some Q's

[B3ndythumbs]

I've seen many problems such as diff. of 2 consec. octagonal numbers or hexagonal numbers or the like. i have derived some formulas for them, but is it worth memorzing? or would u guys just compute it?

if u have like x/80 to a decimal, do u multiply the top by 5/4? i seem to have very bad accuracy when doing these problems -.-

if u have a problem such as 142.857x78 (est.), i divide the 78 by 7 becuz 142 is about 1000/7. but how do u find the number of digits in the answer? would u add the numbero f digits in both numbers and subtract 1?

[stupidityismygam]

1) compute

2) [unparseable or potentially dangerous latex formula]. So just divide by 8 and move the decimal place

3) [unparseable or potentially dangerous latex formula]. Since [unparseable or potentially dangerous latex formula] you add 3 digits to the end of the 11 R 1. So, [unparseable or potentially dangerous latex formula]

[AuSmith]

1) How did your formulas look? It would be good to memorize them, but not by simple rote memory. When you understand them thoroughly enough, memorization is no effort. Like, [unparseable or potentially dangerous latex formula] makes lots of sense after you manipulate Pythagorean theorem some. Here, the following definition for the [unparseable or potentially dangerous latex formula]th
[unparseable or potentially dangerous latex formula]-gonal number is conducive to subtraction because it's a series.



The red dots are the new ones (the consecutive differences that you're looking for). You can tell from the pictures that with each successive figure, there are [unparseable or potentially dangerous latex formula] more red dots than last time: one for each of the red sides. If you add [unparseable or potentially dangerous latex formula] red dots each time and start with only one red dot, it follows that the number of red dots on the [unparseable or potentially dangerous latex formula]st figure is one more than [unparseable or potentially dangerous latex formula] times [unparseable or potentially dangerous latex formula]. Now your formula is

[unparseable or potentially dangerous latex formula].

This also gives way to the formula

[unparseable or potentially dangerous latex formula].

2) Yea, multiply the top by [unparseable or potentially dangerous latex formula] and move the decimal two places back.

3) Substitute [unparseable or potentially dangerous latex formula] for [unparseable or potentially dangerous latex formula], then you still have [unparseable or potentially dangerous latex formula] after dividing by [unparseable or potentially dangerous latex formula]. Just add [unparseable or potentially dangerous latex formula] zeros... to [unparseable or potentially dangerous latex formula] or whatever you got. It's not so much an issue of how many total digits, but how many [unparseable or potentially dangerous latex formula]'s to tack on.

Was all that clear?

[B3ndythumbs]

Ya thanx!

also i was doing some 2006-2007 tmsca numsen tests and saw some weird things.

The graph of y=(x^2-9)/(x^2-5x+6) has a hole at...
it should be 2 AND 3 right? they only had 3 as the answer.

A 5-element set has ??? improper subsets.
Improper is always one, namely the set itself. they said 31. i think they meant proper?

36 has how many integral divisors?
unfortuneately i missed the fact that they did not say positive...so u multiply the PIDs by 2?

how do u find number of PRIME PIDs?

[SumitG]

1. The graph should only have a hole at x=3, since at x=2, there is a vertical asymptote. "Holes" basically occur at points where the denominator of a function equals 0 but the function itself can be simplified to another function, an agreeable function, whose denominator does not equal 0 when the first function did. Here, x^2-9=(x+3)(x-3), and x^2-5x+6=(x-3)(x-2). The x-3's can cancel out to give us an agreeable function, so we have a hole at x=3.

2. They probably meant proper or the key was wrong.

3. Yes, you simply multiply the number of positive integral divisors by 2.

4. I've always just found them intuitively.

[AuSmith]

1) There's a hole at [unparseable or potentially dangerous latex formula] because the limit exists at [unparseable or potentially dangerous latex formula], but

[unparseable or potentially dangerous latex formula].

So, when you're tracing along the graph, everything's hunky dory close to [unparseable or potentially dangerous latex formula], but it's like when Obi Wan couldn't find the planet in the archives... Gravitational forces indicated the existence of a planet, yet the archives showed none. Here [unparseable or potentially dangerous latex formula] tells us what the value of [unparseable or potentially dangerous latex formula] should be (if our function were continuous at [unparseable or potentially dangerous latex formula]), yet when we plug [unparseable or potentially dangerous latex formula] into the function, we get [unparseable or potentially dangerous latex formula]. Thus, we draw a little open circle on the graph to show the hole.

Remark: saying [unparseable or potentially dangerous latex formula] only works for [unparseable or potentially dangerous latex formula]. The graphs are almost identical... only difference being the hole.

4) The number of prime PIDs is the number of distinct primes in the factorization. All the other divisors are combinations of those primes.

[SumitG]

Listen to AuSmith. I just gave a trick, rather than an explanation.

[AuSmith]

Mostly, I wanted to get Obi Wan in there!

[B3ndythumbs]

Lol nice...okays thanks! A few more though :/

65x75-33: would u just have to calculate it?

200log200: ?? I know what log is, but I always miss these questions...

[AuSmith]

1) If I remember, there's a trick for multiplying two-digit numbers that end in [unparseable or potentially dangerous latex formula] and whose tens digits differ by [unparseable or potentially dangerous latex formula]. You should expand [unparseable or potentially dangerous latex formula] to check me.

a. Write down [unparseable or potentially dangerous latex formula].
b. Add [unparseable or potentially dangerous latex formula] to the bigger ten's place and multiply by the smaller. Write that down.

So, here you would write down [unparseable or potentially dangerous latex formula], but you have to subtract [unparseable or potentially dangerous latex formula]. Write down [unparseable or potentially dangerous latex formula]. Then, [unparseable or potentially dangerous latex formula]. Write that down. Admittedly, I don't see any intrinsic trick.

2) I believe the reason that logarithms were invented was to reduce the task of multiplication to simple addition. At least, that's the logarithm's most notable functionality. Take the large exponent problems on the calculator apps test... e.g. [unparseable or potentially dangerous latex formula]. You take the [unparseable or potentially dangerous latex formula] to reduce multiplying [unparseable or potentially dangerous latex formula] together [unparseable or potentially dangerous latex formula] times to adding [unparseable or potentially dangerous latex formula] together [unparseable or potentially dangerous latex formula] times. Then you can truncate and raise [unparseable or potentially dangerous latex formula] to that power and get the answer in scientific notation. Anyway, before high-powered calculators were invented, people had tables of logarithms. If they wanted to multiply [unparseable or potentially dangerous latex formula], they might use

[unparseable or potentially dangerous latex formula][unparseable or potentially dangerous latex formula],

look those up in their table, compute, then raise [unparseable or potentially dangerous latex formula] to that power. Much easier.

The point is that you can force big numbers to be nice to you. Just show them your badge. Especially when you have trailing [unparseable or potentially dangerous latex formula]'s.

[unparseable or potentially dangerous latex formula]

You know [unparseable or potentially dangerous latex formula] because [unparseable or potentially dangerous latex formula] is an increasing function and [unparseable or potentially dangerous latex formula]. Moreover, [unparseable or potentially dangerous latex formula] should be close to [unparseable or potentially dangerous latex formula] (the first bound should be skewed to the left) because [unparseable or potentially dangerous latex formula] is close to [unparseable or potentially dangerous latex formula] (because the second bound is for sure skewed to the left). Here, you can make estimates. You know [unparseable or potentially dangerous latex formula]. From this, you know since [unparseable or potentially dangerous latex formula], then [unparseable or potentially dangerous latex formula]. If you wanted a step further, [unparseable or potentially dangerous latex formula] so [unparseable or potentially dangerous latex formula], which is close. In any case, all this is overkill. You can guess anywhere between [unparseable or potentially dangerous latex formula] and end up correct.

[stupidityismygam]

[datnguyen160]

just wanna say thanks to AuSmith for explaining almost every q's posted in NS forum.

[AuSmith]

You're welcome! I enjoy questions! Stupid's answering a good bit, too... but he makes you work more. Lol.

[SumitG]

And of course, we can't forget the absolutely amazing analogies and allusions.

[B3ndythumbs]

some more now. :D

58^2+55^2-3^2
524^2
22^2+66^2

i hate squares :/

[315937]

a^2 + (3a)^2 = 10 a^2

22^2 + 66^2 =
484 x 10 = 4840

a^2 + b^2 - (a-b)^2 = 2ab

58^2 + 55^2 - 3^2 =
110 x 58 = 6380

524^2 =
square the 4, 16, write 6, remember 1
multiply 52, 4, 2, add 1, write 7, remember 41
square the 52, add 41, write 2745
answer is 274576

524^2 =
square the 24, 576, write 76, remember 5
multiply 5, 24, 2, add 5, write 45, remember 2
square the 5, add 2, write 27
answer is 274576

[was]

For 524^2 you could think (500+24)^2.
(a+b)^2 = a^2 + 2ab + b^2
24^2 = 576, write 76, carry 5
2(5)(24) =240 + 5 = 245, write 45 carry 2
5^2 + 2 = 27
274576