More Amc 8 Problems

8/13/2019 More Amc 8 Problems

1/21


2/21


3/21

Solutions

1. Complete 12 + 12623 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 1262 = 0 we can finally write

12 + (1262)3 = 12


4/21

2. Evaluate 1000000010000090.

Solution: We can write this as:

10000010010000090 = 100000(10090) = 10000010 = 1000000


5/21

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

xy = 2013/3

Using the same approach as in the previous problem (Gaussian elimination), we get

2x= 43 2013 so x= 2

32013 = 1342

and

2y=2

32013 so y=

1

3 2013 = 671


6/21

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 20091445 = 564,and 564 is divisible by 4, our digit is the last digit of the 141-st odd 4-digit number, (which is

1000 + 14121 = 1281). Therefore, the 2009-th digit in that sequence is 1.


7/21

5. Find the last digit of 20092009.

Solution: Since 20092009 = (20010 + 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.


8/21

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

Solution: Since 2c= 27 3a 6b= 3(9a2b),it follows that c is divisible by 3, hencec = 3.We get 3a+ 6b= 21,that is a+ 2b= 7. The only possibility isa = 3 andb = 2.


9/21

7. Find the positive integersa,b,andc, such that

ab= 144, bc= 240, ac= 60

Solution: From the given relations we obtain:

(ab)(bc)(ac) = 14424060

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.


10/21

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.


11/21

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, 42, . . . , 4250,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 4001000 = 40%.


12/21

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.Fork = 13, we have 1314

2 = 13 7 = 91100.

It follows that the number in position 91 is 13 and the number in position 92 is 14, hence inposition 100 is also 14.


13/21

11. The sum of 19 consecutive integers is 209. Find the least of these integers.

Solution: Letx be the least of these 19 numbers. Hence

x+ (x+ 1) +. . .+ (x+ 18) = 19x+ 171 = 209

This implies that x = 2.


14/21

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

Solution: Answer: 9758642031.


15/21

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.


16/21

14. Evaluate 1 + 23 + 4 + 56 +. . .109 + 110111.

Solution: In groups of three, this is equal to

0 + 3 +. . .+ 108 = 3(0 + 1 +. . .+ 36) = 336 37

2 = 1998


17/21

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

679 = 378.


18/21

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 20103015 +1 =133 numbers in the given sequence.


19/21

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 (4a3) + (4b3) = 2010 hence

a+b=2010 + 6

4 = 504

Hencea

34 +

b

34 =

a+b

3 8 =

504

3 8 = 1688 = 160


20/21


21/21

19. Find the greatest number such that if we remove its fractional part, we obtain an integer thatis equal to 56 of the original number.

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

6(x+{x}). This reduces to 6x= 5x+ 5{x}orx= 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.

Documents

More Amc 8 Problems