The prime factorization of intezer N is A x A x B x C where A B and C are

Since A,B,C are the prime no.hence we can apply the formula:
no of factors=(n+1)(m+1)(p+1)......
where n,m,p...are the powers of prime no..here we can write the given expression as A^2 x B^1 x C^1. hence no of factors are(2+1)(1+1)(1+1)=12(Ans)
12 years agoHelpfull: Yes(65) No(2)

Here n is A*A*B*C=A^2*B*C.so the no of factors of n is
(2+1)*(1+1)*(1+1)=3*2*2=12(ans).
12 years agoHelpfull: Yes(34) No(8)
given N=A*A*B*C AND A,B,C ARE PRIME INTEGERS...
SO A,B,C HAVE NO FACTORS EXCEPT 1 AND THEM SELVES...
the other factors of N are formed by selecting product of at least 2 of A*A*B*C
that is product of 2 numbers formed (A*A,A*B,A*C,B*C)=4
product of 3 numbers formed (A*A*B,A*A*C,A*B*C)=3
OTHER KNOWN FACTORS ARE (1,A,B,C,N)=5
SO TOTAL FACTORS OF N ARE 12....
12 years agoHelpfull: Yes(11) No(7)
given N=A*A*B*C AND A,B,C ARE PRIME INTEGERS...
SO A,B,C HAVE NO FACTORS EXCEPT 1 AND THEM SELVES...
the other factors of N are formed by selecting product of at least 2 of A*A*B*C
that is product of 2 numbers formed (A*A,A*B,A*C)=3
product of 3 numbers formed (A*A*B,A*A*C,A*B*C)=3
OTHER KNOWN FACTORS ARE (1,A,B,C,N)=5
SO TOTAL FACTORS OF N ARE 11....
12 years agoHelpfull: Yes(3) No(18)
factors are A,A,B,C,AA,AB,AC,AB,AC,AAB,AAC,AABC,BC,ABC,AAB,AAC,ABC,...(since all are prime factors so the two A's are different) ans is 24
12 years agoHelpfull: Yes(1) No(20)
If n=(p1)a∗(p2)b∗(p3)c∗......where p1,p2,p3 are primes then, the total number of factors of n is (a+1)∗(b+1)∗(c+1)∗........

so, here, a,b,c are primes, and their powers are 2,1,1 respectively.

so, according to the formula,
number of factors of n is (2+1)∗(1+1)∗(1+1) which turns out to be 3∗2∗2=12.
5 years agoHelpfull: Yes(0) No(0)

The prime factorization of intezer N is A x A x B x C, where A, B and C are all distinct prine intezers. How many factors does N have?