ITS QUESTION TIME AGAIN!

useless · Long-term Member

ITS QUESTION TIME AGAIN!

fht · Long-term Member

Aw, the first one is one of the many cherished shortcuts from the old tests.

16) n(n-1) + (8n+1)n + (n+1)(n+4), expand and combine, => (3n+2)^2 => (n=7) 23^2 = 529

21) 228-168=60, 168-132=36, (60,36)=12

24) 1*3367=3367, 2*3367=6734, and, most importantly,

3*3367=10101. So 17*3367=(15+2)*3367=50505+6734=57239

30*) Round it to 30000/130 => 300/1.3=230(13*23=299)

37) The commas are big give-aways because they separate the number into groups of 3. You look at the first group=>1, that means 1 on the hundred's place for the cube root. Then look at the first two groups 1,367. What is the greatest cube that's less than 1367=>11, so it's eleventy something. Look at the last digit=>1, only 1 cubed ends with one(2 is 8, 3 is 7, 4 is 4, 5 is 5, 6 is 6, 7 is 3, 8 is 2, 9 is 9, 0 is 0), so the cube root is 111. Additional evidence include: 1367631 is not much more than 110^3=1331000, and also that 1367631 is a palindrome, which indicate repetition of 1s in its power roots(like 11111^2 is 123454321 and such).

40*) Round to 160000=>400, it'd be 403 (162409)to be more exact.

Best Wishes

kyniemxotxa · Star Member

so that's the shorcut way to do GCD? does it always work?

also for

30*) Round it to 30000/130 => 300/1.3=230(13*23=299)

how did you get the part in bold? why's there a 299?

kyniemxotxa · Long-term Member

Thanks alot Sam! I really like the (3n+2)^2 trick. That's really neat! And the gcd trick is absolutely awesome... I just wish I could do them faster. I think that I'm going to need to practice the 3367's problems, especially when one problem requires you to know 2*3367 and 3*3367 going into it. The cube root, and square root problems make total and complete sense now. Thanks!

And kyniemxotxa, the way Sam posted to do GCD is pretty fast, but you have to make sure that the number you write down is, indeed, a factor. Not so much on the problems with 3 numbers, but on the ones with 2. White has been known to throw a few nasty tricks in his test that require you to think a little before you write.

Also... The part in bold just shows you how he got 230, when he divided by 13. If you have your times tables memorized up to 25, thats an easier way of figuring out the approximate answer, since 13*23 almost equals 300.

I'm gonna take another test in about an hour or so... So there will probably be more questions posted by the end of the day!

- Zack -

kyniemxotxa · Star Member

im not sure i understand what you said about the GCD. what's the algorithm to perform it? does it work with both 2 and 3 numbers? what factors are you referring to? they all need to have 2 as a factor?

useless · Long-term Member

Sorry. I was really unclear in that last post.

To find the GCD of three numbers take the difference between the first two and the last two. Find the GCD of the two differences, and you should have the GCD of the three numbers.

Ex. The GCD of 16, 28, and 84 is ___________________

Difference in first two numbers: (28-16) = 12

Difference in last two numbers: (84-28) = 56.

GCD(56,12) is 4. so... the answer = 4.

When finding the GCD of two numbers... Start by taking the difference between them. This is the maximum GCD you can have. If that number doesnt divide evenly into both numbers, you'll have to take the prime factorization of the number, and see if any of its factors are shared by the other two numbers

Ex. The GCD of 56 and 36 is ___________________

Difference = (56-36) = 20. Does 20 divide evenly into both? No.

20 = 2*2*5.

36 and 56 are both divisible by 2*2 so your answer is 4.

useless · Long-term Member

These are off the 201 B test

20) 103635 / 147 / 3 = _________________.

ARG!

25) (14^2 +3*12+1) / 6 has a remainder of _______________.

I hate these... I never get these right. How are they done?

bradp · Senior Member

for 3367: http://math-magic.com/multiply_numbers/mult_3367.htm

Date: May 21, 2006

to follow up on the GCD, if the difference is divisible into both, then that's the GCD? does this also work with GCF? i dont know the difference between them, i've always thought they're the same.

useless · Long-term Member

Factor... Denominator... Its all the same thing. And yeah... if the difference is divisible into both numbers, that has to be the GCF or GCD.

- Zack -

fht · Long-term Member

Suppose there are many numbers and they have GCD of k. Therefore they are all multiples of k's. Now think about this -- no matter how you subtract those numbers, the difference will still be multiples of k because you can pull k out and distribute. So in some cases you do need a few more subtractions, if you aim to subtract to get a smaller number, the smallest you'd get must be the smallest multiple of k -- k itself.

fht · Long-term Member

#20 is a hard one. There are several lousy ways to do it,

147*3=441, so do 100000/440=1000/4.4=230 (44*23=1012)

Or round it to 105000/441=10500/21/21=5000/21=240 (21*24=504)...

Well, the answer is exactly 235, so, either way.

#25 Do actually do what's in the parenthesis. Consider everything as remainders of 6.

(14^2 +3*12+1) / 6 ===>> (2^2+0+1)=5

Best Wishes

Date: May 23, 2006

Hey guys, long time since I posted.

#20

103635 / 147 / 3

Make the problem 103000 / 450

Make the problem 10300 / 45

Now the problem is 103/45 plus two digits places

For 103 / 45, Do 2 13/45

Make 13/45 into 15/45 or 1/3

2 1/3 is 2.33

Answer 233.

The trick to doing a lot of the estimations is to do an extremely crude estimation and then add or subtract to your answer based on how you initially rounded.

For example, if I am estimating 18 x 20 x 22, I know that the answer will be less than 8000. What I don't know is exactly how much less than 8000 the answer will be. 5 % of 8000 is 400. So, if I put 7600, I should be safe.

That estimation problem was pretty straightforward. There really is no need to do the 5% thing.

Say the problem was 30000/130.

Make the problem 30/13 plus two digits.

30/13 is 2 4/13.

Make 2 4/13 into 2 4/12 or 2 1/3.

This is 233. Because 4/13 is less than 4/12, the answer will be less than 233.

5% of 233 is roughly 10. For your answer, you could put 223 and be safe.

Once again, the 5% is not necessary in this problem. But, the cruder the rounding and estimations used, the more the 5% ending estimation is needed.

One more thing, when estimating using pi, use 22/7 or 3 1/7.

For example, say we had 45 x pi.

45 x 3 = 135; 45 x 1/7 = about 7. Answer is 142.

I lied, one more thing. When writing the answer to an estimation problem ending in many zeros. Say the answer was 40000. Write 40111. This will save time because it is quicker to write a 1 than a zero. Also, there is less chance of making a writing error. It may seem like nothing. But if you are nearly finishing a test, this trick can save you around 3-4 seconds per test.

fht · Long-term Member

The fraction method is very good for people that are fond of and better with fractions, and unfortunately, I am not one of them.

Best Wishes

kyniemxotxa · Star Member

fht wrote:

(14^2 +3*12+1) / 6 ===>> (2^2+0+1)=5

how did u simplified 14^2 = 2^2? and for any other remainder problems similar to that.

useless · Long-term Member

You take the remainder of the number and square it. The remainder when 14 is divided by 6 is 2, so since 14is squared, you also square the remainder. After you do that, you add up the remainders of all the other numbers when 6 is divided into them. It makes the problem easier that way.