start with 2 and keep dividing the number with the subsequent prime numbers

847000=7*11*11*2*2*5*5
it helps to remember that 100=2^2*5^2
so u can separate the 100 like 847000=847*100

Dont know if this insight will work always
but one can see 7 is a factor straight away(12*7=84)
84,7000 =7*121,000=7*11^2*10^3=7*(11^2)*(2^3)*(5^3)

There is no fixed shortcut for such questions.

Answer is 52.

Just break the number 84700 into primes
for this it will be like this= 2^2 x 5^2 x7^1 x 11^2

Just add one to every power, (2+1) x (2+1) x (1+1) x (2+1) = 52 factors.

hope it helps.

Gouman wrote:I tried searching but, I found no answers that satisfied me.

Can anybody offer any tips for breaking up larger numbers into primes

for example a problem in the Manhattan guides for the prime factorization of

84,700. I just got stuck by the size of the number.

Usually I will just start halving until the numbers become more manageable, but, when dealing with a beast like 84,700 I'm freezing up.

I usually use the tree method, and separate the 100 and 847 first.

If you see a high prime factor, use that first, For 847, I see it can be divisible by 7 (7*121)

121 can be broken down into 11*11

the rest of the tree which is 100, can be further factored down into 5's and 2's.

Gouman wrote:I tried searching but, I found no answers that satisfied me.

Can anybody offer any tips for breaking up larger numbers into primes

for example a problem in the Manhattan guides for the prime factorization of

84,700. I just got stuck by the size of the number.

Usually I will just start halving until the numbers become more manageable, but, when dealing with a beast like 84,700 I'm freezing up.

first split the number 847 and 100

the problem here is to factorize 847 because it doe not have 2,3 or 5 as factors.

here is a tip you can use to find whether a number is multiple of 7,11,13 or any other number.

let's say that you want to find whether 847 is multiple of 7

try to find a number x, such 847 - 7x ends in 0

for example 1 ---> 847 - 7(1) = 840. now drop the 0
is 84 multiple of 7? you may not know this but repeat the process

try to find a number x, such 84 - 7x ends in 0
x = 2 ---> 84 - 7(2) = 84 - 14 = 70. now drop the 0

is 7 a multiple of 7? yes so 847 is a multiple of 7.

let's say now that you want to factorize 377

it's clear that 2,3 o 5 are not among its factors.

let's try 7
377 - 7*11 = 300 drop the 0's --> is 3 a multiple of 7? no

try 11
377 - 11*7 = 300 drop the 0's ---> is 3 a multiple of 11? no

try 13
377 - 13*9 = 260 drop the 0 ---> is 26 a multiple of 13? yes so 13 is a factor of 377


a trick to find whether a number is multiple of 11 is the following

let's say that you have a three digit number x = ABC and you want to find out whether x is multiple of 11 or not.
sum and rest its digits in this way
A - B + C
if the result is 0 or a multiple of 11 the number is multiple of eleven
if you have a 4 digit number ABCD
A - B + C - D

for instance
847 --> 8 -4 + 7 = 11 so 847 is multiple of 11

if you are interested in this kind of tricks, I recommend you read "secrets of mental math".

Thanks for the replies.

Coolboy, I like your tip for finding multiples of numbers. I will have to think about that further.

I'm feeling like I just need to practice recognizing primes. I mean it seems like you guys are able to just pick up on certain primes pretty easily.

I can halve things easily. Some numbers I can break into thirds easily. I suppose if you know the primes you can use them as a reference to break numbers down using a trial and error method, if you don't just see something.

Thanks again everybody!