If p, s, and t are positive prime numbers, what is the value

Nice.

I thought both together are sufficient...
This was trapy.

Q Can be rephrased as what is p*q*r?

1) Sufficient
728 = 2^3 * 7 * 13. Hence prime numbers are 2,7,13.

2) Insufficient.

From 1, 728 can be written as\(2^3*13*7\) , so the only combination possible is p=2 and st = 13*7 hence \(p^3*s^3*t^3\)= \(2^3*13^3*7^3\) Sufficient

From 2, t = 13 but we do not know anything about p or s, so insufficient.

Answer A

Bunuel's approach is very good, but let's apply a slightly different approach:
1) p^3st=728

st=728/(p^3) --> here we have to pay attention that the last digit of 728 is 8 which is contained only in the cyclicity of 2 and 8. Lets clarify it:
the cyclicity of 1, 5, 6 are these numbers themselves (regardless of their power)
the cyclicity of 2: 2, 4, 8, 6
the cyclicity of 3: 3, 9, 7, 1
the cyclicity of 4: 4, 6
the cyclicity of 7: 7, 9, 3, 1
the cyclicity of 8: 8, 4, 2, 6
the cyclicity of 9: 9, 1

Hence, 728 is divisible either by 2 or 8.
2^3 = 8 and 8^3 = 512

As you see, 728 is divisible only by 2^3 provided p is a positive prime number.

The result is 728/(2^3) = 91

s × t = 91 can be written as 1 × 91 or 13 × 7. As to the original question p^3 × s^3 × t^3 it does not matter whether we assume s × t is 1 × 91 or vice versa the same is true in case of 13 × 7.

both conditions s × t = 1 × 91 and s × t = 13 × 7 satisfy stmt 1)
because 2^3 × 1 × 91 and 2^3 × 13 × 7 yield 728

Thus, we get the same result regardless whether p^3 × s^3 × t^3 is written as either 2^3 × 1^3 × 91^3 or 2^3 × 13^3 × 7^3

stmt 2) is obviously not sufficient.

Hope, it helps.

Just to touch on some key points on this problem.
This problem is meant to test prime factorization.

So we need to break 728 into primes first.
This will give us (2^3) (13) (7) which is enough to solve the question.

Keep in mind some properties of primes and prime factorization

- All prime numbers except 2 and 5 end in '1','3','7' or '9'
- All prime numbers above 3 are of the form '6n+1' or '6n-1'
- The first prime number is 2 which is also the only even prime
- The Prime Numbers up to 100 are: 2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,89,97)
- The square of any prime number greater than 3 is 1 more than a multiple of 12
- Verifying the primality of a given number 'n' can be done by trial division, that is to say dividing 'n' by all integer numbers smaller than √n

Hope this helps for future questions
Cheers!

J

Question, p^3*s^3*t^3 = ?

1) p^3 * s * t = 728, prime factorization of 728 = 2^3 * 7*13, no other primes so then P^3 is 8^3, and we are given S & T can be either 7,13 or 13,7. However, both possibilities will provide the same result. So 1 is sufficient.

2) t= 13, no information about S or P, so insufficient

If p, s, and t are positive prime numbers, what is the value of \(p^3s^3t^3\)?


(1) \((p^3)*s*t=728\)

\(728 = 2^3 * 7 * 13\)

We're told p, s, and t are positive prime numbers. Thus, p^3 must be 2^3. We can then find the answer. Sufficient.

(2) \(t=13\)

No information on p or s. Insufficient.

Answer is A.