Fast Arithmetic Tips

Fast Arithmetic Tips

Getting the result fast

1. Multiplication by 5
   
   It's often more convenient instead of multiplying by 5 to multiply first by 10 and then divide by 2.
   For example,

   137·5 = 1370/2 = 685.

   
   
   
   
   
2. Division by 5
   
   Similarly, it's often more convenient instead to multiply first by 2 and then divide by 10.
   For example,

   1375/5 = 2750/10 = 275.

   
   
   
   
3. Division/multiplication by 4
   
   Replace either with a repeated operation by 2.
   For example,

   124/4 = 62/2 = 31. Also, 124·4 = 248·2 = 496.

   
   
   
   
4. Division/multiplication by 25
   
   Use operations with 4 instead.
   For example,

   37·25 = 3700/4 = 1850/2 = 925.

   
   
   
   
5. Division/multiplication by 8
   
   Replace either with a repeated operation by 2.
   For example,

   124·8 = 248·4 = 496·2 = 992.

   
   
   
   
6. Division/multiplication by 125
   
   Use operations with 8 instead.
   For example,

   37·125 = 37000/8 = 18500/4 = 9250/2 = 4625.

   
   
   
   

7. Squaring two digit numbers.
   
   

   
   
   1. You should memorize the first 25 squares:

      1 2 3 4 5 6 7 8 9 10 11 12 13 14 1 4 9 16 25 36 49 64 81 100 121 144 169 196 15 16 17 18 19 20 21 22 23 24 25 225 256 289 324 361 400 441 484 529 576 625

      
      
      
   2. Squares of numbers from 26 through 50.
      
      Let A be such a number. Subtract 25 from A to get x. Subtract x from 25 to get, say, a. Then
      A² = a² + 100x. For example, if A = 26, then x = 1 and 1375/5 = 2750/10 = 275. Hence

      262 = 242 + 100 = 676.

      Similarly, if A = 37, then x = 37 - 25 = 12, and a = 25 - 12 = 13. Therefore,

      372 = 132 + 100·12 = 1200 + 169 = 1369.

      
      
      Why does this work?

      (25 + x)2 - (25 - x)2 = [(25 + x) + (25 - x)]·[(25 + x) - (25 - x)]   = 50·2x   = 100x.

      
      
      
      
   3. Squares of numbers from 51 through 99.
      
      The idea is the same as above.

      (50 + x)2 - (50 - x)2 = 100·2x   = 200x.

      
      
      For example,

      32 = 372 + 200·13 = 1369 + 2600 = 3969.

      
      
      
      
   4. Squares of numbers from 51 through 99, second approach (this one was communicated to me by my late father Moisey Bogomolny).
      
      We are looking to compute A², where A = 50 + a. Instead compute 100·(25 + a) and add a². Example: 57². a = 57 - 50 = 7. 25 + 7 = 32. Append 49 = 7². Answer: 57² = 3249.
   5. Squares can be computed sequentially: (a + 1)² = a² + a + (a + 1). For example,

      1112 = 1102 + 110 + 111   = 12100 + 221   = 12321.

      
      
      
   6. In general, a² = (a + b)(a - b) + b². Let a be 57 and, again, we wish to compute 57². Let b = 3. Then

      572 = (57 + 3)(57 - 3) + 32,

      
      
      or

      572 = 60·54 + 9   = 3240 + 9   = 3249.

      
      
      
   
   
   
   
8. Squares of numbers that end with 5.
   
   Let A = 10a + 5. Then

   A2 = (10a + 5)2 = 100a2 + 2·10a·5 + 25 = 100a(a + 1) + 25.

   
   
   For example, compute 115², where a = 11. First compute 11·(11 + 1) = 11·12 = 132 (since 3 = 1 + 2). Next, append 25 to the right of 132 to get 13225!
   Another example: to compute 245² let a = 24. Then

   24·(24 + 1) = 242 + 24 = 576 + 24 = 600.

   
   
   Therefore 245²=60025. Here is another way to compute 24·25:

   24·25 = 2400/4 = 1200/2 = 600.

   
   
   The rule naturally applies to 2-digit numbers as well. 75² = 5625 (since 7·8 = 56).
9. Product of 10a+b and 10a+c where b+c = 10.
   
   Similar to the squaring of numbers that end with 5:

   (10a + b)(10a + c) = 100a2 + 10a·(b + c) + bc = 100a(a + 1) + bc.

   
   
   For example, compute 113×117, where a = 11, b = 3, and c = 7. First compute 11·(11 + 1) = 11·12 = 132 (since 3 = 1 + 2). Next, append 21 (= 3×7) to the right of 132 to get 13221! Another example: compute 242×248, with a = 24, b = 2, and c = 8. Then

   24·(24 + 1) = 242 + 24 = 576 + 24 = 600.

   
   
   Therefore 242×242²=60016.
10. Product of two one-digit numbers greater than 5.
    
    This is a rule that helps remember a big part of the multiplication table. Assume you forgot the product 7·9. Do this. First find the excess of each of the multiples over 5: it's 2 for 7 (7 - 5 = 2) and 4 for 9 (9 - 5 = 4). Add them up to get 6 = 2 + 4. Now find the complements of these two numbers to 5: it's 3 for 2 (5 - 2 = 3) and 1 for 4 (5 - 4 = 1). Remember their product 3 = 3·1. Lastly, combine thus obtained two numbers (6 and 3) as 63 = 6·10 + 3.
    
    
    The explanation comes from the following formula:

    (5 + a)(5 + b) = 10(a + b) + (5 - a)(5 - b)

    
    
    In our example, a = 2 and b = 4.

11. Product of two 2-digit numbers.
    
    

    
    
    1. If the numbers are not too far apart, and their difference is even, one might use the well known formula

       (a + n)(a - n) = a2 - n2.

       
       
       a here is the average of the two numbers.
       For example,

       28·24 = 262 - 22 = 676 - 4 = 672

       
       
       since 26 = (24 + 28)/2. Also,

       19·31 = 252 - 62 = 625 - 36 = 589

       
       
       since 25 = (19 + 31)/2.
    2. If the difference is odd use either n(m + 1) = nm + n or n(m - 1) = nm - n.
       For example,

       7·34 = 37·35 - 37 = 362 - 12 - 37 = 1296 - 1 - 37 = 1258.

       
       
       On the other hand,

       37·34 = 37·33 + 37 = 352 - 22 + 37 = 1225 - 4 + 37 = 1258.

       
       
       
       
    3. Babilonian method: ab = [(a + b)² - (a - b)²]/4. For example,

       33×32 = [652 - 12]/4 = (4225 - 1)/4 = 4224/4 = 1056.

       
       
       
       
    
    
    
    
12. Product of numbers that only differ in units.
    
    If the numbers only differ in units and the sum of the units is 10, like with 53 and 57 or 122 and 128, then think of them as, say 10a + b and 10a + c, where b + c = 10. The product (10a + b)(10a + c) is given by

    (10a + b)(10a + c) = 100a2 + 10a(b + c) + bc = 100a(a + 1) + bc.

    
    
    Thus to compute 53 times 57 (a = 5, b = 3, c = 7), multiply 5×(5 + 1) to get 30. Append to the result (30) the product of the units (3·7 = 21) to obtain 3021. Similarly,

    122·128  =  12·13·100 + 2·8 = 15616.

    
    
    
    
13. Product of numbers close to 100.
    
    
    Say, you have to multiply 94 and 98. Take their differences to 100: 100 - 94 = 6 and 100 - 98 = 2. Note that 94 - 2 = 98 - 6 so that for the next step it is not important which one you use, but you'll need the result: 92. These will be the first two digits of the product. The last two are just 2×6 = 12. Therefore, 94×98 = 9212.The same trick works with numbers above 100: 93×102 = 9486. First find the differences: 100 - 93 = 7 and 100 - 102 = -2. Then subtract one of the differences from the other number, e.g. 93 - (-2) = 95. This intends to show the first two digits of the product, i.e., the number 9500. Add to this the product of 7 and -2, or -14: 9500 - 14 = 9486.This works because

    (100 - a)(100 - b) = (100 - a)·100 - (100 - a)·b   = (100 - a)·100 - 100b + ab   = (100 - a - b)·100 + ab.

    
    
    
    
14. Multiplying by 11.
    
    To multiply a 2-digit number by 11, take the sum of its digits. If it's a single digit number, just write it between the two digits. If the sum is 10 or more, do not forget to carry 1 over.
    For example, 34·11 = 374 since 3 + 4 = 7. 47·11 = 517 since 4 + 7 = 11.
15. Faster subtraction.
    
    Subtraction is often faster in two steps instead of one.
    For example,

    427 - 38 = (427 - 27) - (38 - 27) = 400 - 11 = 389.

    
    
    
    A generic advice might be given as "First remove what's easy, next whatever remains". Another example:

    1049 - 187 = 1000 - (187 - 49) = 900 - 38 = 862.

    
    
    
    
16. Faster addition.
    
    Addition is often faster in two steps instead of one.
    For example,

    487 + 38 = (487 + 13) + (38 - 13) = 500 + 25 = 525.

    
    
    A generic advice might be given as "First add what's easy, next whatever remains". Another example:

    1049 + 187 = 1100 + (187 - 51) = 1200 + 36 = 1236.

    
    
    
    
17. Faster addition, #2.
    
    It's often faster to add a digit at a time starting with higher digits. For example,

    583 + 645 = 583 + 600 + 40 + 5   = 1183 + 40 + 5   = 1223 + 5   = 1228.

    
    
    
    
    
18. Multipliply, then subtract.
    
    When multiplying by 9, multiply by 10 instead, and then subtract the other number. For example,

    
    
    

    23·9 = 230 - 23 = 207.

    
    
    The same applies to other numbers near those for which multiplication is simplified:
    
    
    
    
    
    
    
    
    
    
    
    
    

    	23·51	= 23·50 + 23
    		= 2300/2 + 23
    		= 1150 + 23
    		= 1173.
    	87·48	= 87·50 - 87·2
    		= 8700/2 - 160 - 14
    		= 4350 - 160 - 14
    		= 4190 - 14
    		= 4176.

    
    
    

The Guess Card

This guess card is a puzzle made from 2-base number system. To determine how much card is needed, we must take the range of the age we will guess. For example, we will guess the age of the boy whose range 1-15 years old, so here we need 4 cards. Here we use the relation between 10-base number and 2-base number. With the four first 2-base numbers we can make 24-1 combination. So that cards contain till 15 numbers, and so on. Let’s look like the picture:

If the child who his age is guess say that his age is in the card A, C and D then the child’s age definitely 13. Where is the number from???
The 13 number is the sum of the smallest number in the A, C and D cards. They are 1+4+8 = 13.
Try the other number… :)
Next time I try to make some application to generate this card, so you can play this game. See next…

The Reverse Numbers

Let’s try the following steps:
* Write the number with 2 digits.
* Reverse that number.
* Find the difference between that 2 numbers from step 1 and step 2
* Divide the answer from step 3 by the positive difference between 2 digits number from step 1.
* The result absolutely always 9. Why??
Example:

step 1	97	81	12
step 2	79	18	21
step 3	|97 – 79| = 18	|81 – 18| = 63	|12 – 21| = 9
step 4	18 : |9-7| = 9	63 : |8-1| = 9	9 : |1-2| = 9

The S-methic(Smart Arithmetic)

Here I have an application for the child who learning arithmetic. I made this application using Java. Actually this is my assignment in Java class. Not too bad for the beginner like me. I try to make some option as like addition, substraction, multiplication and division. Also there are the level choice, beginner, intermediate and advance. I mean here this game is for the child who learning arithmetic.

click here for download

About Number

the Interest thing about 10-base number
1 X 9 + 2 = 11
12 X 9 + 3 = 111
123 X 9 + 4 = 1111
1234 X 9 + 5 = 11111
12345 X 9 + 6 = 111111
123456 X 9 + 7 = 1111111
1234567 X 9 + 8 = 11111111
12345678 X 9 + 9 = 111111111

9 X 9 + 7 = 88
98 X 9 + 6 = 888
987 X 9 + 5 = 8888
9876 X 9 + 4 = 88888
98765 X 9 + 3 = 888888
987654 X 9 + 2 = 8888888
9876543 X 9 + 1 = 88888888
98765432 X 9 + 0 = 888888888

999999 X 2 = 11
999999 X 3 = 111
999999 X 4 = 1111
999999 X 5 = 11111
999999 X 6 = 111111
999999 X 7 = 1111111
999999 X 8 = 11111111
999999 X 9 = 111111111

what is the conclusion from these numbers???

Welcome

Welcome back in new blog. actually I'm doing something for my new legal site but cause of many trouble so that's pending. before I use this blog I've wrote in wordpress. when it proceed I found any difficulty so I move the site here.
ok welcome still in GAMEMATICS new version.

I hope your opinion and support. thx