2014 AIME II Exam Solutions

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All of the real AMC 8, AMC 10, AMC 12, and AIME problems in our complete solution collection are used with official legal permission of the Mathematical Association of America (MAA).

1.

Abe can paint the room in 1515 hours, Bea can paint 5050 percent faster than Abe, and Coe can paint twice as fast as Abe. Abe begins to paint the room and works alone for the first hour and a half. Then Bea joins Abe, and they work together until half the room is painted. Then Coe joins Abe and Bea, and they work together until the entire room is painted. Find the number of minutes after Abe begins for the three of them to finish painting the room.

Concepts:ratefraction

Difficulty rating: 1890

Solution:

Abe paints 1900\frac{1}{900} of the room per minute, so Bea paints 321900=1600\frac{3}{2} \cdot \frac{1}{900} = \frac{1}{600} per minute and Coe paints 2900=1450\frac{2}{900} = \frac{1}{450} per minute. In the first 9090 minutes Abe paints 90900=110\frac{90}{900} = \frac{1}{10} of the room.

Abe and Bea together paint 1900+1600=1360\frac{1}{900} + \frac{1}{600} = \frac{1}{360} per minute, and they must bring the total from 110\frac{1}{10} up to 12,\frac{1}{2}, which takes 25360=144\frac{2}{5} \cdot 360 = 144 minutes. All three together paint 1360+1450=1200\frac{1}{360} + \frac{1}{450} = \frac{1}{200} per minute, so the remaining half of the room takes 12200=100\frac{1}{2} \cdot 200 = 100 minutes.

The total time is 90+144+100=33490 + 144 + 100 = 334 minutes.

2.

Arnold is studying the prevalence of three health risk factors, denoted by A,A, B,B, and C,C, within a population of men. For each of the three factors, the probability that a randomly selected man in the population has only this risk factor (and none of the others) is 0.1.0.1. For any two of the three factors, the probability that a randomly selected man has exactly these two risk factors (but not the third) is 0.14.0.14. The probability that a randomly selected man has all three risk factors, given that he has AA and B,B, is 13.\frac{1}{3}. The probability that a man has none of the three risk factors given that he does not have risk factor AA is pq,\frac{p}{q}, where pp and qq are relatively prime positive integers. Find p+q.p + q.

Difficulty rating: 2110

Solution:

Take a population of 100100 men and fill in a Venn diagram. Each of the three exactly-one regions contains 1010 men, and each of the three exactly-two regions contains 14.14. If xx men have all three factors, then the men with both AA and BB number x+14,x + 14, so the given conditional probability says xx+14=13,\frac{x}{x + 14} = \frac{1}{3}, giving x=7.x = 7.

The union of the three sets therefore contains 310+314+7=793 \cdot 10 + 3 \cdot 14 + 7 = 79 men, leaving 2121 with no risk factor. The men with risk factor AA number 10+14+14+7=45,10 + 14 + 14 + 7 = 45, so 5555 men do not have A.A.

The desired probability is 2155,\frac{21}{55}, which is in lowest terms, so p+q=21+55=76.p + q = 21 + 55 = 76.

3.

A rectangle has sides of length aa and 36.36. A hinge is installed at each vertex of the rectangle and at the midpoint of each side of length 36.36. The sides of length aa can be pressed toward each other keeping those two sides parallel so the rectangle becomes a convex hexagon as shown. When the figure is a hexagon with the sides of length aa parallel and separated by a distance of 24,24, the hexagon has the same area as the original rectangle. Find a2.a^2.

Difficulty rating: 2110

Solution:

In the hexagon, each side of length 3636 has folded at its midpoint into two bars of length 18.18. The two sides of length aa are 2424 apart, so each bar spans a vertical distance of 1212 and hence a horizontal distance of 182122=180=65.\sqrt{18^2 - 12^2} = \sqrt{180} = 6\sqrt{5}.

The line through the two midpoint hinges splits the hexagon into two congruent trapezoids with parallel sides aa and a+125a + 12\sqrt{5} and height 12,12, so the hexagon has area 2a+(a+125)212=24a+1445.2 \cdot \frac{a + (a + 12\sqrt{5})}{2} \cdot 12 = 24a + 144\sqrt{5}.

Setting this equal to the rectangle's area 36a36a gives 12a=1445,12a = 144\sqrt{5}, so a=125a = 12\sqrt{5} and a2=720.a^2 = 720.

4.

The repeating decimals 0.ababab0.abab\overline{ab} and 0.abcabcabc0.abcabc\overline{abc} satisfy 0.ababab+0.abcabcabc=3337,0.abab\overline{ab} + 0.abcabc\overline{abc} = \frac{33}{37}, where a,a, b,b, and cc are (not necessarily distinct) digits. Find the three-digit number abc.abc.

Difficulty rating: 2230

Solution:

Writing abab and abcabc for the two- and three-digit numbers, the decimals equal ab99\frac{ab}{99} and abc999.\frac{abc}{999}. Since 99=91199 = 9 \cdot 11 and 999=2737,999 = 27 \cdot 37, the common denominator is 273711=10989,27 \cdot 37 \cdot 11 = 10989, and multiplying the equation by it gives 111ab+11abc=333710989=9801.111 \cdot ab + 11 \cdot abc = \frac{33}{37} \cdot 10989 = 9801.

Modulo 11,11, since 9801=118919801 = 11 \cdot 891 and 1111,111 \equiv 1, this forces 11ab,11 \mid ab, so a=b.a = b. Then ab=11a,ab = 11a, and dividing the equation by 1111 gives 111a+abc=891.111a + abc = 891. Since abc=110a+c,abc = 110a + c, this is 221a+c=891,221a + c = 891, which requires a=4a = 4 and c=7.c = 7.

Thus a=b=4,a = b = 4, c=7,c = 7, and the three-digit number abcabc is 447.447.

5.

Real numbers rr and ss are roots of p(x)=x3+ax+b,p(x) = x^3 + ax + b, and r+4r + 4 and s3s - 3 are roots of q(x)=x3+ax+b+240.q(x) = x^3 + ax + b + 240. Find the sum of all possible values of b.|b|.

Difficulty rating: 2560

Solution:

Both cubics have zero x2x^2 coefficient, so their roots sum to 0:0: the third root of pp is t=rs,t = -r - s, and the third root of qq is (r+4)(s3)=t1.-(r+4) - (s-3) = t - 1. The coefficient of xx is aa in both, so rs+st+tr=(r+4)(s3)+(s3)(t1)+(t1)(r+4),rs + st + tr = (r+4)(s-3) + (s-3)(t-1) + (t-1)(r+4), which simplifies to t=4r3s+13.t = 4r - 3s + 13.

The constant terms give b=rstb = -rst and b+240=(r+4)(s3)(t1),b + 240 = -(r+4)(s-3)(t-1), so 240=rst(r+4)(s3)(t1),240 = rst - (r+4)(s-3)(t-1), i.e. rs4st+3tr3r+4s+12t252=0.rs - 4st + 3tr - 3r + 4s + 12t - 252 = 0. Substituting t=4r3s+13t = 4r - 3s + 13 reduces this to 12[(rs)2+7(rs)8]=0,12\left[(r-s)^2 + 7(r-s) - 8\right] = 0, so rs=1r - s = 1 or rs=8.r - s = -8.

If rs=1,r - s = 1, then t=4r3s+13=r+16t = 4r - 3s + 13 = r + 16 and t=rs=2r+1,t = -r - s = -2r + 1, so r=5:r = -5: the roots are 5,-5, 6,-6, 11,11, and b=rst=330.b = -rst = -330. If rs=8,r - s = -8, then t=r11=2r8,t = r - 11 = -2r - 8, so r=1:r = 1: the roots are 1,1, 9,9, 10,-10, and b=90.b = 90. The requested sum is 330+90=420.330 + 90 = 420.

6.

Charles has two six-sided dice. One of the dice is fair, and the other die is biased so that it comes up six with probability 23,\frac{2}{3}, and each of the other five sides has probability 115.\frac{1}{15}. Charles chooses one of the two dice at random and rolls it three times. Given that the first two rolls are both sixes, the probability that the third roll will also be a six is pq,\frac{p}{q}, where pp and qq are relatively prime positive integers. Find p+q.p + q.

Solution:

The desired conditional probability is Pr(three sixes)Pr(first two are sixes)=12(23)3+12(16)312(23)2+12(16)2,\frac{\Pr(\text{three sixes})}{\Pr(\text{first two are sixes})} = \frac{\frac{1}{2}\left(\frac{2}{3}\right)^3 + \frac{1}{2}\left(\frac{1}{6}\right)^3} {\frac{1}{2}\left(\frac{2}{3}\right)^2 + \frac{1}{2}\left(\frac{1}{6}\right)^2}, since each die is chosen with probability 12\frac{1}{2} and the fair die shows a six with probability 16.\frac{1}{6}.

The numerator is 12(827+1216)=65432\frac{1}{2}\left(\frac{8}{27} + \frac{1}{216}\right) = \frac{65}{432} and the denominator is 12(49+136)=1772,\frac{1}{2}\left(\frac{4}{9} + \frac{1}{36}\right) = \frac{17}{72}, so the probability is 654327217=65102.\frac{65}{432} \cdot \frac{72}{17} = \frac{65}{102}.

Since 65=51365 = 5 \cdot 13 and 102=2317102 = 2 \cdot 3 \cdot 17 share no factor, p+q=65+102=167.p + q = 65 + 102 = 167.

7.

Let f(x)=(x2+3x+2)cos(πx).f(x) = \left(x^2 + 3x + 2\right)^{\cos(\pi x)}. Find the sum of all positive integers nn for which k=1nlog10f(k)=1.\left|\sum_{k=1}^{n} \log_{10} f(k)\right| = 1.

Difficulty rating: 2450

Solution:

Since cos(πk)=(1)k\cos(\pi k) = (-1)^k and k2+3k+2=(k+1)(k+2),k^2 + 3k + 2 = (k+1)(k+2), we have f(k)=[(k+1)(k+2)](1)k.f(k) = \left[(k+1)(k+2)\right]^{(-1)^k}. The sum of the logarithms is the log of the product k=1nf(k),\prod_{k=1}^n f(k), which telescopes: consecutive factors 1(k+1)(k+2)\frac{1}{(k+1)(k+2)} and (k+2)(k+3)(k+2)(k+3) leave only boundary terms.

For even nn the product is 34(n+2)23(n+1)=n+22,\frac{3 \cdot 4 \cdots (n+2)}{2 \cdot 3 \cdots (n+1)} = \frac{n+2}{2}, and for odd nn it is 34(n+1)23(n+2)=12(n+2).\frac{3 \cdot 4 \cdots (n+1)}{2 \cdot 3 \cdots (n+2)} = \frac{1}{2(n+2)}. The absolute value of the log equals 11 exactly when the product is 1010 or 110.\frac{1}{10}.

For even n,n, n+22=10\frac{n+2}{2} = 10 gives n=18;n = 18; for odd n,n, 2(n+2)=102(n+2) = 10 gives n=3.n = 3. The requested sum is 18+3=21.18 + 3 = 21.

8.

Circle CC with radius 22 has diameter AB.\overline{AB}. Circle DD is internally tangent to circle CC at A.A. Circle EE is internally tangent to circle C,C, externally tangent to circle D,D, and tangent to AB.\overline{AB}. The radius of circle DD is three times the radius of circle EE and can be written in the form mn,\sqrt{m} - n, where mm and nn are positive integers. Find m+n.m + n.

Difficulty rating: 2710

Solution:

Let C,C, D,D, EE also name the circles' centers, let ss be the radius of circle E,E, so circle DD has radius 3s,3s, and let FF be the foot of EE on AB.\overline{AB}. Tangency gives CE=2s,CE = 2 - s, DE=3s+s=4s,DE = 3s + s = 4s, and EF=s,EF = s, while DD lies on AB\overline{AB} with DC=23s.DC = 2 - 3s.

Right triangles CEFCEF and DEFDEF give CF=(2s)2s2=44sCF = \sqrt{(2-s)^2 - s^2} = \sqrt{4 - 4s} and DF=(4s)2s2=s15.DF = \sqrt{(4s)^2 - s^2} = s\sqrt{15}. Since FF is on the opposite side of CC from A,A, we have DF=DC+CF,DF = DC + CF, so s15=(23s)+44s.s\sqrt{15} = (2 - 3s) + \sqrt{4 - 4s}.

Moving 23s2 - 3s to the left and squaring gives 24s28s=215s(23s),24s^2 - 8s = 2\sqrt{15}\,s\,(2 - 3s), i.e. 12s4=15(23s);12s - 4 = \sqrt{15}\,(2 - 3s); squaring again yields 9s2+84s44=0,9s^2 + 84s - 44 = 0, so s=14+4153.s = \frac{-14 + 4\sqrt{15}}{3}. The radius of circle DD is 3s=41514=24014,3s = 4\sqrt{15} - 14 = \sqrt{240} - 14, and m+n=240+14=254.m + n = 240 + 14 = 254.

9.

Ten chairs are arranged in a circle. Find the number of subsets of this set of chairs that contain at least three adjacent chairs.

Difficulty rating: 2760

Solution:

The full set of 1010 chairs qualifies; count the others by locating each maximal run of at least three adjacent chosen chairs at its clockwise start. Any such subset contains a block of four consecutive chairs that is empty-chosen-chosen-chosen. There are 1010 positions for this block, and the remaining 66 chairs are free, giving 1026=640.10 \cdot 2^6 = 640.

This counts once for each maximal run of length at least 3.3. Two such runs require at least 3+33 + 3 chosen chairs plus two gaps, so three runs are impossible, and subsets with exactly two runs are counted twice. To have two runs, place two disjoint empty-chosen-chosen-chosen blocks: 1032=15\frac{10 \cdot 3}{2} = 15 ways (the second block fits in 33 positions among the remaining 66 chairs), with the last 22 chairs free, for 1522=6015 \cdot 2^2 = 60 subsets.

The total is 1+64060=581.1 + 640 - 60 = 581.

10.

Let zz be a complex number with z=2014.|z| = 2014. Let PP be the polygon in the complex plane whose vertices are zz and every ww such that 1z+w=1z+1w.\frac{1}{z+w} = \frac{1}{z} + \frac{1}{w}. Then the area enclosed by PP can be written in the form n3,n\sqrt{3}, where nn is an integer. Find the remainder when nn is divided by 1000.1000.

Difficulty rating: 2560

Solution:

Multiplying 1z+w=1z+1w\frac{1}{z+w} = \frac{1}{z} + \frac{1}{w} by zw(z+w)zw(z+w) gives zw=(z+w)2,zw = (z+w)^2, i.e. z2+zw+w2=0.z^2 + zw + w^2 = 0. Multiplying by zwz - w yields z3w3=0,z^3 - w^3 = 0, so w=ωzw = \omega z or w=ω2z,w = \omega^2 z, where ω\omega is a primitive cube root of unity (and both indeed satisfy the original equation).

Thus PP is the equilateral triangle with vertices z,z, ωz,\omega z, ω2z,\omega^2 z, inscribed in the circle of radius 2014.2014. Its area is 334(2014)2=3100723,\frac{3\sqrt{3}}{4}\,(2014)^2 = 3 \cdot 1007^2 \sqrt{3}, so n=310072=3042147.n = 3 \cdot 1007^2 = 3042147.

The remainder when nn is divided by 10001000 is 147.147.

11.

In RED,\triangle RED, RD=1,RD = 1, DRE=75\angle DRE = 75^\circ and RED=45.\angle RED = 45^\circ. Let MM be the midpoint of segment RD.\overline{RD}. Point CC lies on side ED\overline{ED} such that RCEM.\overline{RC} \perp \overline{EM}. Extend segment DE\overline{DE} through EE to point AA such that CA=AR.CA = AR. Then AE=abc,AE = \frac{a - \sqrt{b}}{c}, where aa and cc are relatively prime positive integers, and bb is a positive integer. Find a+b+c.a + b + c.

Difficulty rating: 3060

Solution:

Since RDE=1807545=60,\angle RDE = 180^\circ - 75^\circ - 45^\circ = 60^\circ, place D=(0,0)D = (0,0) with EE on the positive xx-axis, so R=(12,32).R = \left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right). The law of sines gives DE=sin75sin45=3+12,DE = \frac{\sin 75^\circ}{\sin 45^\circ} = \frac{\sqrt{3}+1}{2}, and M=(14,34).M = \left(\frac{1}{4}, \frac{\sqrt{3}}{4}\right).

The slope of EMEM is 3/4143+12=31+23,\frac{\sqrt{3}/4}{\frac{1}{4} - \frac{\sqrt{3}+1}{2}} = \frac{-\sqrt{3}}{1 + 2\sqrt{3}}, so line RCRC has slope 1+233.\frac{1 + 2\sqrt{3}}{\sqrt{3}}. Descending from RR by 32\frac{\sqrt{3}}{2} to the xx-axis moves us left by 3/21+23=63322,\frac{3/2}{1 + 2\sqrt{3}} = \frac{6\sqrt{3} - 3}{22}, so C=(c,0)C = (c, 0) with c=1263322=73311.c = \frac{1}{2} - \frac{6\sqrt{3} - 3}{22} = \frac{7 - 3\sqrt{3}}{11}.

For A=(t,0),A = (t, 0), the condition CA=ARCA = AR reads (tc)2=(t12)2+34,(t - c)^2 = \left(t - \frac{1}{2}\right)^2 + \frac{3}{4}, which is linear in t:t: t=1c212c=9+4311.t = \frac{1 - c^2}{1 - 2c} = \frac{9 + 4\sqrt{3}}{11}. Then AE=t3+12=18+831131122=73322=72722,AE = t - \frac{\sqrt{3}+1}{2} = \frac{18 + 8\sqrt{3} - 11\sqrt{3} - 11}{22} = \frac{7 - 3\sqrt{3}}{22} = \frac{7 - \sqrt{27}}{22}, so a+b+c=7+27+22=56.a + b + c = 7 + 27 + 22 = 56.

12.

Suppose that the angles of ABC\triangle ABC satisfy cos(3A)+cos(3B)+cos(3C)=1.\cos(3A) + \cos(3B) + \cos(3C) = 1. Two sides of the triangle have lengths 1010 and 13.13. There is a positive integer mm so that the maximum possible length for the remaining side of ABC\triangle ABC is m.\sqrt{m}. Find m.m.

Difficulty rating: 2990

Solution:

Using 1cos3A=2sin23A21 - \cos 3A = 2\sin^2\frac{3A}{2} and cos3B+cos3C=2cos3(B+C)2cos3(BC)2,\cos 3B + \cos 3C = 2\cos\frac{3(B+C)}{2}\cos\frac{3(B-C)}{2}, together with 3(B+C)2=2703A2\frac{3(B+C)}{2} = 270^\circ - \frac{3A}{2} so that cos3(B+C)2=sin3A2,\cos\frac{3(B+C)}{2} = -\sin\frac{3A}{2}, the condition becomes 0=2sin3A2(sin3A2+cos3(BC)2)=2sin3A2(cos3(BC)2cos3(B+C)2)=4sin3A2sin3B2sin3C2.0 = 2\sin\tfrac{3A}{2}\left(\sin\tfrac{3A}{2} + \cos\tfrac{3(B-C)}{2}\right) = 2\sin\tfrac{3A}{2}\left(\cos\tfrac{3(B-C)}{2} - \cos\tfrac{3(B+C)}{2}\right) = 4\sin\tfrac{3A}{2}\sin\tfrac{3B}{2}\sin\tfrac{3C}{2}.

For an angle XX of a triangle, 3X2\frac{3X}{2} lies strictly between 00^\circ and 270,270^\circ, so sin3X2=0\sin\frac{3X}{2} = 0 exactly when X=120.X = 120^\circ. Hence one angle of the triangle is 120.120^\circ.

The remaining side is longest when the 120120^\circ angle sits between the sides of lengths 1010 and 1313 (if 120120^\circ were opposite one of them, the remaining side would be shorter than that side). By the law of cosines its length is 102+132+1013=399,\sqrt{10^2 + 13^2 + 10 \cdot 13} = \sqrt{399}, so m=399.m = 399.

13.

Ten adults enter a room, remove their shoes, and toss their shoes into a pile. Later, a child randomly pairs each left shoe with a right shoe without regard to which shoes belong together. The probability that for every positive integer k<5,k \lt 5, no collection of kk pairs made by the child contains the shoes from exactly kk of the adults is mn,\frac{m}{n}, where mm and nn are relatively prime positive integers. Find m+n.m + n.

Difficulty rating: 3060

Solution:

The child's pairing matches left shoe jj with right shoe π(j)\pi(j) for a uniformly random permutation π\pi of {1,,10}.\{1, \ldots, 10\}. A collection of kk pairs uses kk left and kk right shoes, so it involves exactly kk adults precisely when those adults' indices are closed under π\pi — that is, when the collection is a union of cycles of π.\pi. The condition therefore says π\pi has no cycle of length less than 5.5.

The cycle lengths must partition 1010 into parts of size at least 5:5: either one 1010-cycle or two 55-cycles. There are 9!9! ten-cycles, and 12(105)(4!)2=9!5\frac{1}{2}\binom{10}{5}(4!)^2 = \frac{9!}{5} permutations that are products of two 55-cycles.

The probability is 9!+159!10!=1+1510=325,\frac{9! + \frac{1}{5} \cdot 9!}{10!} = \frac{1 + \frac{1}{5}}{10} = \frac{3}{25}, so m+n=3+25=28.m + n = 3 + 25 = 28.

14.

In ABC,\triangle ABC, AB=10,AB = 10, A=30,\angle A = 30^\circ, and C=45.\angle C = 45^\circ. Let H,H, D,D, and MM be points on line BC\overline{BC} such that AHBC,\overline{AH} \perp \overline{BC}, BAD=CAD,\angle BAD = \angle CAD, and BM=CM.BM = CM. Point NN is the midpoint of segment HM,\overline{HM}, and point PP is on ray ADAD such that PNBC.\overline{PN} \perp \overline{BC}. Then AP2=mn,AP^2 = \frac{m}{n}, where mm and nn are relatively prime positive integers. Find m+n.m + n.

Solution:

Let ray ADAD meet the circumcircle of ABC\triangle ABC again at E.E. Since ADAD bisects angle A,A, the point EE is the midpoint of arc BC,BC, so EE lies on the perpendicular bisector of BC\overline{BC} and projects onto line BCBC at M.M. The projections of the collinear points A,A, P,P, EE onto line BCBC are H,H, N,N, M,M, and projection preserves ratios along a line; since NN is the midpoint of HM,\overline{HM}, point PP is the midpoint of AE.\overline{AE}.

Here B=105,\angle B = 105^\circ, and CBE=CAE=15\angle CBE = \angle CAE = 15^\circ (both subtend arc CECE), so ABE=120.\angle ABE = 120^\circ. Also AEB=ACB=45\angle AEB = \angle ACB = 45^\circ (both subtend arc ABAB). The law of sines in ABE\triangle ABE gives AE=ABsinABEsinAEB=10sin120sin45=56.AE = AB \cdot \frac{\sin \angle ABE}{\sin \angle AEB} = 10 \cdot \frac{\sin 120^\circ}{\sin 45^\circ} = 5\sqrt{6}.

Therefore AP=12AE=562,AP = \frac{1}{2} AE = \frac{5\sqrt{6}}{2}, so AP2=752AP^2 = \frac{75}{2} and m+n=75+2=77.m + n = 75 + 2 = 77.

15.

For any integer k1,k \ge 1, let p(k)p(k) be the smallest prime which does not divide k.k. Define the integer function X(k)X(k) to be the product of all primes less than p(k)p(k) if p(k)>2,p(k) \gt 2, and X(k)=1X(k) = 1 if p(k)=2.p(k) = 2. Let {xn}\{x_n\} be the sequence defined by x0=1,x_0 = 1, and xn+1X(xn)=xnp(xn)x_{n+1} X(x_n) = x_n p(x_n) for n0.n \ge 0. Find the smallest positive integer tt such that xt=2090.x_t = 2090.

Difficulty rating: 3270

Solution:

List the primes in order as ρ0=2,\rho_0 = 2, ρ1=3,\rho_1 = 3, ρ2=5,.\rho_2 = 5, \ldots. Every xnx_n is squarefree, so it is described by the set of primes dividing it, and we claim this set encodes nn in binary: if n=idi2in = \sum_i d_i 2^i with di{0,1},d_i \in \{0, 1\}, then xn=iρidi.x_n = \prod_i \rho_i^{d_i}.

Indeed, suppose xn=iρidix_n = \prod_i \rho_i^{d_i} and let jj be the smallest index with dj=0.d_j = 0. Then p(xn)=ρj,p(x_n) = \rho_j, and X(xn)=ρ0ρ1ρj1X(x_n) = \rho_0 \rho_1 \cdots \rho_{j-1} is exactly the product of the primes for the trailing 11-bits (with X(xn)=1X(x_n) = 1 when j=0j = 0). So xn+1=xnρjρ0ρ1ρj1x_{n+1} = \frac{x_n \, \rho_j}{\rho_0 \rho_1 \cdots \rho_{j-1}} removes the trailing ones and inserts ρj\rho_j — precisely adding 11 in binary. Since x0=1x_0 = 1 corresponds to 0,0, induction proves the claim.

Now 2090=251119=ρ0ρ2ρ4ρ7,2090 = 2 \cdot 5 \cdot 11 \cdot 19 = \rho_0 \rho_2 \rho_4 \rho_7, which corresponds to binary digits at positions 0,0, 2,2, 4,4, 7.7. Hence t=20+22+24+27=149.t = 2^0 + 2^2 + 2^4 + 2^7 = 149.