Similar Triangles II

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Weird but Delicious Math Questions

Check out Mathcounts here--the best competition math for middle school mathletes.

Problem: (Solution below)
#1: 1993 Mathcounts National Team Round #4 :The teacher whispers positive integer A to Anna, B to Brett, and C to Chris. The students don't know one another's numbers but they do know that the sum of their numbers is 14. Anna says: "I know that Brett and Chris have different numbers." Then Brett says: "I already knew that all three of our numbers were different." Finally, Chris announces: "Now I know all three of our numbers." What is the product ABC?

#2: 2000 AMC10 #22: One morning each member of Angela’s family drank an 8-ounce mixture of coffee with milk. The amounts of coffee and milk varied from cup to cup, but were never zero. Angela drank a quarter of the total amount of milk and a sixth of the total amount of coffee. How many people are in the family?

Solution: 

#1: For Anna to know that Brett and Chris have different numbers, she must have an odd number
because 14 - odd = odd and you can only get odd sum of two numbers if they are different, one odd
one even.

For Brett to say that he already know all three have different numbers, he not only must have an odd number, but also his number has to be larger or equal to 7 and is not the same as what A have.

Otherwise, A-B-C could be 1-1-12; 3-3-8; or 5-5-4.
It would exceed 14 if you have 7-7-__. (All numbers are positive) so Brett"s and Anna's numbers must
be different.

If Brett has 7, then the numbers could be A-B-C = 1-7-6 ; 3-7-4 or 5-7-2.

If Brett has 9, then the numbers could be A-B-C = 1-9-4; or 3-9-2.

If Brett has 11, then then numbers could be A-B-C = 1-11-2.

From the above possibilities you know Chris has to have 6 for him to be sure he knows all the numbers. 

So A-B-C = 1-7-6 and the product of ABC = 1 x 7 x 6 = 42

#2: Let there be m cups of mild, c cups of coffee. and x people in Angela's family.
According to the given, you can set up the following equation: 
  m + c = = 8. Two ways to solve this equation.

Solution I:

m + c =   Both sides times 12 x

Both m and c need to be positive so the only x that works is when x = 5.

Solution II: 

m + c  = 8, Both sides times 12 and you have 

3m + 2c = 96; m + c is a multiple of 8.

30      3

28      6

26      9

.

.

.

2       45;  from  (30 + 3 ) = 33 to (2 +45) = 47 only 40 is a multiple of 8

and 40 divided by 8 = 5 so 5 is the answer.

 

Mental Math Tricks I

Multiples of 11:

11 x 14 = 1 ___ 4 , the middle number is (1 + 4) so the answer is 154

Let's try a few calculations mentally.

11 x 33 = 3 __ 3, the middle number is ( 3 + 3 ) so the answer is 363

11 x 52 = 5 __ 2, the middle number is ( 5+2 ) so the answer is 572

11 x 27 = 2 __ 7 , the middle number is ( 2 + 7 ) so the answer is 297

What about if the sum of the middle number is > 9? Then you carry over the "1" to the digit on its left
as you are doing the addition.

Examples:

11 x 67 = 6 __ 7, the middle number is (6 + 7) = 13 so the answer is 737

11 x 89 = 8 __ 9, the middle number is (8 + 9) = 17 so the answer is 979

11 x 47 = 4 __ 7, the middle number is (4 + 7) = 11 so the answer is 517

Harder trick:

11 x 223 = ?  2 _ _3, Write the two digits on the far left and right. Now add two numbers together,
starting with the unit digit. When the sum is  larger than 9,carry over to the next digit to the left.
2 _5(3+2) 3;  2 4 (2 + 2) 53 so the answer is 2453.

11 x 40532 = 4_ _ _ _ 2; 4_ _ _ 5 (2 +3) 2;  4_ _ 8(5+3)52;  4_ 5(0+5)852;  44(4 +0)5852,
so the answer is 445852  (This is fun!!)

Other interesting pattern:

11 x 11 =  121
111 x 111 = 12321
1111 x 1111 = 1234321
etc… till  111111111 x 111111111 = 12345678987654321

Divisibility rules for 11: 

The difference  of the sum of alternative digits is a multiple of 11, including “0”                                               (0 x 11 = 0, a multiple of 11.)

Example:

61985   (6 + 9 + 5) – (1 + 8) = 11   The number is divisible by 11.

7469     (7 + 6) – (4 + 9) = 0   The number is divisible by 11.

Try these mentally:(Answers below.)

#1: 11 x 23 =

#2: 11 x 72 =

#3: 11 x 97 =

#4: 11 x 55 =

#5: 11 x 76 =

#6: 11 x 60 =

#7: Sum of the first multiples of 11 smaller than 150.

#8:11 x 3421 =

#9: 11 x 452360 =

#10:  11 x 204673=

#11: 11111 x 11111=

#12: 1111111 x 1111111=

#13: What is the sum of the digits of 11111111 x 11111111?

#14: What is n if 45732n is divisible by 11?

#15: How many solutions for distinct numbers A and B if 4A8B is divisible by 11?

#16: How many solutions for distinct numbers A and B if A7B2 is divisible by 11?  What is their sum?

#17: How many solutions for distinct numbers A and B if A3B41 is divisible by 11?  

Answer key:
#1: 253
#2: 792
#3: 1067
#4: 605
#5: 836
#6: 660
#7 1001 The easiest way to do this is to see that this is an arithmetic sequence, starting with 11, 22, 33...
143 (11 x 13). There are 13 terms and the median is 77 so 13 x 77 or 13 x 7 x 11 = 1001
To do Mathcounts well, you need to know 7 x 11 x 13 =1001 by heart.
#8: 37631
#9: 4975960
#10: 2251403
#11:  123454321
#12:  1234567654321
#13:  The number is 123456787654321 so the sum of the digit is (4 x 7) x 2 + 8 = 64
#14:  n = 5
#15:  A + B = 12 (9, 3); (8, 4); (7, 5); skip (6, 6) because A and B are distinct (5, 7); (4, 8), (3, 9)
#16:  90, 81, 72, 63, 54, 45, 36, 27, 18;  A can’t be “0” so there are 9 pairs and the numbers are equally spaced, an arithmetic sequence. Thus the sum is median times how many numbers.                                      54 x 9 = 486
#17: A + B + 1 = 7 so A + B = 6 ; There are (6, 0);(5, 1); (4, 2); (2, 4); (1, 5) 
A + B + 1 = 7 + 11 = 18; A + B = 17; There are (9, 8) and (8, 9) so total 7 pairs

Mathcounts strategies: Some sums

The following sequences are all arithmetic sequences and for any arithmetic sequences, the sum is 

always average times the terms (how many numbers). 

To find the average, you can use 

a. sum divided by how many numbers.
b. average of the first and the last term, the second first and the second last term, or the third first and the third last term, etc...

Sum of the first consecutive natural numbers:

1 + 2 + 3 + 4 + 5 + ... + ( n -1 ) + n = \( \dfrac {n\left( n+1\right) } {2}\) 

Examples : 

#1: 1 + 2 + 3 + ... + 100 = \(\dfrac {100\left ( 101\right) } {2}=5050\)

#2: 1 + 2 + 3 + ... + 27 = \( \dfrac {27\left( 28\right) } {2}=378\)

#3: 4 + 8 + 12 + ... + 80 = 4 (1 + 2 + 3 + ... + 20) = \(\dfrac {4\times 20\times 21} {2}=840\)

Sum of the first consecutive natural even numbers: Proof without words 

Have you noticed in this sequence, every number is double the numbers in the first example,

so you don't need to divide by 2.  

\(2+4+6+\ldots +2n=n\left( n+1\right) \)

Examples: 

#1: 2 + 4 + 6 + ... 100 = 50 * 51 = 2550

#2: 2 + 4 + 6 + 8 + 10 + 12 = 6 x 7 = 42 

In this case, you can also find the midpoint, which is 7 and then 7 x 6 = 42

#3: 2 + 4 + 6 + ...420 = 210 * 211 = 44,310

Sum of the first consecutive odd numbers: Proof without words

In this special case, the mean is the same as how many numbers.

So it's easier if you find the mean by averaging the first and the last term and then square the mean.

1 + 3 + 5 + ... + ( 2n -1) = \(n^{2}\)

Examples: 

#1: 1 + 3 + 5 + ... + 39 = \(\left[ \dfrac {\left( 39+1\right) } {2}\right] ^{2}\) = (20^{2}\)

#2: 1 + 3 + 5 + ... + 89 = \(\left[ \dfrac {89+1} {2}\right] ^{2}=45^{2}=2025\)

#3: 1 + 3 + 5 + ... + 221 = \(\left[ \dfrac {221+1} {2}\right] ^{2}= 111^{2}=12321 \)

Applicable problems:

#1: What is the sum of the first 40 consecutive positive integers? 

#2: What is the sum of the first 40 consecutive positive even integers?

#3: What is the sum of the first 40 consecutive positive odd integers? 

#4: How many times does a 12-hour clock strikes in one day if it strikes once on one o'clock, twice on two o'clock, etc...?

#5: At a game show,you win $100 for the first correct answer and $200 for the second correct answer, etc. How much do you win if you answer 8 questions in a row correctly? 

Answer key:

#1: 820

#2: 1640

#3: 1600

#4: 156

#5: 100 + 200 + 300 ... + 800 = 100 (1 + 2 + 3 + ...8) = 100* \(\dfrac {8\times 9} {2}\) =  3600 

Painted Cube Problems: Beginning Level

Please refrain yourself from checking the answers too soon. Use Lego blocks, unit cubes, or Rubik's cubes to help you think. If you are really stuck, check out this link on painted cube problems.

Please comment and help me make my blog more user friendly. Thanks a lot!! 

Painted Cube Problem, competition math, problem solving by Mrs. Lin

Answer Key to the Painted Cube Problem