2015 Kansas MAA Undergraduate Mathematics Competition: Solutions! 1! + 2! + 3! ! 1! + 2! + 3! ! mod 10 = 1! + 2! + 3! + 4! mod 10 = 3.

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1 215 Kasas MAA Udergraduate Mathematcs Competto: Solutos! 1. Show that the equato x x + 216y 3 = 3 has o teger solutos x, y Z. Soluto: Gve ay tegers x, y Z, t s clear that 216y 3 wll be a eve teger. Further, f x s eve the x x s the sum of two eve tegers, ad hece eve, whle f x s odd the x x s the sum of two odd umbers, ad hece also eve. I ay case, the sum x x + 216y 3 wll be eve for ay x, y Z, ad hece the gve equato ca have o teger solutos. 2. Fd the frst dgt (the oes dgt the sum 1! + 2! + 3! ! Soluto: Notce that the oes dgt the above sum s gve by the sum mod 1. Sce! = mod 1 for all 5, t follows that 1! + 2! + 3! ! mod 1 = 1! + 2! + 3! + 4! mod 1 = 3. Thus, the oes dgt of the above sum s Determe whether there exsts a fte sequece (a of postve real umbers such that the seres a 1 + a a coverges. =1 Soluto: Sce all of the a are postve, t follows that, for each N, a 1 + a a a 1 ad hece the seres dverges by the comparso test. 4. Let f(x be a strctly postve cotuous fucto. Evaluate the tegral f(x f(x + f(4 x dx. Soluto: Lettg I deote the value of the above tegral, we have I = f(x + f(4 x f(4 x f(x + f(4 x dx = 4 f(4 x f(x + f(4 x dx.

2 Mag the substtuto y = 4 x we fd f(4 x f(x + f(4 x dx = f(y 4 f(4 y + f(y dy = I so that, combg wth the prevous detty, we fd I = 4 I I = Suppose that f, g : N N, f s oto, g s oe-to-oe, ad f( g( for all N. Prove that f( = g( for all N. Soluto 1: Sce f s oto, there exsts a 1 N such that f( 1 = 1. Sce g( 1 f( 1 ad g( 1 N, t follows that g( 1 = 1 = f( 1 Furthermore, sce g s oe-to-oe, t follows that 1 s the uque atural umber wth ths property. Smlarly, sce f s oto there exsts a 2 N such that f( 2 = 2. Sce g( 2 f( 2 t follows that g( 2 {1, 2}. Usg that g s oeto-oe, t follows that g( 2 1 (sce 1 2 ad hece that g( 2 = 2. Thus, g( 2 = f( 2 = 2 ad, aga by the fact that g s oe-to-oe, 2 s the uque atural umber wth ths property. Cotug by ducto, t follows that, for each N there exsts a uque N such that f( = g( =. It ow remas to show that { } =1 = N, but ths s clear sce f m N \ { } =1 the g(m N ad hece, by above, there exsts a uque N such that g( = f( = g(m. Sce g s oe-to-oe, t follows that m = ad hece that { } =1 = N, as clamed. Together the, t follows that f( = g( for every N. Soluto 2: By cotradcto. Assume there exsts m N wth g(m < f(m. Sce f s oto, there exsts m 1 N wth f(m 1 = g(m. Sce g s oe-to-oe, ad g(m = f(m 1, g(m 1 < f(m 1 = g(m. Now assume that there exsts postve tegers m, m 1,... m wth g(m < g(m 1 < < g(m. Sce f s oto, there exsts m +1 N wth f(m +1 = g(m. Sce g s oe-to-oe, g(m +1 < f(m +1 = g(m. By ducto, we obta a sequece of postve tegers {m } wth the property that {g(m } s a strctly decreasg sequece. Sce {g(m } N, ths s a cotradcto. 6. For each N, show that ( = = ( 3 + = 7.

3 Soluto: Frst, rewrte the sum as = ( 3 = ( 3. By the bomal theorem, we ow ( = 3 = (1 + 3 = 4 ad hece, usg the bomal theorem aga, the gve sum s equal to as clamed. = ( 3 4 = (4 + 3 = 7, 7. Let f : [, 1] R be a tegrable fucto (ot ecessarly cotuous. Prove that f f(t 2 for all t [, 1] the there exsts a uque soluto x [, 1] of the equato 3x 1 = x f(tdt. Soluto: Defe the fucto g : [, 1] R by g(x = 3x 1 x f(tdt. Sce f s tegrable, t follows by the fudametal theorem of calculus that g s cotuous o [, 1]. Now, otce that g( = 1 ad that g(1 = 2 f(tdt 2 2 dt =. If g(1 =, the x = 1 s a soluto to the equato. If g(1 >, by the termedate value theorem, we have that there exsts some c (, 1 such that g(c =. To see that the soluto s uque, we clam that g s a strctly creasg fucto o [, 1]. Ideed, gve ay a, b [, 1] wth a < b we have g(b g(a = 3(b a b a f(tdt 3(b a b a 2 dt = b a >. Thus, g s strctly creasg o [, 1] ad hece c s uque, as clamed. 8. Suppose f : [, R s a cotuous, o-egatve fucto. Suppose that f(x + 1 = (1/2f(x for all x, ad tegral f(x dx exsts ad fd ts value. f(x dx = 1. Show that the

4 Soluto: Frst, we clam that the gve mproper tegral coverges. Ideed, ote that sce f s o-egatve t follows that the fucto g : R [, gve by g(t = t f(xdx s a o-decreasg fucto ad, furthermore, gve ay t 1 we have mples g(t t = t f(xdt = = f(x + dx t = (1/2 f(xdx 1 (1/2 = 2. Thus, g s a mootoe o-decreasg fucto that s bouded above. It follows that lm t g(t exsts, whch proves the gve mproper tegral coverges, as clamed. Now, to evaluate the tegral, t suffces to calculate the lmt lm t g(t. Sce we ow the lmt exst, we ca tae the lmt alog the atural umbers. To ths ed, for each N otce that g( = f(x + dx = (1/2 f(xdx = 1 (1/2 ad hece that It follows that = = = lm g( = 1 (1/2 = 2. = f(xdx = Determe, wth proof, all polyomals satsfyg satsfyg P ( = ad P (x = (P (x for all x. Soluto: We clam the oly polyomal satsfyg the gve propertes s P (x = x. To see ths, frst otce that the codto P ( = mples so that P (1 = 1. Smlarly, we fd P ( = P ( = 1. P ( = P ( = P (( = P ( = ( =

5 By ducto, t follows that f we defe the recursve sequece a 1 = 1, a +1 = a for all N the P (a = a for all N. Furthermore, the sequece {a } =1 s strctly creasg sce for all N we have a +1 a = a a = (a 1/ /4 >. It follows that the fucto G(x = P (x x s a polyomal wth ftely may dstct real roots. Sce a o-trval polyomal of degree m ca have at most m real roots by the fudametal theorem of algebra, t follows that G(x = for all x R,.e. P (x = x for all x R, as clamed. 1. Suppose that you have a 2 2 grd wth a sgle 1 1 square removed. Prove that the remag squares ca be tled wth L-shaped tles cosstg of three 1 1 tles - that s, a 2 2 tle wth a sgle 1 1 square removed. Soluto: We prove ths by ducto o. Clearly, ths s possble f = 1, sce a 2 2 grd wth exactly oe square removed s precsely the shape of a gve L-shaped tle. Now, suppose that, for some gve N t s ow that that ths s possble for ay gve 2 2 grd wth exactly oe square removed ca be covered by such L-shaped tles. Cosder the a grd ad otce ths ca be decomposed to a sub-grds a uque way. Furthermore, we ca cosder the mddle 2 2 sub-grd of the grd made up of exactly oe square from each of the four 2 2 sub-grds. Sce there s exactly oe square mssg from the gve grd, t follows that exactly oe of these 2 2 sub-grds has exactly oe square mssg (by hypothess ad hece ca be tled by such L-shaped tles by the ducto hypothess. After coverg ths 2 2 subgrd, t follows that the ceter 2 2 grd has exactly oe square mssg, ad hece ca be tled wth exactly oe L-shaped tle. The remag sub-grds ow each have exactly oe square covered, ad hece ca each be tled by the gve L-shaped tles by the ducto hypothess. Thus, the gve grd ca be tled by such L-shaped tles. The proof s thus complete by mathematcal ducto.