I'm an 8th grader and got into algebra. My question is how to simplify problems like 5•99

Question

Other examples:

 

2•(473•5)

12•199

4•27•5•3

they do some different things like 

5*99

5*(100+(-1))

500+(-5)

=495

Answer 1

2*(473*5)=(2*5)*473=10*473=4730

12*199=12*(200-1)=12*200-12*1=2400-12=2388

4*27*5*3=4*5*3*27=20*3*27=60*27=60(30-3)=1800-180=1620

you are using the commutative, associative, and distributive properties

commutative: 2*3=3*2 or 4+5=5+4

associative: (2*3)*4=2*(3*4) or (5+6)+7=5+(6+7)

distributive: 5(2+3)=5*2+5*3, 6(7-4)=6*7-6*4, 6*99=6(100-1)=6*100-6*1

 

Answer 2

Fortunately, there is a formula for working out problems like this. The formula is called the "Order of Operations". The best way to remember the order is by using PEMDAS.

 

P
Parentheses first--do everything inside a set of Parentheses (---) OR Brackets [----] AND Braces {----} first.

E
Exponents--anything raised to a power or is a root.

MD
Multiplication and Division--starting from the left side and going right. Just like reading a sentence.

 

AS
Addition and Subtraction--again starting from the left and going right.

 

For example: A + B - (C * D) -  E + F²

 

you would start by squaring the F, then multiple C and D, then do the addition and subtraction.

Answer 3

The idea is to reduce a complex set of multiplications to a set of easier ones and a few easy additions or subtractions.  In the first problem , for example, you could multiply 2*473, then multiply that by 5 which is hard to do in your head.  However, it's easier if you realize that 2*5 is 10 and 10*473 is 4730 (just add a zero to the end of 473). It's quicker and easier than doing the problem the formal way.

 

2*(473*5) = 2*5*473 = 10*473 = 4730             

 

12*199 = 12(200 - 1) = 2400 - 12 = 2388

 

4*27*5*3 = 4*(25+2)*15 = (100 + 8)*15 = 1500 + (10-2)*15 = 1500 +150 - 30 = 1650 - 30 = 1620