More AMC 8 Problems

Dr. Titu Andreescu

November 9, 2013

Problems

1. Complete 12 + 12÷ 6− 2 · 3 with one set of brackets ( ) in order to obtain 12.

2. Evaluate 10000000− 100000 · 90.

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?

4. Write the sequence of odd numbers without separating them:

13579111315171921 . . .

Find the digit that occupies position 2009.

5. Find the last digit of 20092009.

6. Find all primes a, b, c satisfying3a + 6b + 2c = 27

7. Find the positive integers a, b, and c, such that

ab = 144, bc = 240, ac = 60

8. A merchant offers a large group of items at 30% off. Later, the merchant takes 20% off the saleprices and claims the final price of the items is 50% off the original price. What is the actualtotal discount?

9. What percent of the numbers 1, 2, . . . , 1000 are divisible by at least one of the numbers 4 and5?

10. Consider the following sequence

1, 2, 2, 3, 3, 3, 4, 4, 4, 4, 5, 5, 5, 5, 5, . . .

Find the integer in the 100th position.

11. 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 adjacentdigits differ by 1.

13. If the average of three different positive integers is 70, what is the greatest possible value of oneof these integers?

14. Evaluate 1 + 2− 3 + 4 + 5− 6 + . . . + 109 + 110− 111.

15. How many perfect squares divide 211 · 313 · 517?

16. How many numbers are in the following sequence?

30, 45, 60, . . . , 2010

17. If the numbers 4a− 3 and 4b− 3 add up to 2010, find the sum of the numbers a3− 4 and b

3− 4.

18. The number of different pairs (m,n) of positive integers such thatm2 − n2 = 2012 is

(A) 0 (B) 1 (C) 2 (D) 3 (E) 4

19. Find the greatest number such that if we remove its fractional part, we obtain an integer thatis equal to 5

6of the original number.

Solutions

1. Complete 12 + 12÷ 6− 2 · 3 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 wenotice that 12÷ 6− 2 = 0 we can finally write

12 + (12÷ 6− 2) · 3 = 12

2. Evaluate 10000000− 100000 · 90.

Solution: We can write this as:

100000 · 100− 100000 · 90 = 100000 · (100− 90) = 100000 · 10 = 1000000

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 =4

3· 2013 so x =

2

3· 2013 = 1342

and

2y =2

3· 2013 so y =

1

3· 2013 = 671

4. Write the sequence of odd numbers without separating them:

13579111315171921 . . .

Find the digit that occupies position 2009.

Solution: The first 5 digits come from the five odd single-digit numbers. Next come the 90digits from the 45 odd 2-digit numbers. They are followed by 3 · 450 = 1350 digits from the450 odd 3-digit numbers. The total so far is 5 + 90 + 1350 = 1445. Since 2009 − 1445 = 564,and 564 is divisible by 4, our digit is the last digit of the 141-st odd 4-digit number, (which is1000 + 141 · 2− 1 = 1281). Therefore, the 2009-th digit in that sequence is 1.

5. Find the last digit of 20092009.

Solution: Since 20092009 = (200 · 10 + 9)2009 = 10k + 92009, the last digit of this integer is givenby the last digit of 92009. Since 2009 is odd, it follows that the last digit is 9.

6. Find all primes a, b, c satisfying3a + 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.

7. Find the positive integers a, b, and c, such that

ab = 144, bc = 240, ac = 60

Solution: From the given relations we obtain:

(ab)(bc)(ac) = 144 · 240 · 60

This relation is equivalent to:(abc)2 = 14402

hence we get abc = 1440. 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 theserelations are:

a = 6, b = 24, and c = 10

Alternatively, from the second and the third equation, we find that b = 4a. When we use thiswith the first equation, we find 4a2 = 144, i.e., a = 6. Then b = 24 and c = 10.

8. A merchant offers a large group of items at 30% off. Later, the merchant takes 20% off the saleprices and claims the final price of the items is 50% off the original price. What is the actualtotal discount?

Solution: The first discount means that the customer will pay 70% of the original price. Thesecond discount means a selling price of 80% of the discounted price. Because 0.80 ·0.70 = 56%,the customer pays 56% of the original price and thus receives a 44% discount.

9. What percent of the numbers 1, 2, . . . , 1000 are divisible by at least one of the numbers 4 and5?

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, . . . , 5 · 200. However, we overcounted– we counted the numbers divisible by both 4 and 5 twice. Those are the multiples of 20:20 · 1, 20 · 2, . . . , 20 · 50. It follows that we have 250 + 200− 50 = 400 numbers divisible by 4 or5 out of the 1000 given numbers, so the percentage is 400

1000= 40%.

10. Consider the following sequence

1, 2, 2, 3, 3, 3, 4, 4, 4, 4, 5, 5, 5, 5, 5, . . .

Find the integer in the 100th position.

Solution: The sequence1, 2, 2, 3, 3, 3, . . . , k, k, . . . , k︸ ︷︷ ︸

k times

contains 1 + 2 + 3 + · · ·+ k = k(k+1)2

terms.

For k = 13, we have 13·142

= 13 ·7 = 91 < 100, and for k = 14, we have 14·152

= 7 ·15 = 105 > 100.It follows that the number in position 91 is 13 and the number in position 92 is 14, hence inposition 100 is also 14.

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 + 171 = 209

This implies that x = 2.

12. Write the greatest possible number such that there are no repeated digits and no two adjacentdigits differ by 1.

Solution: Answer: 9758642031.

13. If the average of three different positive integers is 70, what is the greatest possible value of oneof these integers?

Solution: The sum of the three numbers is equal to 210. Note that the other two can eitherbe 1 or 2. Hence the maximum of the third number is 207. Clearly, 208 does not work becauseit would require one of the other numbers to be 0 or both of them equal. The same reasoningworks for 209 and 210.

14. Evaluate 1 + 2− 3 + 4 + 5− 6 + . . .− 109 + 110− 111.

Solution: In groups of three, this is equal to

0 + 3 + . . . + 108 = 3 · (0 + 1 + . . . + 36) = 3 · 36 · 37

2= 1998

15. How many perfect squares divide 211 · 313 · 517?

Solution: A perfect square that divides 211 · 313 · 517 is of the form 22a · 32b · 52c 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

6 · 7 · 9 = 378.

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 2010−3015

+1 =133 numbers in the given sequence.

17. If the numbers 4a− 3 and 4b− 3 add up to 2010, find the sum of the numbers a3− 4 and b

3− 4.

Solution: We are given that (4a− 3) + (4b− 3) = 2010 hence

a + b =2010 + 6

4= 504

Hencea

3− 4 +

b

3− 4 =

a + b

3− 8 =

504

3− 8 = 168− 8 = 160

18. The number of different pairs (m,n) of positive integers such thatm2 − n2 = 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)

19. Find the greatest number such that if we remove its fractional part, we obtain an integer thatis equal to 5

6of the original number.

Solution: Let bxc and {x} be the integer and fractional part of x, respectively. We havex = bxc+ {x} and bxc = 5

6(bxc+ {x}). This reduces to 6bxc = 5bxc+ 5{x} or bxc = 5{x}. It

follows that {x} = 0, 0.2, 0.4, 0.6, or 0.8. The greatest x is obtained for {x} = 0.8 and is equalto 4.8.