The 2nd Texas A&M at Galveston Mathematics Olympiad. September 24, Problems & Solutions

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Alyson Holmes
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1 nd Math Olympiad Solutions Problem #: The nd Teas A&M at Galveston Mathematics Olympiad September 4, 00 Problems & Solutions Written by Dr. Lin Qiu A runner passes 5 poles in 30 seconds. How long will he take to pass 0 poles? The consecutive poles are o equal distance rom each other. Since it takes the runner 30 seconds to cover 4 spaces between the 5 poles, the rate o covering space is 30/4 seconds. There are 9 spaces between 0 holes, 30 4 so the time o passing 9 holes is seconds. Problem #: A hour clock shows o clock now. Eactly when will the hour and minute hands net coincide? Every minute the hour hand passes o the distance between o clock & 60 o clock positions, which is o a ull rotation, whereas the minute hand passes o the ull rotation around the clock. Ater o clock, assume the hour hand 60 and the minute hand net coincide minutes ater o clock. Then the raction o the distance around the clock rom the vertical position) is or the

2 nd Math Olympiad Solutions minute hand, while it is a raction 60 o a ull rotation. Thus which simpliies to, Solution or gives So the hour hand and the minute hand will net coincide eactly 5 5 minutes ater o clock,or eactly at : Problem #3: There were originally 7 pieces o papers in a pile. Somebody picked up a piece and cut it into 7 pieces and put them back into the pile again. This was repeated many times. Then, somebody counted the pieces o the papers in the pile and got the number 00. Prove that he made a mistake counting. There were originally 7 pieces o papers. Every time somebody cut and put back papers, he added 6 pieces o papers. So, Ater his st cut and put back, he got pieces; Ater his nd cut and put back, he got pieces; Ater his 333 rd cut and put back, he got pieces; - -

3 nd Math Olympiad Solutions Ater his 334 th cut and put back, he got pieces. So it is impossible to get 00 pieces. A simpler eplanation: Since is not divisible by 6, so it is wrong. An even simpler eplanation: the numbers o papers should always be odd. Problem #4: 00 a) What is the last digit o? 00 b) What is the irst digit o? a) Let s observe the powers o : The last digits o powers o are, 4, 8, 6,, with the cycle o 4. And 00 has 00 the remainder when it is divided by 4. So the last digit o is b) Write in the scientiic notation, we have n c 0, where c is a decimal between and 0, 00 and n is a positive integer. Let a be the irst digit o integer 00 a can be,,,9), so a c < a. Then - 3 -

4 nd Math Olympiad Solutions n 00 n a 0 a ) 0 Take the common logarithm, log < n a ) ) n 00 a 0 ) log ) < log 0 n log a 00 log < n log a ) Estimate 00 log So, n 605, rom the whole part o the number above. And, a, since the decimal part o the number above is between log 0 and log So, the irst digit o is. Problem #5: Prove that n n )n ) 6 is an integer, or any integer n. Proving n n )n ) 6 is an integer is equivalent as proving n n )n ) is divisible by 6. And because 6 3, we will just need to prove n n )n ) is divisible by both and 3. a) Proo o divisibility o : i n is even n is divisible by ), then the product n n )n ) is o course divisible by, we are done. I n is odd, then n divisible by. So the product n n )n ) is divisible by. b) Proo o divisibility o 3: i n is divisible by 3, then we are done. I n has the remainder when divided by 3, then by The Remainder Theorem, we have n 3 k, where k is an integer is

5 nd Math Olympiad Solutions Then, n 3k ) 6k 3 6 k ) is divisible by 3. So that n n )n ) is divisible by 3. I n has the remainder when divided by 3, then by The Remainder Theorem, we have Then, n 3 k, where k is an integer. n 3k 3k 3 3 k ) is divisible by 3. So n n )n ) is divisible by 3. Problem #6: Let o ) ), n )... ))) is the nth composition. 99. Find ) Composition or superposition o two unctions g )) means: plug g) as an argument into ).) Let s observe the irst several compositions: ) ) - 5 -

6 nd Math Olympiad Solutions )) ) )) ) We can use mathematical induction to show that ) n n Then, 0 00 ) 99 ) 99. Problem #7: Numbers p and 8 p are both primes. Prove 8 p p is prime. Prime number is an integer > which is divisible only by and itsel.) When 3 p, 73 83) 8 p is prime. So ) 8 p p is prime. Net, we will prove there are no greater choices o prime p satisying the conditions.

7 nd Math Olympiad Solutions p is prime and p 3, then p isn t divisible by 3. That means p must have the remainder or when divided by 3. First, i p had the remainder when divided by 3, then by the Remainder Theorem, Then, p 3 k, where k is an integer 8 p 83k ) 89k 6k ) 7k 48k 9 34k 6k 3) is divisible by 3, and hence it s not prime. Second, i p had the remainder when divided by 3, then by the Remainder Theorem, Then, p 3k, where k is an integer 8 p 83k ) 89k k 4) 7k 96k 33 34k 3k ) is divisible by 3, and hence it s not prime. Problem #8: Write down the irst 3 signiicant digits o with a hundred 9 s. Show your work. Signiicant digits: the digits which start rom the irst non-zero digit ater the decimal point.) Given by Dr. Luemburg) We will use the ollowing two acts about inequalities: 8.) i and then - 7 -

8 nd Math Olympiad Solutions 8.) i and then First o all, we need to show that: 8.3) I a number is such that then 8.4) Let us irst prove that 8.5) Indeed, assume the contrary i.e. that 8.6) Then, multiplying both sides o 8.6) by we get and using property 8.) with 8.7) > or. However, due to 8.6), thereore, since the rule 8.) implies that 8.8) which contradicts the assumption that. This contradiction proves 8.5). Multiplying both parts o inequality 8.5) by, and using the rule 8.) we get 8.9) Thereore 8.4) ollows rom 8.8) and 8.9). From 8.5) it ollows that decimal representation starts with 0. ollowed by some digits. in Let now with a hundred 9s). Then. I in the irst 00 digits o there is at least one which is less than 9 then which contradicts 8.9). Thereore, the irst 00 digits o are all 9s, the same as or. A simpler proo: We know that < <, so - 8 -

9 nd Math Olympiad Solutions < <, i.e < < Hence the irst three signiicant digit o with a hundred 9 s is 999. Problem #9: Given any 50 integers, prove that you can choose some o them so that their sum is divisible by 50. Let s name the given integers a, a, a3,..., a50. Consider the ollowing 50 sums: S a S a a S 3 a a a3... S 50 a a a3... a50 I one o S, S, S3,..., S50 is divisible by 50, we are done. Now assume none o S, S, S3,..., S50 is divisible by 50, then each o S, S, S3,..., S50 has a non-zero remainder when divided by 50. But there are only 49 non-zero remainders:,,3,,49. By the Dirichlet Principle, there must be two sums S k and S t assume k < t ) among S, S, S3,..., S50 having the same remainder r. This means So, S k 50 k r and S t 50 k r S S t k 50k r) 50k r) 50k 50k 50 k k) - 9 -

10 nd Math Olympiad Solutions is divisible by 50. And S a, a, a,..., a. t S k ak ak... at is a sum o some o 3 50 Helpul Notes: <The Dirichlet Principle> Suppose we have at least n people living in n houses. The Dirichlet principle says that in this case, there should be a house with at least people in it. <The Remainder Theorem> For any positive integers a b, we can ind unique integers k and r such that a kb r, where 0 r < b.eg, when dividing 05 by 3, we will have 68 as the quotient and as the remainder, that means

Math-3 Lesson 8-5. Unit 4 review: a) Compositions of functions. b) Linear combinations of functions. c) Inverse Functions. d) Quadratic Inequalities

Math-3 Lesson 8-5. Unit 4 review: a) Compositions of functions. b) Linear combinations of functions. c) Inverse Functions. d) Quadratic Inequalities Math- Lesson 8-5 Unit 4 review: a) Compositions o unctions b) Linear combinations o unctions c) Inverse Functions d) Quadratic Inequalities e) Rational Inequalities 1. Is the ollowing relation a unction