More AMC 8 Problems. Dr. Titu Andreescu. November 9, 2013

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1 More AMC 8 Problems Dr. Titu Andreescu November 9, 2013

2 Problems 1. Complete with one set of brackets ( ) in order to obtain Evaluate I am thinking of two numbers. Their sum is 2013 and their difference is a third of their sum. What numbers am I thinking of? 4. Write the sequence of odd numbers without separating them: Find the digit that occupies position Find the last digit of Find all primes a, b, c satisfying 3a + 6b + 2c = Find the positive integers a, b, and c, such that ab = 144, bc = 240, ac = A merchant offers a large group of items at 30% off. Later, the merchant takes 20% off the sale prices and claims the final price of the items is 50% off the original price. What is the actual total discount? 9. What percent of the numbers 1, 2,..., 1000 are divisible by at least one of the numbers 4 and 5? 10. Consider the following sequence Find the integer in the 100 th position. 1, 2, 2, 3, 3, 3, 4, 4, 4, 4, 5, 5, 5, 5, 5, The sum of 19 consecutive integers is 209. Find the least of these integers. 12. Write the greatest possible number such that there are no repeated digits and no two adjacent digits differ by If the average of three different positive integers is 70, what is the greatest possible value of one of these integers? 14. Evaluate How many perfect squares divide ? 16. How many numbers are in the following sequence? 30, 45, 60,..., If the numbers 4a 3 and 4b 3 add up to 2010, find the sum of the numbers a 3 4 and b The number of different pairs (m, n) of positive integers such that m 2 n 2 = 2012 is (A) 0 (B) 1 (C) 2 (D) 3 (E) Find the greatest number such that if we remove its fractional part, we obtain an integer that is equal to 5 of the original number. 6

3 Solutions 1. Complete with one set of brackets ( ) in order to obtain 12. Solution: This is a fairly simple problem, so we can use the trial-and-error approach. Once we notice that = 0 we can finally write 12 + (12 6 2) 3 = 12

4 2. Evaluate Solution: We can write this as: = (100 90) = =

5 3. I am thinking of two numbers. Their sum is 2013 and their difference is a third of their sum. What numbers am I thinking of? Solution: If the two numbers are x and y, the equations are x + y = 2013 x y = 2013/3 Using the same approach as in the previous problem (Gaussian elimination), we get 2x = so x = = and 2y = so y = = 671 3

6 4. Write the sequence of odd numbers without separating them: Find the digit that occupies position Solution: The first 5 digits come from the five odd single-digit numbers. Next come the 90 digits from the 45 odd 2-digit numbers. They are followed by = 1350 digits from the 450 odd 3-digit numbers. The total so far is = Since = 564, and 564 is divisible by 4, our digit is the last digit of the 141-st odd 4-digit number, (which is = 1281). Therefore, the 2009-th digit in that sequence is 1.

7 5. Find the last digit of Solution: Since = ( ) 2009 = 10k , the last digit of this integer is given by the last digit of Since 2009 is odd, it follows that the last digit is 9.

8 6. Find all primes a, b, c satisfying 3a + 6b + 2c = 27 Solution: Since 2c = 27 3a 6b = 3(9 a 2b), it follows that c is divisible by 3, hence c = 3. We get 3a + 6b = 21, that is a + 2b = 7. The only possibility is a = 3 and b = 2.

9 7. Find the positive integers a, b, and c, such that Solution: From the given relations we obtain: This relation is equivalent to: ab = 144, bc = 240, ac = 60 (ab)(bc)(ac) = (abc) 2 = hence we get abc = Since bc = 240, from relation a(bc) = 1440, it follows that a = 6. From ab = 144, we get b = 24, and from ac = 60, we get c = 10. The integers satisfying these relations are: a = 6, b = 24, and c = 10 Alternatively, from the second and the third equation, we find that b = 4a. When we use this with the first equation, we find 4a 2 = 144, i.e., a = 6. Then b = 24 and c = 10.

10 8. A merchant offers a large group of items at 30% off. Later, the merchant takes 20% off the sale prices and claims the final price of the items is 50% off the original price. What is the actual total discount? Solution: The first discount means that the customer will pay 70% of the original price. The second discount means a selling price of 80% of the discounted price. Because = 56%, the customer pays 56% of the original price and thus receives a 44% discount.

11 9. What percent of the numbers 1, 2,..., 1000 are divisible by at least one of the numbers 4 and 5? Solution: The given array contains 250 numbers that are divisible by 4: 4 1, 4 2,..., 4 250, and 200 numbers that are divisible by 5: 5 1, 5 2,..., However, we overcounted we counted the numbers divisible by both 4 and 5 twice. Those are the multiples of 20: 20 1, 20 2,..., It follows that we have = 400 numbers divisible by 4 or 5 out of the 1000 given numbers, so the percentage is 400 = 40%. 1000

12 10. Consider the following sequence Find the integer in the 100 th position. Solution: The sequence 1, 2, 2, 3, 3, 3, 4, 4, 4, 4, 5, 5, 5, 5, 5,... 1, 2, 2, 3, 3, 3,..., k, k,..., k }{{} k times contains k = k(k+1) terms. 2 For k = 13, we have = 13 7 = 91 < 100, and for k = 14, we have = 7 15 = 105 > It follows that the number in position 91 is 13 and the number in position 92 is 14, hence in position 100 is also 14.

13 11. The sum of 19 consecutive integers is 209. Find the least of these integers. Solution: Let x be the least of these 19 numbers. Hence x + (x + 1) (x + 18) = 19x = 209 This implies that x = 2.

14 12. Write the greatest possible number such that there are no repeated digits and no two adjacent digits differ by 1. Solution: Answer:

15 13. If the average of three different positive integers is 70, what is the greatest possible value of one of these integers? Solution: The sum of the three numbers is equal to 210. Note that the other two can either be 1 or 2. Hence the maximum of the third number is 207. Clearly, 208 does not work because it would require one of the other numbers to be 0 or both of them equal. The same reasoning works for 209 and 210.

16 14. Evaluate Solution: In groups of three, this is equal to = 3 ( ) = = 1998

17 15. How many perfect squares divide ? Solution: A perfect square that divides is of the form 2 2a 3 2b 5 2c where 2a = 0, 2, 4, 6, 8, 10 (6 possibilities) 2b = 0, 2, 4, 6, 8, 10, 12 (7 possibilities) 2c = 0, 2, 4, 6, 8, 10, 12, 14, 16 (9 possibilities) Hence the total number of perfect squares that divide our given number is = 378.

18 16. How many numbers are in the following sequence? 30, 45, 60,..., 2010 Solution: The numbers are 15 units apart from each other so there will be a total of = 133 numbers in the given sequence.

19 17. If the numbers 4a 3 and 4b 3 add up to 2010, find the sum of the numbers a 3 4 and b 3 4. Solution: We are given that (4a 3) + (4b 3) = 2010 hence Hence a + b = = 504 a b 3 4 = a + b 8 = = =

20 18. The number of different pairs (m, n) of positive integers such that m 2 n 2 = 2012 is (A) 0 (B) 1 (C) 2 (D) 3 (E) 4 Solution: The answer is (B). Because (m n) + (m + n) = 2m is an even number, (m n) and (m + n) are both even or both odd. They cannot be odd, because (m n)(m + n) = 2012 an even number. Hence m n = 2a and m + n = 2b for some positive integers a and b, a < b. It follows that (2a)(2b) = 2012, implying ab = 503, a prime number. Hence a = 1 and b = 503, yielding the unique pair (m, n) = (504, 502)

21 19. Find the greatest number such that if we remove its fractional part, we obtain an integer that is equal to 5 of the original number. 6 Solution: Let x and {x} be the integer and fractional part of x, respectively. We have x = x + {x} and x = 5 ( x + {x}). This reduces to 6 x = 5 x + 5{x} or x = 5{x}. It 6 follows that {x} = 0, 0.2, 0.4, 0.6, or 0.8. The greatest x is obtained for {x} = 0.8 and is equal to 4.8.

Grade VIII. Mathematics Formula Notes. #GrowWithGreen

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