2004 AIME II 考试题目

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1.

A chord of a circle is perpendicular to a radius at the midpoint of the radius. The ratio of the area of the larger of the two regions into which the chord divides the circle to the smaller can be expressed in the form aπ+bcdπef,\frac{a\pi + b\sqrt{c}}{d\pi - e\sqrt{f}}, where a,a, b,b, c,c, d,d, e,e, and ff are positive integers, aa and ee are relatively prime, and neither cc nor ff is divisible by the square of any prime. Find the remainder when the product abcdefa \cdot b \cdot c \cdot d \cdot e \cdot f is divided by 1000.1000.

Answer: 592
Concepts:sectorchordcircle area

Difficulty rating: 2050

Solution:

Scale so the radius is 2.2. The chord lies at distance 11 from the center, so each radius to an endpoint of the chord makes a 6060^\circ angle with the bisected radius, and the two endpoint radii form a central angle of 120.120^\circ. The isosceles triangle they cut off has area 1222sin120=3,\frac{1}{2} \cdot 2 \cdot 2 \sin 120^\circ = \sqrt{3}, and the whole disk has area 4π.4\pi.

The smaller region is the 120120^\circ sector minus the triangle, 4π33,\frac{4\pi}{3} - \sqrt{3}, and the larger region is the rest, 8π3+3.\frac{8\pi}{3} + \sqrt{3}. The ratio is 8π3+34π33=8π+334π33,\frac{\frac{8\pi}{3} + \sqrt{3}}{\frac{4\pi}{3} - \sqrt{3}} = \frac{8\pi + 3\sqrt{3}}{4\pi - 3\sqrt{3}}, which has the required form with (a,b,c,d,e,f)=(8,3,3,4,3,3).(a, b, c, d, e, f) = (8, 3, 3, 4, 3, 3).

The product is 833433=2592,8 \cdot 3 \cdot 3 \cdot 4 \cdot 3 \cdot 3 = 2592, whose remainder upon division by 10001000 is 592.592.

2.

A jar has 1010 red candies and 1010 blue candies. Terry picks two candies at random, then Mary picks two of the remaining candies at random. Given that the probability that they get the same color combination, irrespective of order, is mn,\frac{m}{n}, where mm and nn are relatively prime positive integers, find m+n.m + n.

Answer: 441

Difficulty rating: 2180

Solution:

The combinations match exactly when both draw two reds, both draw two blues, or both draw one candy of each color. The probability that Terry draws two reds is (102)(202)=45190=938,\frac{\binom{10}{2}}{\binom{20}{2}} = \frac{45}{190} = \frac{9}{38}, after which 88 reds and 1010 blues remain, so Mary draws two reds with probability (82)(182)=28153.\frac{\binom{8}{2}}{\binom{18}{2}} = \frac{28}{153}. That case has probability 93828153=14323,\frac{9}{38} \cdot \frac{28}{153} = \frac{14}{323}, and by symmetry two blues each is also 14323.\frac{14}{323}.

For mixed draws, Terry succeeds with probability 1010(202)=1019,\frac{10 \cdot 10}{\binom{20}{2}} = \frac{10}{19}, leaving 99 of each color, and Mary with probability 99(182)=917,\frac{9 \cdot 9}{\binom{18}{2}} = \frac{9}{17}, for 1019917=90323.\frac{10}{19} \cdot \frac{9}{17} = \frac{90}{323}.

The total is 14+14+90323=118323.\frac{14 + 14 + 90}{323} = \frac{118}{323}. Since 118=259118 = 2 \cdot 59 and 323=1719,323 = 17 \cdot 19, the fraction is in lowest terms, and m+n=118+323=441.m + n = 118 + 323 = 441.

3.

A solid rectangular block is formed by gluing together NN congruent 11-cm cubes face to face. When the block is viewed so that three of its faces are visible, exactly 231231 of the 11-cm cubes cannot be seen. Find the smallest possible value of N.N.

Answer: 384

Difficulty rating: 2110

Solution:

Let the block measure p×q×r.p \times q \times r. A cube is hidden exactly when it touches none of the three visible faces, so the hidden cubes form a (p1)×(q1)×(r1)(p-1) \times (q-1) \times (r-1) block, giving (p1)(q1)(r1)=231=3711.(p-1)(q-1)(r-1) = 231 = 3 \cdot 7 \cdot 11.

The ways to write 231231 as a product of three positive integers are 3711,3 \cdot 7 \cdot 11, 1377,1 \cdot 3 \cdot 77, 1733,1 \cdot 7 \cdot 33, 11121,1 \cdot 11 \cdot 21, and 11231,1 \cdot 1 \cdot 231, giving blocks 4×8×12,4 \times 8 \times 12, 2×4×78,2 \times 4 \times 78, 2×8×34,2 \times 8 \times 34, 2×12×22,2 \times 12 \times 22, and 2×2×232,2 \times 2 \times 232, with volumes 384,384, 624,624, 544,544, 528,528, and 928.928.

The smallest is N=384.N = 384.

4.

How many positive integers less than 10,00010{,}000 have at most two different digits?

Answer: 927

Difficulty rating: 2300

Solution:

All 9999 positive integers below 100100 qualify. A qualifying 33-digit number is either a repdigit (99 of them) or uses a leading digit a1a \ge 1 together with a second value bab \ne a in some of the last two positions: 221=32^2 - 1 = 3 patterns, each realized in 999 \cdot 9 ways (99 choices for a,a, then 99 for bb), for 9+381=2529 + 3 \cdot 81 = 252 numbers.

Similarly a qualifying 44-digit number is a repdigit (99) or has bab \ne a appearing in a nonempty subset of the last three positions: 231=72^3 - 1 = 7 patterns, each in 999 \cdot 9 ways, for 9+781=5769 + 7 \cdot 81 = 576 numbers.

The total is 99+252+576=927.99 + 252 + 576 = 927.

5.

In order to complete a large job, 10001000 workers were hired, just enough to complete the job on schedule. All the workers stayed on the job while the first quarter of the work was done, so the first quarter of the work was completed on schedule. Then 100100 workers were laid off, so the second quarter of the work was completed behind schedule. Then an additional 100100 workers were laid off, so the third quarter of the work was completed still further behind schedule. Given that all workers work at the same rate, what is the minimum number of additional workers, beyond the 800800 workers still on the job at the end of the third quarter, that must be hired after three-quarters of the work has been completed so that the entire project can be completed on schedule or before?

Answer: 766

Difficulty rating: 2390

Solution:

Measure time so that 10001000 workers complete a quarter of the job in 11 unit; the schedule allows 44 units in all. With 900900 workers the second quarter takes 109\frac{10}{9} units, and with 800800 workers the third quarter takes 108=54\frac{10}{8} = \frac{5}{4} units. The time used so far is 1+109+54=12136,1 + \frac{10}{9} + \frac{5}{4} = \frac{121}{36}, leaving 412136=23364 - \frac{121}{36} = \frac{23}{36} of a unit for the last quarter.

The last quarter requires 10001000 worker-units of labor, so the workforce ww must satisfy w23361000,w \cdot \frac{23}{36} \ge 1000, that is w36000231565.2,w \ge \frac{36000}{23} \approx 1565.2, so at least 15661566 workers are needed.

Since 800800 remain on the job, at least 1566800=7661566 - 800 = 766 additional workers must be hired.

6.

Three clever monkeys divide a pile of bananas. The first monkey takes some bananas from the pile, keeps three-fourths of them, and divides the rest equally between the other two. The second monkey takes some bananas from the pile, keeps one-fourth of them, and divides the rest equally between the other two. The third monkey takes the remaining bananas from the pile, keeps one-twelfth of them, and divides the rest equally between the other two. Given that each monkey receives a whole number of bananas whenever the bananas are divided, and the numbers of bananas the first, second, and third monkeys have at the end of the process are in the ratio 3:2:1,3 : 2 : 1, what is the least possible total for the number of bananas?

Answer: 408
Solution:

Say the first monkey takes 8x8x bananas, keeping 6x6x and giving xx to each of the others; the second takes 8y,8y, keeping 2y2y and giving 3y3y to each; the third takes 24z,24z, keeping 2z2z and giving 11z11z to each. All divisions are whole numbers exactly when x,x, y,y, zz are positive integers. The final amounts are 6x+3y+11z,6x + 3y + 11z, x+2y+11z,x + 2y + 11z, and x+3y+2z.x + 3y + 2z.

The ratio 3:2:13 : 2 : 1 says the first amount is triple the third and the second is double the third: 6x+3y+11z=3(x+3y+2z)3x+5z=6y,6x + 3y + 11z = 3(x + 3y + 2z) \quad\Longrightarrow\quad 3x + 5z = 6y, x+2y+11z=2(x+3y+2z)x+4y=7z.x + 2y + 11z = 2(x + 3y + 2z) \quad\Longrightarrow\quad x + 4y = 7z. Substituting x=7z4yx = 7z - 4y into the first equation gives 26z=18y,26z = 18y, so 9y=13z.9y = 13z. Thus y=13ny = 13n and z=9nz = 9n for a positive integer n,n, and then x=63n52n=11n.x = 63n - 52n = 11n.

The total is 8x+8y+24z=(88+104+216)n=408n,8x + 8y + 24z = (88 + 104 + 216)n = 408n, least when n=1:n = 1: the answer is 408.408.

7.

ABCDABCD is a rectangular sheet of paper that has been folded so that corner BB is matched with point BB' on edge AD.\overline{AD}. The crease is EF,\overline{EF}, where EE is on AB\overline{AB} and FF is on CD.\overline{CD}. The dimensions AE=8,AE = 8, BE=17,BE = 17, and CF=3CF = 3 are given. The perimeter of rectangle ABCDABCD is mn,\frac{m}{n}, where mm and nn are relatively prime positive integers. Find m+n.m + n.

Answer: 293
Solution:

Folding reflects BB to BB' across the crease, so BE=BE=17.B'E = BE = 17. In right triangle AEB,AEB', AB=17282=15,AB' = \sqrt{17^2 - 8^2} = 15, and AB=AE+EB=25.AB = AE + EB = 25. Place A=(0,0),A = (0, 0), B=(25,0),B = (25, 0), B=(0,15).B' = (0, 15).

Points on the crease are equidistant from BB and B,B', so EF\overline{EF} is perpendicular to BB.\overline{BB'}. Since BBBB' has slope 35,-\frac{3}{5}, the crease through E=(8,0)E = (8, 0) has slope 53,\frac{5}{3}, and it meets the line CDCD (at height h=BCh = BC) at x=8+3h5.x = 8 + \frac{3h}{5}. The condition CF=3CF = 3 gives 25(8+3h5)=3,soh=703.25 - \left(8 + \frac{3h}{5}\right) = 3, \qquad \text{so} \qquad h = \frac{70}{3}.

The perimeter is 2(25+703)=2903,2\left(25 + \frac{70}{3}\right) = \frac{290}{3}, so m+n=290+3=293.m + n = 290 + 3 = 293.

8.

How many positive integer divisors of 200420042004^{2004} are divisible by exactly 20042004 positive integers?

Answer: 54

Difficulty rating: 2450

Solution:

Since 2004=223167,2004 = 2^2 \cdot 3 \cdot 167, we have 20042004=24008320041672004,2004^{2004} = 2^{4008} \cdot 3^{2004} \cdot 167^{2004}, so its divisors are N=2i3j167kN = 2^i 3^j 167^k with i4008i \le 4008 and j,k2004.j, k \le 2004. Such an NN has (i+1)(j+1)(k+1)(i+1)(j+1)(k+1) divisors, so we need (i+1)(j+1)(k+1)=2004.(i+1)(j+1)(k+1) = 2004.

Every ordered triple of positive integers with product 20042004 yields admissible exponents, since each factor is at most 2004.2004. Counting prime by prime: the exponent 22 of the prime 22 is split among the three factors in (2+22)=6\binom{2+2}{2} = 6 ways by stars and bars, and each of the primes 33 and 167167 goes to one of the 33 factors.

The count is 633=54.6 \cdot 3 \cdot 3 = 54.

9.

A sequence of positive integers with a1=1a_1 = 1 and a9+a10=646a_9 + a_{10} = 646 is formed so that the first three terms are in geometric progression, the second, third, and fourth terms are in arithmetic progression, and, in general, for all n1,n \ge 1, the terms a2n1,a_{2n-1}, a2n,a_{2n}, and a2n+1a_{2n+1} are in geometric progression, and the terms a2n,a_{2n}, a2n+1,a_{2n+1}, and a2n+2a_{2n+2} are in arithmetic progression. Let ana_n be the greatest term in this sequence that is less than 1000.1000. Find n+an.n + a_n.

Answer: 973
Solution:

Let a2=r.a_2 = r. The geometric condition gives a3=r2,a_3 = r^2, the arithmetic condition gives a4=2r2r=r(2r1),a_4 = 2r^2 - r = r(2r-1), then a5=(2r1)2,a_5 = (2r-1)^2, and so on: inductively a2k+1=(kr(k1))2,a2k+2=(kr(k1))((k+1)rk).a_{2k+1} = \bigl(kr - (k-1)\bigr)^2, \qquad a_{2k+2} = \bigl(kr - (k-1)\bigr)\bigl((k+1)r - k\bigr). In particular a9=(4r3)2a_9 = (4r-3)^2 and a10=(4r3)(5r4),a_{10} = (4r-3)(5r-4), so a9+a10=(4r3)(9r7)=646.a_9 + a_{10} = (4r-3)(9r-7) = 646. Expanding gives 36r255r625=0,36r^2 - 55r - 625 = 0, which factors as (r5)(36r+125)=0,(r - 5)(36r + 125) = 0, so r=5.r = 5.

With r=5r = 5 we get kr(k1)=4k+1,kr - (k-1) = 4k + 1, so a2k+1=(4k+1)2a_{2k+1} = (4k+1)^2 and a2k+2=(4k+1)(4k+5);a_{2k+2} = (4k+1)(4k+5); the sequence is increasing. Since a17=332=1089>1000a_{17} = 33^2 = 1089 \gt 1000 while a16=2933=957,a_{16} = 29 \cdot 33 = 957, the greatest term below 10001000 is a16=957.a_{16} = 957.

Therefore n+an=16+957=973.n + a_n = 16 + 957 = 973.

10.

Let S\mathcal{S} be the set of integers between 11 and 2402^{40} whose binary expansions have exactly two 11's. If a number is chosen at random from S,\mathcal{S}, the probability that it is divisible by 99 is pq,\frac{p}{q}, where pp and qq are relatively prime positive integers. Find p+q.p + q.

Answer: 913
Solution:

The set S\mathcal{S} consists of the (402)=780\binom{40}{2} = 780 numbers 2a+2b2^a + 2^b with 0a<b39.0 \le a \lt b \le 39. Since 2a2^a is coprime to 9,9, we have 92a(2ba+1)9 \mid 2^a(2^{b-a} + 1) exactly when 2ba1(mod9).2^{b-a} \equiv -1 \pmod{9}. The powers of 22 modulo 99 cycle through 2,4,8,7,5,12, 4, 8, 7, 5, 1 with period 6,6, so 2d812^d \equiv 8 \equiv -1 exactly when d3(mod6).d \equiv 3 \pmod{6}.

For each difference d=bad = b - a there are 40d40 - d pairs, so the number of multiples of 99 in S\mathcal{S} is d=3,9,,39(40d)=37+31+25+19+13+7+1=133.\sum_{d = 3, 9, \ldots, 39} (40 - d) = 37 + 31 + 25 + 19 + 13 + 7 + 1 = 133.

The probability is 133780,\frac{133}{780}, and since 133=719133 = 7 \cdot 19 while 780=223513,780 = 2^2 \cdot 3 \cdot 5 \cdot 13, it is in lowest terms. Thus p+q=133+780=913.p + q = 133 + 780 = 913.

11.

A right circular cone has a base with radius 600600 and height 2007.200\sqrt{7}. A fly starts at a point on the surface of the cone whose distance from the vertex of the cone is 125,125, and crawls along the surface of the cone to a point on the exact opposite side of the cone whose distance from the vertex is 3752.375\sqrt{2}. Find the least distance that the fly could have crawled.

Answer: 625

Difficulty rating: 2990

Solution:

The slant height is 6002+(2007)2=360000+280000=800.\sqrt{600^2 + (200\sqrt{7})^2} = \sqrt{360000 + 280000} = 800. Cutting the cone along the ruling through the starting point and unrolling gives a sector of radius 800800 whose arc has the base circumference 2π600=1200π;2\pi \cdot 600 = 1200\pi; since a full circle of radius 800800 has circumference 1600π,1600\pi, the central angle is 34360=270.\frac{3}{4} \cdot 360^\circ = 270^\circ. A point on the exact opposite side of the cone is halfway around, which in the unrolled sector is 135135^\circ away.

The shortest crawl is the straight segment between the two points, at radii 125125 and 3752375\sqrt{2} with a 135135^\circ angle between them. By the law of cosines, d2=1252+(3752)221253752cos135=15625+281250+93750=390625.d^2 = 125^2 + (375\sqrt{2})^2 - 2 \cdot 125 \cdot 375\sqrt{2} \cos 135^\circ = 15625 + 281250 + 93750 = 390625.

Thus d=625.d = 625.

12.

Let ABCDABCD be an isosceles trapezoid, whose dimensions are AB=6,AB = 6, BC=5=DA,BC = 5 = DA, and CD=4.CD = 4. Draw circles of radius 33 centered at AA and B,B, and circles of radius 22 centered at CC and D.D. A circle contained within the trapezoid is tangent to all four of these circles. Its radius is k+mnp,\frac{-k + m\sqrt{n}}{p}, where k,k, m,m, n,n, and pp are positive integers, nn is not divisible by the square of any prime, and kk and pp are relatively prime. Find k+m+n+p.k + m + n + p.

Answer: 134
Solution:

Dropping perpendiculars from CC and DD shows each leg of length 55 spans a horizontal offset of 642=1,\frac{6 - 4}{2} = 1, so the height of the trapezoid is 251=24.\sqrt{25 - 1} = \sqrt{24}. By symmetry the inner circle's center OO lies on the vertical axis through the midpoints EE of AB\overline{AB} and FF of CD.\overline{CD}. If its radius is x,x, external tangency gives OA=x+3OA = x + 3 and OC=x+2,OC = x + 2, so with AE=3AE = 3 and CF=2,CF = 2, OE=(x+3)29=x2+6x,OF=(x+2)24=x2+4x.OE = \sqrt{(x+3)^2 - 9} = \sqrt{x^2 + 6x}, \qquad OF = \sqrt{(x+2)^2 - 4} = \sqrt{x^2 + 4x}.

Since OE+OF=24,OE + OF = \sqrt{24}, moving one radical across and squaring gives 24(x2+4x)=12x,\sqrt{24(x^2 + 4x)} = 12 - x, and squaring again yields 24x2+96x=14424x+x2,24x^2 + 96x = 144 - 24x + x^2, that is 23x2+120x144=0.23x^2 + 120x - 144 = 0.

The positive root is x=120+14400+1324846=120+96346=60+48323,x = \frac{-120 + \sqrt{14400 + 13248}}{46} = \frac{-120 + 96\sqrt{3}}{46} = \frac{-60 + 48\sqrt{3}}{23}, so k+m+n+p=60+48+3+23=134.k + m + n + p = 60 + 48 + 3 + 23 = 134.

13.

Let ABCDEABCDE be a convex pentagon with ABCE,\overline{AB} \parallel \overline{CE}, BCAD,\overline{BC} \parallel \overline{AD}, ACDE,\overline{AC} \parallel \overline{DE}, ABC=120,\angle ABC = 120^\circ, AB=3,AB = 3, BC=5,BC = 5, and DE=15.DE = 15. Given that the ratio between the area of triangle ABCABC and the area of triangle EBDEBD is mn,\frac{m}{n}, where mm and nn are relatively prime positive integers, find m+n.m + n.

Answer: 484

Difficulty rating: 3160

Solution:

By the law of cosines, AC2=32+52235cos120=49,AC^2 = 3^2 + 5^2 - 2 \cdot 3 \cdot 5 \cos 120^\circ = 49, so AC=7.AC = 7. Let FF be the intersection of AD\overline{AD} and CE.\overline{CE}. Since AFBCAF \parallel BC and CFAB,CF \parallel AB, quadrilateral ABCFABCF is a parallelogram, so FF lies at the same distance hh from line ACAC as B,B, on the opposite side, where [ABC]=127h.[ABC] = \frac{1}{2} \cdot 7h.

Since ACDE,\overline{AC} \parallel \overline{DE}, triangles FACFAC and FDEFDE are similar with ratio AC:DE=7:15,AC : DE = 7 : 15, so the distance from FF to line DEDE is 15h7,\frac{15h}{7}, with DEDE on the far side of FF from AC.AC. The distance from BB to line DEDE is therefore h+h+15h7=29h7,h + h + \frac{15h}{7} = \frac{29h}{7}, giving [EBD]=121529h7=435h14.[EBD] = \frac{1}{2} \cdot 15 \cdot \frac{29h}{7} = \frac{435h}{14}.

Thus [ABC][EBD]=7h/2435h/14=49435,\frac{[ABC]}{[EBD]} = \frac{7h/2}{435h/14} = \frac{49}{435}, which is in lowest terms since 435=3529.435 = 3 \cdot 5 \cdot 29. The answer is m+n=49+435=484.m + n = 49 + 435 = 484.

14.

Consider a string of nn 77's, 777777,7777\ldots77, into which ++ signs are inserted to produce an arithmetic expression. For example, 7+77+777+7+7=8757 + 77 + 777 + 7 + 7 = 875 could be obtained from eight 77's in this way. For how many values of nn is it possible to insert ++ signs so that the resulting expression has value 7000?7000?

Answer: 108

Difficulty rating: 3270

Solution:

Dividing by 77 turns the summands into 1,1, 11,11, or 111111 (no longer summand fits, since 1111>10001111 \gt 1000) and the target into 1000.1000. If x,x, y,y, zz count the summands of each size, then x+11y+111z=1000x + 11y + 111z = 1000 and n=x+2y+3z.n = x + 2y + 3z. Subtracting gives n=10009(y+12z),n = 1000 - 9(y + 12z), so the possible nn correspond exactly to the attainable values of v=y+12z.v = y + 12z.

The constraints are z9z \le 9 and 0y1000111z110 \le y \le \frac{1000 - 111z}{11} (then x0x \ge 0 follows). For z=0,1,,9z = 0, 1, \ldots, 9 the value v=y+12zv = y + 12z ranges over the intervals [0,90],[0, 90], [12,92],[12, 92], [24,94],[24, 94], [36,96],[36, 96], [48,98],[48, 98], [60,100],[60, 100], [72,102],[72, 102], [84,104],[84, 104], [96,106],[96, 106], and {108}\{108\} (when z=9,z = 9, only y=0y = 0 fits). Their union is every integer from 00 to 106106 together with 108;108; only 107107 is unattainable.

So vv takes 107+1=108107 + 1 = 108 values, and n=10009vn = 1000 - 9v takes 108108 values.

15.

A long thin strip of paper is 10241024 units in length, 11 unit in width, and is divided into 10241024 unit squares. The paper is folded in half repeatedly. For the first fold, the right end of the paper is folded over to coincide with and lie on top of the left end. The result is a 512512 by 11 strip of double thickness. Next, the right end of this strip is folded over to coincide with and lie on top of the left end, resulting in a 256256 by 11 strip of quadruple thickness. This process is repeated 88 more times. After the last fold, the strip has become a stack of 10241024 unit squares. How many of these squares lie below the square that was originally the 942942nd square counting from the left?

Answer: 593

Difficulty rating: 3500

Solution:

After ff folds the strip is 210f2^{10-f} squares long and 2f2^f layers thick, so the positions LL from the left and RR from the right satisfy L+R=210f+1,L + R = 2^{10-f} + 1, and the positions BB from the bottom and TT from the top satisfy B+T=2f+1.B + T = 2^f + 1. When the right half is folded over onto the left, a square in the left half keeps its LL and B,B, while a square in the right half is flipped: its new LL is its old R,R, and its new TT is its old B.B.

The 942942nd square starts at (L,B)=(942,1).(L, B) = (942, 1). Applying the rule through the ten folds gives (83,2), (83,2), (83,2), (46,15), (19,18), (14,47), (3,82), (3,82), (2,431), (1,594).(83, 2),\ (83, 2),\ (83, 2),\ (46, 15),\ (19, 18),\ (14, 47),\ (3, 82),\ (3, 82),\ (2, 431),\ (1, 594). For example, at the fourth fold the strip has length 128128 and L=83>64,L = 83 \gt 64, so the new LL is 128+183=46128 + 1 - 83 = 46 and the new TT is the old B=2,B = 2, making B=16+12=15.B = 16 + 1 - 2 = 15.

In the final stack of 10241024 squares, this square sits at height 594594 from the bottom, so 5941=593594 - 1 = 593 squares lie below it.