Hehehe,

Thanks for coming to the forum.  You never know when someone else might benefit from your questions.  It sounds like you need to do a lot of work on approximation problems.  First of all, realize that there are MANY ways of doing all sorts of problems.  Another reason I like the forum is because you might get different ideas.  Let me see if I can help.  Just realize my ways may not be the easiest, other people feel free to chime in if you have a different approach.

Let's look at how to multiply 2 numbers together first (approximating).  The first thing I do is make one of the numbers into a number that is easy to multiply and close to the original number.  Then take the difference and add or subtract it back to the second number:

Ex [1] 389 x 423 = _______________

We can change 389 to 400.  To do that we must add 11 to 389.  So we should subtract 11 from 423 and get 412.  Now we can multiply 400 x 412 = 164800.  Notice the range needs to be between 156320 and 172774.  We are almost right in between.  The actual answer is only 164547.  Look how close we are to the actual answer.  Just remember if you add numbers to the first number, you must compensate by subtracting numbers from the remaining and vice versa.

Ex [2] 272 x 411 = _______________

It's easier to change 411 to 400 in this case.  So we subtract 11 from 411 and add 11 to 272 and end up with 283 x 400.  In this case, we don't have to multiply 283 but we can change it to 280.  So we have 280 x 400 = 11200.  The range is between 106203 and 117381.  Not bad!

I will let you do the others.  Work on them.  They get easier with practice.

Now let's look at multiply 3 numbers:

In your example it is easier to just multiply them:  25 x 24 x 34 = 600 x 34 = 20400.  But in general take a look at my website on multiplying 3 numbers.  You can find that here:

multiplying 3 numbers

For changing base 9 to base 3 visit:  Base 9 to 3

For approximating e^n visit: Powers of e  (e^7 is approximately equal to 1100)

Now for the last question:

I assume the question read if n/4 has a remainder of 3, then 3n/4 has a remainder of ____.  On this one you can separate it out.  n/4 has a remainder of 3.  We know this part.  So now we need to find the remainder of 3n/4.  3n/4 = 3 * n/4.  n/4 has remainder of 3.  So 3*3 = 9 from the substition rule.  9/4 has a remainder of 1.  The answer is 1.  You really need to read up on Modulos.  I think this will help.  Do some research on the internet for modulos. 

If this fails, you can always plug in a number that will work.  For example, we know that n/4 has remainder of 3.  This will work for n=7.  So we need to know the remainder of 3n/4.  Plugging in 7 we get 3(7)/4 = 21/4 which has a remainder of 1.

Now a word of advice.  SKIPPING IS OKAY!!!!  If you can do the very next problem and get it right, you can skip the problem and get an extra point on your score.  If you skip every other problem on the test and get what you did right, you still make a 40.  BUT!!!!!  If you skip a problem, find out how to do it when the test is over and do it again and again until you have mastered the method.  This is a good way of improving your score.

Hope this helps,

Webmaster