2005 Lassiter Invitational Varsity Solutions

1. (D) Remembering that in any triangle with sides A,B,C : for any sides A,B,C. Thus, if you plug in all sides in this equation you get the following inequalities: . Thus, satisfies all these conditions so x=4 is smallest possible integer.

2. (B) Let the number be ab. Thus, 10a + b = 2 (a + b). So, 8a = b. The only digits that satisfy this are (1,8). So 18 is the answer.

3. (C) Upon observation of drawing a tangent line of length 2, we see that the hypotenuse of the right triangle formed is . This length is the radius of a circle that encompasses the circle of radius 2 and the desired region. Thus is our desired area.

4. (E) Factor this expression into with x and y ==> . Just like completing the square, you must add –15 to both sides. . Each factor (x+5) or (y-3) must be a factor of 144. From factorization ==> . Thus it has (4+1)(2+1) = 15 factors. Each of (x+5) and (y-3) can be anyone of these factors, and considering both being negative leaves 15(2) = 30 solutions

5.(D) ==> . From this, or . Thus if a number, a, is rounded down to these values, it fits the requirements for (D).

6. (C) To find the zeroes at the end of a number, consider how many 10’s are in the factorization. Since 10=5(2), and there are plenty of 2’s in a factorial, count how many 5’s are in the factorial: .

7. (C) The maximum number of digits the number , n, can have is 5 (). So we are looking for five digits so that the sum is 8 (options are 0,1,2). So it could be or any permutation giving 5C1 = 5. Also, the form works, giving 5C2 = 10. The total is thus 10 + 5= 15

8. (D), ==> .

9.(B) Harmonic average (used for speeds) is . So, .

, thus

10.(A) Let . Then . Subtracting the equations yields:

Using the sum of a geometric series: . Thus, .

11. (C) Consider the three types of surface cubes: corner(8), edge(24), face(24). They have probability of landing face up as follows: . Thus the probability is

12.(B) Since diagonals in rhombus are perpendicular bisectors, the side is . Also, the diameter of the inscribed circle serves as a height of the rhombus. Thus . .

13.(B) Let (log x) =y. Thus, . So (y+1)=0. Thus, y=-1 > So .

14. (E) Remembering that we have =.

15.(D). Find the altitude by area formulas or by noticing that when an altitude is dropped, 14 can be split into (5+9). This forms (9,12,15) and (5,12,13) right triangles. Thus, the altitude is 12. And by observation that the median splits the side into (7+7), the base of the wanted triangle is 2. So

16. (B) 465 factors into . To find the sum of all factors of a number () you do the following: S = ==> For this case, notice the following: . Thus, the number p equals = 200

17. (A) By the given information, we can form a right triangle so that one leg is 1 and the hypotenuse is . Thus the other leg is . So

18. (E) Upon consideration, we are looking for multiples of 1 between 0-59, then multiples of 2, etc. until 12. So we have . However, we must add 12 because each number (1-12) evenly divides 0, so 188.

19. (D) If we make a list of possible scores after each round, we find that the last number impossible to obtain is 23: (1) 5,7 (2) 10,12,14 (3) 15,17,19,21 (4) 20,22,24,26,28 (5) 25,27,29,31,33,35, ….

20. (B) Let the sum of the five numbers be S, thus or . So S= 20 or -20

21. (A) The smallest values for (a,b) would be (2,3) yielding which has (3+1)(2+1)= 12 factors. However, upon consideration (2,4) yields which has (8+1) = 9 factors!

22. ( C ) Using change of base and inverting and multiplying we obtain:

23. (C) The only way to obtain a number that fits this criteria ( 2 mod 4) is to multiply an odd number, odd number, and an even number not divisible by 4. Or odd,odd,even. For odds we have 5 choices, and for the even we have 3 (2,6,10). So . But we can arrange this = 3. So 75(3)= 225. There are 10(10)10= 1000 possibilities. Thus, the probability is

24.(E) In order to pass, he must get a 3, 4, or 5 which have probabilities of respectively. Therefore, the probability that he passes is for each test. So the probability he passes all 3 is:

25. (A) by the given equation. Now, . So

By substitution, . So,

26. (). Consider a case-by-case scenario of what number Mrs.Poss rolls. If she rolls a 1, then a chances it will evenly divide. A 2 yield , etc. Since the probability that rolling any one number is the same: .

27. (6) To find special properties of the reciprocals of roots in any polynomial, simply swap all the coefficients from front to back : . Now, to find the sum of the squares (S2) , first find the sum of the roots (S1) : . Now, to find the S2 utilize the coefficients once again: . OR Find the roots , so the sum of the squares of the reciprocals is .

28. () By investigation, the original line is . The translation results in . The equation that fits this pattern is or . Now, the region is a quarter of a circle with area : .

29. (81) Label the different puzzle pieces as 1,2,3,4. Then consider each bag as a set. To complete the puzzle, we are looking for a complete set, such as so that (Set A) U (Set B) = All pieces (universal set). Upon investigation, we can put each piece in (BagA, BagB, or both Bag A and B) leaving 3 options. The total possibilities are we can put each piece in or out of each bag giving 2x2 = 4 possibilities. The probability for (n) pieces is thus . This problem entails n=4, so

30.(62) Consider factorials in groups of 5 since they all have the same number of zeroes at the end. So for (0-4), there are 0 zeroes, for (5-9) there is 1 zero, for (10-14) there are 2 zeroes ,etc. Thus, the answer is as follows: .