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[H] Algebra

Blogs > kdog3683
Post a Reply
kdog3683
Profile Blog Joined January 2007
United States916 Posts
Last Edited: 2009-02-02 02:21:33
February 02 2009 02:16 GMT
#1
Simplify to the form a^b

729 = + Show Spoiler +
3^6


2048 = + Show Spoiler +
2^11


How would you do this quickly? 15seconds, no calculator.
What is the process you go through?

I divide by small #s, 2,3,4,5... If 2 fits, i divide by 2 again, else I try 3. Else, 4, etc, but this is slow.

*
Multiply your efforts.
Durak
Profile Blog Joined January 2008
Canada3684 Posts
February 02 2009 02:23 GMT
#2
Just so you know, you're actually asking for a quick way of factoring numbers.
Abydos1
Profile Blog Joined October 2008
United States832 Posts
February 02 2009 02:25 GMT
#3
On February 02 2009 11:16 kdog3683 wrote:
Simplify to the form a^b


LOL, thats not simplifying it.
"...perhaps the greatest joy possible in Starcraft, being accused of being a maphacker" - Day[9]
naonao
Profile Blog Joined November 2008
United States847 Posts
Last Edited: 2009-02-02 02:41:30
February 02 2009 02:26 GMT
#4
Thats memorization. powers of 2 and powers of 3.

but otherwise i do a branch thingy like for instance 144 would be like
144
/ \
12 12
/ \ / \
3 4 3 4
| / \ | / \
3 2 2 3 2 2

Also some tips to know when numbers are divisible by others:
2: number is even
3: sum of the digits is a multiple of 3
4: last two digits are divisible by 4
5: ends in 0 or 5
6: even and the digits add up to a multiple of 3
7: NOTHING
8: last 3 digits are divisible by 8
9: sum of the digits is a multiple of 9
paper
Profile Blog Joined September 2004
13196 Posts
February 02 2009 02:27 GMT
#5
you'd just have to be familiar with the numbers lol D:

but i think there are some tricks for some numbers
Hates Fun🤔
Malongo
Profile Blog Joined November 2005
Chile3472 Posts
February 02 2009 02:52 GMT
#6
Any example not solved easily? i mean those you posted are "trivial" after some trainning, and your method isnt that slow, just try only prime numbers (not 4). Also take in mind that you wont face powers over 5 for most numbers, 2 and 3 are suitable but 5^5 = 625*5= 3125, 6^5=7776. Actually powers of 5 end in 5, powers of 6 end in 6.
Help me! im still improving my English. An eye for an eye makes the whole world blind. M. G.
foppa
Profile Blog Joined July 2008
Canada451 Posts
February 02 2009 02:59 GMT
#7
eventually you get really good at making an accurate guess... its just practice... and after a while the numbers start to look familiar; it only gets easier
i can take you
micronesia
Profile Blog Joined July 2006
United States24701 Posts
February 02 2009 03:02 GMT
#8
Of all the recreational mathematics... the only stuff I ever find people asking for help with is terrible!
ModeratorThere are animal crackers for people and there are people crackers for animals.
Plexa
Profile Blog Joined October 2005
Aotearoa39261 Posts
February 02 2009 06:17 GMT
#9
What naonao posted is the best way to solve these problems although there is a method to test if a number is dividable by 7.. its just so complicated that its quicker to just brute force it haha
Administrator~ Spirit will set you free ~
Pwert
Profile Blog Joined September 2008
United States59 Posts
Last Edited: 2009-02-03 02:20:30
February 02 2009 06:40 GMT
#10
On February 02 2009 11:26 naonao wrote:
Thats memorization. powers of 2 and powers of 3.

but otherwise i do a branch thingy like for instance 144 would be like
144
/ \
12 12
/ \ / \
3 4 3 4
| / \ | / \
3 2 2 3 2 2

Also some tips to know when numbers are divisible by others:
2: number is even
3: sum of the digits is a multiple of 3
4: last two digits are divisible by 4
5: ends in 0 or 5
6: even and the digits add up to a multiple of 3
7: NOTHING
8: last 3 digits are divisible by 8
9: sum of the digits is a multiple of 9


Actually, I know of this interesting (but time consuming) way of finding out if a number is divisible by 7. You take the digit farthest to the right, double it, and subtract it from the original number over and over until you have a number from 9 to negative 9. If it turns out to be 7, 0, or negative 7 then the number is divisible by 7.

Example:

182 becomes (18 - 2) which equals 14.

14 becomes (1 - 8) which equals negative 7.

182 is divisible by 7.

Example:

6881 becomes (688 - 2) which equals 686.

686 becomes (68 - 12) which equals 56.

56 becomes (5 - 12) which equals negative 7.

6881 is divisible by 7

Example:

531 becomes (53 - 2) which equals 51.

51 becomes (5 - 2) which equals 3.

531 is not divisible by 7.
hehehe...
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