Three different numbers have a product that is greater than

Question

GMATH  practice exercise (Quant Class 16)

The product of three different numbers is greater than 1000. What is the value of the largest among them?

(1) The numbers are primes.
(2) The product of the numbers is less than 1310.

DavidTutorexamPAL · Answer

1) and 2) are clearly insufficient: there are endless possible results. 
Let's try combined: we'll make a short list of prime numbers, starting with the lowest, and see if we can find more than one trio whse product is between 1000 and 1310:
2, 3, 5, 7, 9, 11, 13, 17... 
the product of 9*11*13 is 99*13 - definitely in the range (even without calculating).
can we change this at all? 
if we replace 13 with the next largest prime (17), it's too large: 99*17>1300
if we replace 9 with the next smallest (7): 77*13 = 770+210+21=980+21=1001 - in the range!
two possible solutions, not sufficient!
 answer E

fskilnik · Answer

a < b < c

abc > 1000

? = c

\left( 1 ight)\,\,{m{primes}}\,\,\,\left\{ \matrix{
 \,{m{Take}}\,\,\left( {a,b,c} ight) = \left( {7,11,13} ight)\,\,\,\,\left \,\,\,\,\, \Rightarrow \,\,\,? = 13 \hfill \cr 
 \,{m{Take}}\,\,\left( {a,b,c} ight) = \left( {7,11,17} ight)\,\,\,\,\, \Rightarrow \,\,\,\,\,? = 17 \hfill \cr} ight.

\left( 2 ight)\,\,abc < 1310\,\,\,\,\,\,\left\{ \matrix{
 \,\left( {{\mathop{m Re}
olimits} } ight){m{Take}}\,\,\left( {a,b,c} ight) = \left( {7,11,13} ight)\,\,\,\, \Rightarrow \,\,\,? = 13 \hfill \cr 
 \,\left( {{\mathop{m Re}
olimits} } ight){m{Take}}\,\,\left( {a,b,c} ight) = \left( {7,11,17} ight)\,\,\,\,\left \,\,\,\,\, \Rightarrow \,\,\,? = 17 \hfill \cr} ight.\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\left( {m{E}} ight)

We follow the notations and rationale taught in the  GMATH  method.

Regards, 
Fabio.

IanStewart · Answer

9 is not a prime number.

IanStewart · Answer

To make calculation easy here, you can set two of the primes to 2 and 5. Then we'll know the information is not sufficient if we can find at least two different prime numbers between 100 and 131. And there are a few primes in that range (101 and 103, for example).

DrHuber · Answer

Statement (1) does not provide enough information to determine the value of the largest number. For example, the three prime numbers could be 3, 5, and 7, in which case the largest number is 7. However, they could also be 7, 11, and 13, in which case the largest number is 13.

Statement (2) is more helpful. If the product of the three numbers is less than 1310, then the largest number must be less than the cube root of 1310, which is approximately 11. Therefore, the largest number must be one of the prime numbers less than or equal to 11.

Combining the two statements, we know that the three numbers are prime and that the largest number is less than or equal to 11. Therefore, the largest number must be either 5, 7, or 11. However, we still cannot determine which of these is the largest number.

Therefore, the answer is (E) the information provided is not sufficient to determine the value of the largest number

IanStewart · Answer

If the product of three positive numbers is less than 1310, that ensures that the  smallest  of those three numbers must be less than the cube root of 1310, not the largest. For example, 2*19*29 = 1102, but only one of those three numbers is less than 11. We also know from the question stem that the product of the three numbers is greater than 1000, which guarantees that the largest of the three numbers is greater than the cube root of 1000, so must be greater than 10 (the largest number here cannot be 5 or 7). But as I pointed out above, there are many possible values for the largest prime here and the answer is still E.

Three different numbers have a product that is greater than

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