yellowho wrote:Is the prime number q equal to 29 ?
(1) q -1 has 6 factors
2) q+1 has 2 and 3 as prime factors
Here is an efficient approach, if you know the following method for counting the number of positive factors of an integer:
1) Prime-factorize the integer
2) Add 1 to each exponent
3) Multiply
Since 200 = 2^3 * 5^2, it has (3+1)*(2+1) = 12 factors.
Statement 1:
If q-1 = 2 * 3^2 = 18 -- which yields 2*3 = 6 factors -- then q=19.
If q-1 = 2^2 * 7 = 28 -- which yields 3*2 = 6 factors -- then q=29.
Insufficient.
Statement 2:
If q+1 = 2*3 = 6 -- so it has 2 and 3 as prime factors -- then q=5.
If q+1 = 2*3*5 = 30 -- so it has 2 and 3 as prime factors -- then q=29.
Insufficient.
Statements 1 and 2 together:
We have already seen that q=29 satisfies both statements.
Out of every 3 consecutive integers, exactly 1 will be a multiple of 3. Thus, if q+1 is a multiple of 3, then q and q-1 cannot be multiples of 3. So to satisfy statement 2, q-1 cannot have 3 as a prime factor.
We also know that q-1 must be even, since prime number q must be odd.
Thus, q-1 cannot have 3 as a prime factor and must have 2 as a prime factor:
If q-1 = 2^2 * 11 = 44 -- which yields 3*2 = 6 factors -- then q=45, which is not prime.
If q-1 = 2^2 * 13 = 52 -- which yields 3*2 = 6 factors -- then q=53 and q+1=54, satisfying all the conditions in the problem.
Since q=29 and q=53 both work, insufficient.
The correct answer is
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