How many ordered triples of pairwise distinct, positive integers (a,b,c) are there such that abc=10^6?
10^6 = (2x5)^6
= (2^6)(5^6)
Possible cases:
1x64x15628 --- arrange in 6 ways
2x32x15625 --- arrange in 6 ways
4x16x15625 --- 6 more
8x8x15625 --- arrange in 3 ways
64x5x3125 ---- 6 ways
64x25x625 ----- 6 ways
64x125x125 ---- 3 ways
I count 36 of them
check my logic to make sure I didn't miss any
e.g. for 64x25x625 we could have
(64,25,625) , (64,625,25) , (25,64,625), (25,625,64), (625,25,64), (625,64,25)
false
Do ur hw ur self .....
To find the number of ordered triples (a, b, c) such that abc = 10^6, we first need to determine the prime factorization of 10^6.
10^6 can be written as (2^6)(5^6), as 10 is equal to 2 * 5.
Now, let's consider the possible values for a, b, and c based on their prime factorizations.
For a triple (a, b, c), a can be any positive integer divisors of 2^6, b can be any positive integer divisors of 5^6, and c can be any positive integer divisors of (2^6)(5^6).
To determine the number of divisors for a prime factorization, we add 1 to the exponent of each prime and multiply them together.
For 2^6, we have (6 + 1) = 7 divisors. Similarly, for 5^6, we have (6 + 1) = 7 divisors. Finally, for (2^6)(5^6), we have (6 + 1) * (6 + 1) = 49 divisors.
Since a, b, and c must be pairwise distinct, we need to remove the case where all three values are equal. In this case, the number of ordered triples is the number of divisors of (2^6)(5^6) minus 1.
Therefore, the number of ordered triples (a, b, c) such that abc = 10^6 and a, b, and c are pairwise distinct positive integers is 49 - 1 = 48.