[H] Algebra

Blogs > kdog3683

Post a Reply

kdog3683

United States916 Posts

February 02 2009 02:16 GMT

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.

Durak

Canada3685 Posts

February 02 2009 02:23 GMT

Just so you know, you're actually asking for a quick way of factoring numbers.

Abydos1

United States832 Posts

February 02 2009 02:25 GMT

On February 02 2009 11:16 kdog3683 wrote:
Simplify to the form a^b

LOL, thats not simplifying it.

naonao

United States847 Posts

February 02 2009 02:26 GMT

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

13196 Posts

February 02 2009 02:27 GMT

you'd just have to be familiar with the numbers lol D:

but i think there are some tricks for some numbers

Malongo

Chile3472 Posts

February 02 2009 02:52 GMT

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.

foppa

Canada451 Posts

February 02 2009 02:59 GMT

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

micronesia

United States24755 Posts

February 02 2009 03:02 GMT

Of all the recreational mathematics... the only stuff I ever find people asking for help with is terrible!

Plexa

Aotearoa39261 Posts

February 02 2009 06:17 GMT

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

Pwert

United States59 Posts

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.

Please or register to reply.

[H] Algebra

Completed

Ongoing

Upcoming