Saturday, June 23, 2012

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.

#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?

#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

Thursday, June 21, 2012

Dimentional Change Questions III: Similar Shapes

There are numerous similar triangle questions on Mathcounts.
Here are the basics:

If two triangles are similar, their corresponding angles are congruent and their corresponding sides will have the same ratio or proportion.

= their height ratio = their perimeter ratio

Once you know the linear ratio, you can just square the linear ratio to get the area ratio and cube the linear ratio to get the volume ratio.

Practice Similarity of Triangles here.  Read the notes as well as work on the practice problems.  There is
instant feedback online.

Some students have trouble solving this problem when the two similar triangles are superimposed.

Just make sure you are comparing smaller triangular base with larger triangular base and smaller triangular side with corresponding larger triangular side, etc... In this case:

= their corresponding height ratio = their corresponding perimeter ratio.

Questions to ponder (Solutions below)
#1: Find the area ratio of Δ ABC to trapezoid BCDE to DEGF to FGIH. You can easily get those ratios using similar triangle properties. All the points are equally spaced and line
$\overline{BC}$ // $\overline{DE}$ // $\overline{FG}$ // $\overline{HI}$

#2: Find the volume of the cone ABC to Frustum BCDE to DEGF to FGIH. Again, you can use the similar cone, dimensional change property to easily get those ratios.Same conditions as the previous question.

#1:

#2:

For Young Mathletes 5 : Problems and Solutions

Please wait a little bit for the problems and solutions to appear.
Send me feedback through comments to make the problems more user friendly.
You are more than welcome to submit your questions with answers and solutions provided so I can include
them in my future problem sets.Thanks a bunch!!
Answer Key to for Young Mathletes 5

Wednesday, June 20, 2012

Mathcounts State Preparations: Number Sense

Check out Mathcounts: the best competition math program up to the national level.

Problems: (Solutions below)
#1: 2005 Chapter Team-- A standard deck of playing cards with 26 red cards and 26 black cards is split into two piles, each having at least one card. In pile A there are six times as many black cards as red cards. In pile B, the number of red cards is a multiple of the number of black cards. How many red cards are in pile B?

#2:  2000 State sprint #30. Joe bought a pumpkin that cost $10$ cents more per pound than his sister's. Together, the two pumpkins weighed $20$ pounds, but Joe's pumpkin was heavier. Joe paid $\ 3.52$ dollars and his sister paid $48$ cents. How many pounds did Joe's pumpkin weigh?

Solutions :
#1: You know the total cards in pile A is a multiple of 7 because there are six times as many black cards
as the red cards. (given)
6 Black, 1 Red on pile A gives you 20 Black and 25 Red cards on pile B. (doesn't work)
12 Black and 2 Red cards on pile A gives you 14 Black and 24 Red cards on pile B. (doesn't work)
18 Black and 3 Red cards on pile A gives you 8 Black and 23 Red cards on pile B. (doesn't work)
24 Black and 4 Red cards on pile A gives you 2 Black and 22 Red cards on pile B. Yes!!
The answer is 22 Red cards.

#2: Let x dollars be the cost per pound for Joe's sister's pumpkin and x + .1 dollars are the cost per pound for Joe's pumpkin. Since the pounds of each pumpkin is the cost $\div$ cost per pound, we have

Solution II:  Make a list:
Joe's sister        Joe
1 lb.                 19 lb.       (doesn't work since 19 x 58 cents are too much)
2 lbs.                18 lbs      (doesn't work)
3 lbs.                17 lbs      (No)17 x (48/3 + 10) = 442 (still too much)
4 lbs                 16 lbs      16 x (48/4 + 10)= 352 (yes)