. And don t foget that telescoping can be used with n n + poducts as well. You can always tun poducts into sums with logaithms. You don t eally need to lean the calculus of diffeences to handle ecuence sequence poblems. Almost always, low-tech methods (usually involving telescoping!) suffice. Even though it is eally a calculus topic, you should be able to pove that the hamonic seies +/2+/3+ /4 + diveges (without calculus). Likewise, you should be able to pove that + /4 + /9 + /6 + conveges, again, without calculus. The sum of the fist n odd numbes is equal to n 2. This is a emakably fuitful fact. 6 Miscellaneous. Floo and ceiling functions Floo functions and counting multiples; lattice points. Absolute value Logaithms: Be able to pove the all-impotant, and easy-to-memoize fomula log a blog b c = log a c. Complex numbes: Cis fom, sums of oots of unity, take absolute value of both sides. Mostly Algeba 2004:7. Let C be the coefficient of x 2 in the expansion of the poduct Find C. 2 2008 II:. Find the emainde when is divided by 000. ( x)( + 2x)( 3x) ( + 4x)( 5x). 00 2 + 99 2 98 2 97 2 + 96 2 + + 4 2 + 3 2 2 2 2 3 Fo which intege n is /n closest to p,000,000 p 999,999? 4 BAMM 2000. Let x = 3p 000 3p 999. What intege is closest to /x? 5 Solve x 4 + x 3 + x 2 + x + = 0. 6 2005 II: 7. Let Find (x + ) 48. x = 4 ( p 5 + )( 4p 5 + )( 8p 5 + )( 6p 5 + ).
7 Facto z 5 + z +. 8 Solve z 6 + z 4 + z 3 + z 2 + = 0. 9 2004:3. The polynomial P(x)=( + x + x 2 + + x 7 ) 2 x 7 has 34 complex oots of the fom z k = k [cos(2pa k )+isin(2pa k )],k =,2,3,...,34, with 0 < a apple a 2 apple a 3 apple apple a 34 < and k > 0. Given that a +a 2 +a 3 +a 4 +a 5 = m/n, whee and ae elatively pime positive integes, find m + n. 0 (USAMO 975) If P(x) denotes a polynomial of degee n such that P(k)=k/(k + ) fo k = 0,,2,...,n, detemine P(n + ). Find the minimum value of xy + yz + xz, given that x,y,z ae eal and x 2 + y 2 + z 2 =. No calculus, please! 2 Find all intege solutions (n,m) to n 4 + 2n 3 + 2n 2 + 2n + = m 2. 3 If x 2 + y 2 + z 2 = 49 and x + y + z = x 3 + y 3 + z 3 = 7, find xyz. 4 2008 II:7. Let,s,t be the oots of 8x 3 + 00x + 2008 = 0. Find ( + s) 3 +(s +t) 3 +(t + ) 3. 5 Find all eal values of x that satisfy (6x 2 9) 3 +(9x 2 6) 3 =(25x 2 25) 3. 6 2002 II:9. Given that is a complex numbe such that z + z = 2cos3, find the least intege that is geate than z 2000 +. z 2000 7 (Cux Mathematicoum, June/July 978) Show that n 4 20n 2 + 4 is composite when n is any intege. 8 2008:4. Thee exist unique positive integes x and y that satisfy the equation x 2 + 84x + 2008 = y 2. Find x + y. 9 2005 II:. Let m be a positive intege, and let a 0,a,...,a m be a sequence of eal numbes such that a 0 = 37,a = 72,a m = 0, and a k+ = a k 3 a k fo k =,2,...,m. Find m. 20 Pove that s is ational. 729 22 7 2 + 22 355 7 3 2 + 355 3 729 2 2 2007I:4. A sequence is defined as follows: a = a 2 = 3, and fo n 2 a n+ a n = a 2 n + 2007. Find the geatest intege that does not exceed a2 2006 +a2 2007 a 2006 a 2007 22 (E. Johnston) Let S be the set of positive integes which do not have a zeo in thei base-0 epesentation; i.e., S = {,2,...,9,,2,...,9,2,...}. Does the sum of the ecipocals of the elements of S convege o divege? 0000 23 Let A = Â n= p. Find bac without a calculato. n
24 2006II:5. Given that x,y,z ae eal numbes that satisfy: x = y 2 6 + z 2 6 y = z = z 2 25 + x 2 36 + x 2 25 y 2 36 and that x + y + z = m/ p n, whee m and n ae positive integes and n is not divisible by the squae of any pime, find m + n. Numbe Theoy Toolkit Pimes and Divisibility FTA (Fundamental Theoem of Aithmetic) Evey positive intege can be factoed uniquely into pimes. This factoization is often called the PPF (pime-powe factoization). GCD and LCM Geatest common diviso and least common multiple, espectively. If GCD(a, b) = we say that a and b ae elatively pime and denote this by a? b. If g a and g b then g (a + b),g (a b), etc. In fact, g (ax + by) fo any integes x,y. The expession ax + by is called a linea combination of a and b. Consequently, evey pai of consecutive positive integes is elatively pime. Same is tue of consecutive odd integes. Consecutive even integes have a GCD of 2. Linea combination ule: The GCD of a and b is a linea combination of a and b. In fact, it is smallest positive linea combination of a and b. Thus if a? b, thee exist integes x,y such that ax + by =. The coefficients x, y above can be found by pefoming the Euclidean algoithm backwads. In pactice, they can be found by inspection fo small numbes. Fo example, 5? 7 and 3 5 +( 2) 7 =. (Don t woy if you don t know the Euclidean algoithm.) Numbe of divisos is denoted by t(n) and includes and n in the count. So t(p)=2 fo any pime p and in geneal, t(p a q b )=(a + )(b + ) when the numbe is in PPF fom. (Some books use d(n) instead of t(n) but Geek lettes ae so much moe sophisticated.) Sum of divisos is denoted by s(n) and includes and n. The fomula is athe simple. Make sue you can see why it s tue (ty simple examples such as n = 2 and n = 300): s(p a q b )=( + p + p 2 + + p a )( + q + q 2 + + q b ). 2 Modula Aithmetic Conguence notation The notation x y (mod m) means the following equivalent things. All of them ae woth intenalizing. x y is a multiple of m x has the emainde y when divided by m, povided that 0 apple y < m. x = mk + y fo some intege K Thus x 3 (mod 4) means that x has a emainde of 3 when divided by 4, and it also means that you can wite x in the fom 4k + 3. It is common to use small negative numbes in conguences, especially. Fo example, 99 (mod 4). We could have witten 99 3 (mod 4), but the fome is just as tue, and often moe useful. Conguence algeba Conguence notation is incedibly useful because you can add, subtact, and multiply (but not divide) both just as you can with odinay equality. Fo example,
If a b (mod m) and x y (mod m), then a + x b + y (mod m) and ax by (mod m) 00000 0 (mod ), since 0 (mod ) and thus 0 5 ( ) 5 (mod ). Divisibility Rules You should know the divisibility ules fo 2, 3, 4, 5, 6, 8, 9, and, using conguences o othe methods. Mod n analysis is a cucial stat of any poblem. Put it into mod 2, 3, and 4 filtes, sometimes moe. Fo example, you should veify that all pefect squaes ae conguent to 0 o (mod 4) only. (A mod 2 analysis is also called a paity analysis.) Femat s Little Theoem states that if p is a pime, and a is not a multiple of p, then a p (mod p). 0 Fo example, 4 2 (mod 3). 3 Diophantine Equations Use of paity, mod 3, mod 4 analysis Linea equations x 2 y 2 = n. The AIME uses this humble equation in endless ways. Become intimate with it. Pythagoean equation x 2 + y 2 = n Mostly Numbe Theoy Show that if a 2 + b 2 = c 2, then 3 ab. 2 If x 3 + y 3 = z 3, show that one of the thee must be a multiple of 7. 3 Make sue that you know why 00! ends in 24 zeos and 000! ends in 249 zeos. Can n! end with n/4 zeos? 4 Find the smallest positive intege n such that t(n)=0. 5 Find the emainde when 2 000 is divided by 3. (This was an AHMSE poblem when I was in high school.) 6 2006II:3. Let P be the poduct of the fist 00 positive odd integes. Find the lagest intege k such that P is divisible by 3 k. 7 BAMO 999. Pove that among any 2 consecutive positive integes thee is at least one which is smalle than the sum of its pope divisos. (The pope divisos of a positive intege n ae all positive integes othe than and n which divide n. Fo example, the pope divisos of 4 ae 2 and 7.) 8 BAMO 2000. Pove that any intege geate than o equal to 7 can be witten as a sum of two elatively pime integes, both geate than. (Two integes ae elatively pime if they shae no common positive diviso othe than. Fo example, 22 and 5 ae elatively pime, and thus 37 = 22+5 epesents the numbe 37 in the desied way.) 9 What kind of numbes can be witten as the sum of two o moe consecutive integes? Fo example, 0 is such a numbe, because 0 = + 2 + 3 + 4. Likewise, 3 = 6 + 7 also woks. 2000II:2. A point whose coodinates ae both integes is called a lattice point. How many lattice points lie on the hypebola x 2 y 2 = 2000 2? 2 2000II:4. What is the smallest positive intege with six positive odd intege divisos and twelve positive even divisos?
3 BAMM 999. How many odeed pais (x,y) of integes ae solutions to xy x + y = 99? 4 99. Find x 2 + y 2 if x,y 2 N and xy + x + y = 7, x 2 y + xy 2 = 880. 5 985:3. The numbes in the sequence 0,04,09,6,... ae of the fom a n = 00 + n 2, whee n =,2,3,.... Fo each n, let d n be the geatest common diviso of a n and a n+. Find the maximum value of d n as n anges though the positive integes. 6 200II:0. How many positive intege multiples of 00 can be expessed in the fom 0 j 0 i, whee i, j ae integes and 0 apple i < j apple 99? 7 2003II:0. Two positive integes diffe by 60. The sum of thei squae oots is the squae oot of an intege that is not a pefect squae. What is the maximum possible sum of the two integes? 8 Find the last thee digits of 7 9999. 9 Let {a n } n 0 be a sequence of integes satisfying a n+ = 2a n +. Is thee an a 0 so that the sequence consists entiely of pime numbes? 20 (USAMO 979) Find all non-negative integal solutions (n,n 2,...,n 4 ) to n 4 + n 4 2 + + n 4 4 =,599. 2 2005:2. Fo positive integes let t(n) denote the numbe of positive intege divisos of n including and n. Define S(n)=t()+t(2)+ + t(n). Let a denote the numbe of positive integes n apple 2005 with S(n) odd, and let b denote the numbe of positive integes n apple 2005 with S(n) even. Find a b. 22 2007II: 7. Fo a cetain intege k thee ae exactly 70 positive integes n,n 2,...,n 70 such that and k divides n i fo all i such that apple i apple 70. Find the maximum value of n i /k fo apple i apple 70. k = b 3p n c = b 3p n 2 c = = b 3p n 70 c 23 2004II:0. Let S be the set of integes between and 2 40 whose binay expansions have exactly two s. If a numbe is chosen at andom fom S, the pobability that it is divisible by 9 is m/n whee m and n ae elatively pime positive integes. Find m + n. 24 2007II: 3. A tiangula aay of squaes has one squae in the fist ow, two in the second, and in geneal, k squaes in the kth ow fo apple k apple. With the exception of the bottom ow, each squae ests on two squaes in the ow immediately below (illustated in diagam). In each squae of the eleventh ow, a 0 o a is placed. Numbes ae then placed into the othe squaes, with the enty fo each squae being the sum of the enties in the two squaes below it. Fo how many initial distibutions of 0 s and s in the bottom ow is the numbe in the top squae a multiple of 3? 25 2008II:5. Find the lagest intege satisfying the following conditions: (i) n 2 can be expessed as the diffeence of two consecutive cubes; (ii) 2n + 79 is a pefect squae.