High School Math Contest
University of South Carolina
February 5, 2011
Solutions
1. (b) The left-hand side is an even number unlessmornis zero. Ifm= 0, we have2n
≥1and
2m
−2n
≤0. Ifn = 0, then2m = 64, i.e.m= 6. Thus(6,0)is the only such pair.
2. (d) If the relationab = 1then eithera =
±1, orb = 0(anda 6= 0). The equation 2x2−5 3 = 1
yields solutions x = ±2. If 2x2−5
3 = −1, then x = ±1, but in this case x 2
−2xis odd, so
2x2
−5
3
x2
−2x
=−1. From the last option we obtainx2 −2x= 0, i.e. x = 0orx= 2. Thus
we have three solutions: x= 0,±2.
3. (d) We can write x51+ 51 = Q(x)
·(x+ 1) +R, where Q(x)is the quotient and R is the remainder. Whenx=−1, this givesR= (−1)51+ 51 = 50.
4. (d) Letxbe the total number of people. Then the total number of handshakes is 52x, where we have to divide by 2since every handshake is counted twice. Thus 52x = 60, which leads to x= 24.
5. (b) Every power of six ends in6. Also note that324= 816. Since the last digit of81is1, any
power of it will also end in 1. (Alternatively, for the powers of3we have 31 = 3, 32 = 9,
33 = 27, 34 = 81, 35 = 243, ... – the units digits repeat in cycles of length 4. Since 24is
divisible by4, the final digit of324is the same as the final digit of34 = 81, i.e. 1.) Thus the
units digit of625
−324is6
−1 = 5.
6. (c) Obviously, x > y. If r is the radius of the circle, then x = 2πr andy = 4√2r. Hence x
y = π
2√2 < 3.2
2.8 ≈1.14< 3 2.
7. (c) The product has 2011 factors. In order for it to be negative, we need to have an odd number of negative factors (notice that all the exponents are odd) and, hence, an even number of positive factors. But the number of positive factors is simply equal to x. Since there are
8. (c) Draw the perpendicular DO from D onto the side BC. Then BO = OC = AD. Let X be the intersection point of P RandDO, andY be the intersection of QS andDO. Then the triangles △DXR and△DOC are similar, so that XROC = DR
11. (a) LetF(n)be the number of ways to seatnstudents in a row ofnchairs as described in the problem. The student who has the ticket to seat numbern has two options. He/she can either sit in his/her own chair, in which case the remaining n−1students can arrange themselves inF(n−1)ways. Alternatively, if this student takes the(n−1)st seat, the student with the ticket to that chair is forced to take the nth seat, and the othern
15. (a) Denote byvA,vB, andvT the speeds of Allan, Bill, and the train, respectively. The relative speeds of the train with respect to Allan an Bill are thenvT −vAandvT −vB. It follows from the condition that the train passes Allan in10seconds and Bill in9seconds that vT−vA
vT−vB =
9 10.
The relative speed of Allan with respect to Bill isvA−vB. We have vA−vB
vT −vB =
vT −vB vT −vB −
vT −vA
vT −vB = 1−
9 10 =
1 10.
Thus, Allan’s speed with respect to Bill is ten times smaller than the train’s, so, in order to reach Bill, it would take Allan ten times the time it would take the train, i.e. 200 minutes.
16. (b) We have 1020 = 103·6+2. Since 103 = 1001−1, we find that 103·6 = (1001 −1)6 =
1001 ·k + 1, where k is some integer. Indeed, when we multiply out (1001−1)6, all the
terms will be divisible by 1001 except for the last one which is (−1)6 = 1. Then 1020 =
(1001·k+ 1)·100 = 1001·100k+ 100, in other words, it has remainder100when divided by1001.
17. (b) We see thatf(−1/2) = 1. Next,f(−1) = (3/2)−32 <1andf(−3/2) = 2−2 = 1/4 <1.
Therefore, options (c) and (d) are excluded. We are left with f(0) = (1/2)−12 = √2 and
f(1/3) = (1/6)−16 = 6 √
6which are both greater than1. To compare these two, notice that
6<8, hence,616 <8 1 6 = 2
1 2.
18. (c) It is easy to see (e.g., by performing long multiplication) that the smallest such number is
91. Indeed,90·1111 = 99990has5digits, while91·1111 = 101101 is a six-digit number. Thus, the answer is9 + 1 = 10.
19. (c) Since∠ACB +∠AP B = 90◦+ 90◦ = 180◦, the quadrilateral AP BC can be inscribed
in a circle. But then∠P CB = ∠P AB = 45◦ since they are inscribed angles with the same
20. (d) For the number to be divisible by9the sum of its digits has to be divisible by9. Thus, any nine-digit number with all digits equal is divisible by 9(this gives9 possibilities). A three-digit or a six-three-digit number would have to have the common three-digit be either3,6, or9(2·3 = 6
possibilities). For all other numbers of digits (1, 2, 4,5,7,8) the only available choice of the common digit is9– this yields6more numbers. Thus, all together we have 9 + 6 + 6 = 21
such numbers.
21. (e) It is easy to see that the triangles with specified areas are similar to each other and to
△XY Z. Then, since the ratio of their areas is 9 : 25 : 16, the ratio of the lengths of their bases parallel to the sideXZ is3 : 5 : 4. Denote these bases by3x,5x, and4x, respectively. The three unmarked subregions are parallelograms. Therefore, one can see that XZ = 3x+ 5x+ 4x = 12x. Since△XY Z is similar to any of the smaller triangles, say the one of area
25, its area is 125xx2·25 = 122 = 144.
22. (a) Label the points with numbers from1to6. LetAj be the event that all six points can be covered by some arc of length 1and point number j is the first one in this arc if one counts clockwise, j = 1, . . . ,6. The eventAj happens exactly if all the remaining5points fall onto the semicircle of length1which “starts” at thejth point. The probability of each point being placed onto this semicircle is 1
2. The probability ofAj is therefore 1 25 = 1
32. It is also easy to
see that the events Aj are mutually exclusive. Hence, the probability in question is equal to P6
j=1P(Aj) = 6 32 =
3 16.
23. (c) Label the vertices of the triangle A, B, and C and denote the radius of the circles by r. Let O1 and O2 be the centers of the circles tangent to the side AB, and let O1P and
O2Q be the perpendiculars drawn from these centers to the side AB. Then AP = QB =
r·cot 30◦ = √3r, andP Q = O
1O2 = 2r sinceO1O2QP is a rectangle. Thusa = AB =
AP +P Q+QB = 2√3r+ 2r, so
r= √a = √
3−1
24. (e) This expression is a quadratic polynomial of x. At the same time, it is equal to 1 at three different points: x = a, b, andc. This can happen only if this quadratic polynomial is identically equal to1.
25. (b) Denote the area of△XY Z byx. Observe that the areas of triangles△AY R,△BZP, and
△CXQare the same, denote them byy. The areas of the quadrilateralsAY ZP,BZXQ, and CXY R are also equal, denote them by z. Then we havex+ 3y+ 3z = 1. Since triangles
△ACP and△P CB have the same height drawn from point C, we see that the area of the former is twice the area of the latter, i.e. 2z+x+y= 2·(2y+z), which implies thatx= 3y. By the Law of Cosines
CP2 =CB2+BP2 −2·CB·CP ·cos 60◦
= (3BP)2+BP2
−3·BP2 = 7BP2.
It is easy to see that △ABRis similar to △ZBP, since they have one common angle and ∠BP Z = ∠BRA. SinceAR = CP = √7·BP, the area of△ABRis7times the area of
△ZBP, i.e. 2y+z = 7y, orz = 5y. From the equation x+ 3y+ 3z = 1 we now obtain
3y+ 3y+ 15y= 1, i.e.y= 1
21. Hencex= 3y= 1 7.
26. (d) Notice that forn≥5the numbern!is divisible by both2and5. Hence it is divisible by ten and ends with a0. Therefore, we only need to find the last digit of the sum1! + 2! + 3! + 4! = 1 + 2 + 6 + 24 = 33, which is3.
27. (d) We have4!·5!·6! = (2·3·22)
·(2·3·22
·5)(·2·3·22
·5·(2·3)) = 210
·34
·52. The
prime factorization of a perfect square dividing4!·5!·6!has to consist of even powers of the primes 2, 3, and5, and the exponents should not exceed the corresponding exponents in the factorization of 4!·5!·6!. Thus, there are6choices for the exponent in the power of 2(0, 2,
28. (d) Leta,b,c, anddbe the lengths of the perpendiculars drawn from the pointO to the sides AB, BC, CD, and DA, respectively. Then, by the Pythagorean theorem, we have OA2 =
a2+d2,OB2 =a2+b2,OC2 =b2+c2, andOD2 =c2+d2. Adding the third equation to the
first, and the fourth to the second, we find thatOA2+OC2 =OB2+OD2 =a2+b2+c2+d2.
Thus102+ 52 = 92+x2, andx=√100 + 25
−81 = √44.
29. (b) All together there are27ways to color the vertices garnet or black – we shall refer to them
as “patterns”. Let us count how many patterns correspond to the same coloring.
Assume first that the coloring is non-trivial (not all vertices black or all garnet). In this case, a rotation of a pattern necessarily produces a different pattern – this follows from the fact7is a prime number.
Indeed, if a rotation of a pattern by k vertices (k < 7) yields the same pattern, then you can continue this process, each time obtaining the same pattern. Sincekand7are relatively prime, there existsnsuch thatnkhas remainder1when divided by7. This means that we have now rotated the initial pattern by one vertex and still obtained the same pattern. Thus,allrotations of the pattern have to be the same. But this is only possible if the coloring is monochromatic (all vertices are of the same color).
So we see that each non-trivial coloring corresponds to7non-trivial patterns. There are27
−2
such patterns, and, therefore, 27−2
7 = 18 colorings. In addition we have two trivial colorings
(all black or all garnet). Thus, the total number of colorings is18 + 2 = 20.
30. (c) The total area of the four small circles is 4 × π4 = π – equal to the area of the large
