21I guess i will hav to start frm scratch :
very easy eg. : no = 24 = 233
3 factors
2,2,6
2,3,4
2,1,12
8,1,3
24,1,1
total = 5 [provivded i m not forgetting ne case]
The other day we solved to find the no. of ways in wich a given no. can be expressed as a product of two nos!!!
I wanted to know if v can hav such a generalization for expressing a no. as product of 3integers
Note** It may be impossible to give a general formula for unique integers, so we will first try to find the number of ordered sets
(a,b,c) and then number of sets {a,b,c}
For those who dont know, (1,2) is different from (2,1)
while in sets they are one and the same
P.S: Edited by Nishant!
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30 Answers
2121But
wat I wrote in da abov post wud b used to find ways WITH PERMUTATIOONS (i.e. ORDERED)
but wat v want is non ordered......
eg. 7^3 * 5^4
now v wanna know in how many ways can this b split as 3-factor Product??
21OH okie!!!!
the this wud b jus lyk how we do different permutations for making a 5-lettered word using alphabets of a given 7 or 8 digit word............
I hope I m riter wid THIS ^^^
this gets interedsting.... [12]
I'll giv it a shot and post ma result
62you missed out 5,5,6
by the way, the final answer will be divided by three or six
depending on wheter the (a,b,c) are of the form
(a,a,b) or all different!
Now can you give it a shot ??
(btw I should have been more careful in writing the above post!)
21BTW iska koi upay hai kya? ya fir COUNTING ZINDABAD!!
if v want jus the total cases, i.e. NON-ordered cases only if v want then???
213*5^2*2 = 150
1,1,150
1,3,50
1,10,15
2,3,25
2,5,15
*** 5,5,6****
cases = 5+1
formula : [ (1+3-1)C2 ]2 * 4C2
[11] [11] [5]
ye kya hua?? DISASTER [17]
62yes tapan..
Good work
that is a tricky part.
see here the issue will be that there will be a lot of repetitions
and each repetition will not be unique...
iF you take a number where none of the powers is a multiple of 3 thenit will work with a lot of modification divide by six and three and so on!
Now can you try some more :)
211 more DOUBT!!!
This gives us no. of ordered trios na....
BUT THEN if v want jus the total cases, i.e. NON-ordered cases only if v want then???
LIKE in da example of 24 = 2^3 * 3
The answer in dat case wud be 5; but wen v consider the abov formula we get 5C2 = 10
21OH K!! thnk U sir for the method!!!
But Sir can u pl. tell me the rational for doing this :
3=a1+a2+a3
no of non -ve solutions is given by
5C2
siilary for 2=b1+b2+b3
we get 4C2 solutions
62lets take an example
73132
find th enumber of 3 sets (a,b,c) such that abc=73132
now
3=a1+a2+a3
no of non -ve solutions is given by
5C2
siilary for 2=b1+b2+b3
we get 4C2 solutions
Number of such ordered pairs is given by
5C2 x 4C2
21Sir, I cant find 2 factor case wala post, can u pl. tell me the link/forum sub-section wer i can find it
21Sir, isnt that a bit different from this un
coz dat deals wid divisibility by 2^n
but here we wish to express it as a prodct of 3nos.
62no tapan.
may be u did not understand my hint properly :(
Or i messed up with the notations!
Try to figure the same reasoning with 2 factor case!
21BTW is dis trrue ?
"ya I guess the a1 in ur equation represents : (a1 + a2 + a3 +.... an)"
par sir aapne bata t0o diya..... [11]
x1+y1+z1=a1
21hmmmm.....
ya I guess the a1 in ur equation represents : (a1 + a2 + a3 +.... an)
and this is coz after all the sum of powers in the form of 3 nos. which wen multiplied giv da final NUMBER
21wud it be sumthing lyk :
GROUPING (a1 + a2 + a3 +.... an) items into 3 GRPs?
62Hint is same as the question whose generalization this is
write the number as
N= p_1^{a_1}p_2^{a_2}....p_n^{a_n}
Now try.
21Okie........
so that means we can hav a genralization!!!! [12]
62oh that can also be done
sorry i din see your post
I will make this a QOD.. and then WIll solve this tomorrow..
I wonder why this never came to my mind :)
21jaise ki
100 = 2^2*5^2;
2*2*25
2*10*5
and all the cases wer it can b expressed as a prod of two nos. eacdh multiplied wid one (5cases)
and also 100*1*1
in all 2+5+1 = 8 cases............. [if i dint miss ne]