Finding the last digits of massive exponents.
A fair six-sided die is rolled repeatedly until a 6 is rolled. What is the probability that the sum of all the rolls (including the final 6) is a multiple of 3? Solution: Let P0cap P sub 0
N≡33(mod35)cap N triple bar 33 space open paren mod space 35 close paren Mathcounts National Sprint Round Problems And Solutions
Digit: 0 → 0 (product becomes 0, which is multiple of 8 — wait! Zero is divisible by any number. So if any digit is 0, product = 0 → multiple of 8. So those are favorable , not excluded.)
The MATHCOUNTS National Competition is the pinnacle of middle school mathematics in the United States. Among its four intense rounds—Sprint, Target, Team, and Countdown—the is often the first major test of a student’s speed, accuracy, and mental endurance. Finding the last digits of massive exponents
N=35(9x+4)+33cap N equals 35 open paren 9 x plus 4 close paren plus 33 N=315x+140+33cap N equals 315 x plus 140 plus 33 N=315x+173cap N equals 315 x plus 173 The general solution is . The smallest positive integer solution is when , which gives , it satisfies all conditions of the problem. Problem 3: Geometry (Power of a Point & Right Triangles) Problem: In right triangle ABCcap A cap B cap C . A circle is tangent to side ABcap A cap B and passes through the midpoint of hypotenuse ACcap A cap C . If the circle intersects side BCcap B cap C at a second point , find the length of segment BDcap B cap D
How many ways to arrange the letters in “MATHCOUNTS” such that vowels are in alphabetical order? Solution: Total arrangements 10!/(2!*2!) due to T and A repeated? Wait, M,A,T,H,C,O,U,N,T,S: T twice, all others once except A once? Actually A once, vowels: A,O,U (3 distinct). For permutations where vowels appear in order A,U,O? It says alphabetical: A,O,U. Number of permutations of all letters = 10!/(2! for T). Then divide by 3! because vowels can be in any order, but only 1 order valid. So = 10!/(2! * 3!) = 302400. Solution: Let P0cap P sub 0 N≡33(mod35)cap N
Because the die is fair, look at the transitions when the game does not end (rolling a 1, 2, 3, 4, or 5). There is a 15one-fifth chance of rolling a 3, which keeps you in the same state. There is a 25two-fifths chance of rolling a 1 or 4, moving you to the next state. There is a 25two-fifths chance of rolling a 2 or 5, moving you to the other state.
Since (b>0), (3a-17 >0 \Rightarrow a \ge 6). Also integer: (3a-17) divides (17a). Use division: (17a = 17/3*(3a-17) + 289/3) – messy. Instead, rewrite: (b = \frac17a3a-17 = 5 + \frac853a-17) after polynomial division? Let’s check: Divide 17a by (3a-17): quotient 5 (since 5*(3a-17)=15a-85, remainder 2a+85? No, do carefully: (17a) / (3a-17) = 5 + (2a+85)/(3a-17)? That doesn’t help. Better: Set (k = 3a-17), then (a = (k+17)/3), substitute into b: (b = \frac17(k+17)/3k = \frac17k+2893k = \frac173 + \frac2893k). For b integer, (3k) must divide 289 = 17^2. Thus (3k) is a divisor of 289: 1, 17, 289. But 3k positive, k = (3a-17) >0. 3k=1→k=1/3 no. 3k=17→k=17/3 no. 3k=289→k=289/3 no. So no integer k? That means I made an algebraic slip.
( (10a + b) + (10b + c) = 10a + 11b + c ) = perfect square, say ( k^2 ).