Number properties: Prime Factors ...

Question

Hi All,

Is there a quick way of working out all the "distinct" prime factors of 3528, instead of continuously dividing by 2, and then the next higher prime number etc.

Thanks.
II

eschn3am · Answer

Here are the rules we have:

2: if the integer is even
3: if the sum of the integer's digits is divisible by 3
4: if the integer is divisible by 2 twice
5: if the integer ends in 0 or 5
6: if the integer is divisible by both 2 and 3
7: no rule
8: if the integer is divisible by 2 three times
9: if the sum of the integer's digits is divisible by 9
10: if the integer ends in 0

so for 3528 we know it's divisible by 2, 3, 4, 6, 8 and 9. to find other ones like 7, 11, 13, etc I think you just have to test them.

walker · Answer

3+5+2+8=18 divisible by 9 and number with last two digits (28) is divisible by 4

So, you can start from 3*3*4=36
3528=36*98=36*49*2=7*7*3*3*2*2*2

II · Answer

so there is no real short cut then ?

The only way you can work this out is by the following calculation:
3528/2 = 1764
1764/2 = 882
882/2 = 441
441/3 = 147
147/3 = 49
49/7 = 7

So the distinct prime factors are 2, 3, and 7.

Thanks.
II

AtifS · Answer

Guys! as far as the rule for dividing any number by 7 is considered, I thought and about it and tried the method that came into my mind on some numbers and guess what it does work, tried till 4-digit (didn't try for 5-digit yet). Maybe there is a proof for this method out there and also maybe mentioned by someone but I just did work this out in my mind and thought I should share that with you guys. Please! Do correct me if I do wrong

Here it is what I tried. Let's take the above number 3528 Now the last digit is '8' and we know that if we multiply '7' with 4 we get '8' as a unit digit. So I multiplied '7' with '4' and got 2 8 then I took 10th digit which is '2' and subtracted it by the remaining number without the unit digit '352' and after subtraction got '350', then checked whether the resultant number can be divided by '7' or not, now '0' is as a unit digit and again we know multiplying '7' with 0 gives us 0 but here we don't have any 10th digit so now we don't have to subtract anything from the remaining number after removing the unit digit, now we are left with the number '35' now check whether this is divisible by '7' or not and we can see that it is clearly divisible by '7' after dividing it with '7' we get 5 . Now we can get the answer by putting factors (highlighted in blue color) in a same order we divided them. First one is 4 which will be placed at unit-digit location and then 0 on 10th-digit location and in the end 5 on 100th-digit location. And we get the answer 504 Also you can check the answer whether it is divisible by '7' or not. By using above method we can get 2 as a unit-digit then subtracting '1' from remaining number '50' we get '49' after dividing '49' by '7' we get 7 as 10th-digit so we get the answer by putting factors (in green color) in an order we get 72 which we can see that isn't further divisible by '7'. So that means "3528" is two times divisible by 7 which means it is divisible by 7^2 that is '49'. You can check other easy numbers as well; Let's take 126 now '6' is a unit-digit and when we multiply '7' with 8 we get ' 5 6' and '6' as a unit digit now subtracting the bold digit 5 from the remaining number after removing unit-digit '12' we get '7' and after dividing it with 7 we get 1 . Now we get answer after putting red colored 8 as unit digit and 1 as 10th digit--> 18 . You can try this method on other numbers. Do forgive me for my bad explanation (I am not good at teaching lol!). If any of you can understand my explanation and can re-write in simple & understandable words then please! do so and share that with the community. Cheers, Atif P.S. Once you understand it and do it mentally you'll know that it's easy and fast. It looks lengthy the way I explained hehe.

tarun · Answer

Interesting approach. How did you figure this out?

AtifS · Answer

Don't know, when I saw this thread I took some number and experimented with them in my mind and after 5-10 minutes I figured it out. It all came automatically in my mind (Just experiments with numbers)

. The same I did when I figured out how to take square of any number especially useful to take square of 2-digit or 3-digit numbers as I was too lazy to cram squares of numbers till 25 as mentioned here somewhere, so I figured that out too and also did figure out the formula to prove. It is written in my GMAT MATH Notes thread and also added it to the Math notes & tips collection. By the way happy to know that you got my point (I think I am really bad at explaining

)

tarun · Answer

If you think you are bad at explaining, you better sort it out soon enough, you wud need the clarity of expression to sell your story in the app essays.

AtifS · Answer

Yea! of course, gonna get my hands on it. That's why, I am always trying to explain in anyway I can

. And with practice, I believe I'll be able to explain clearly and succinctly.

nsp007 · Answer

nice work ! thanks !!

gmatready7 · Answer

For 7 you double the last digit and subtract it from the remaining digits (repeating if needed). So, for 3528: First you double the 8 to get 16, then 352-16 = 336...It's too tough to quickly determine if 336 is divisible by 7 so repeat the process with the new number. Using 336, double the 6 to get 12, then 33-12 = 21..21 is divisible by 7 so 3528 is as well.

AtifS · Answer

Yea, that's a nice method and it is described in the file "GMAT Quant Notes_28-03-2010" uploaded by me in another thread.
Thank you! for mentioning it here.
Kudos!

Number properties: Prime Factors ...

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