The product of all the prime numbers

[VJesus12]

The product of all the prime numbers less than 20 is closest to which of the following powers of 10?

(A) 10^9
(B) 10^8
(C) 10^7
(D) 10^6
(E) 10^5

The OA is the option C.

How can I find the correct answer here? Experts, can you show me how would you solve this PS question?

Thanks in advanced.

[Brent@GMATPrepNow]

The product of all the prime numbers less than 20 is closest to which of the following powers of 10?

(A) 10^9
(B) 10^8
(C) 10^7
(D) 10^6
(E) 10^5

The numbers are very spread apart (each answer choice is 10 times greater than the next answer choice). This means we can be quite AGGRESSIVE with our estimation.

One approach:

We have the product (2)(3)(5)(7)(11)(13)(17)(19)

Let's see if we can group the numbers to get some approximate powers of 10

First (2)(5)=10, so we get (2)(3)(5)(7)(11)(13)(17)(19) = (10)(3)(7)(11)(13)(17)(19)

Next, 11 is close enough to 10, so we get: (10)(3)(7)(11)(13)(17)(19) = (10)(3)(7)(10)(13)(17)(19) [approximately]

Next, (7)(13)=91, which is pretty close to 100.
So we get (10)(3)(7)(10)(13)(17)(19) = (10)(3)(100)(10)(17)(19) [approximately]

Finally, 3(17)=51, and (51)(19) is very close to (51)(20), which is very close to 1000
So,(10)(3)(100)(10)(17)(19) = (10)(1000)(100)(10)= 10,000,000 [approximately]

Since 10,000,000 = 10^7, the best answer is C

Cheers,
Brent

[Scott@TargetTestPrep]

The product of all the prime numbers less than 20 is closest to which of the following powers of 10?

(A) 10^9
(B) 10^8
(C) 10^7
(D) 10^6
(E) 10^5

The OA is the option C.

How can I find the correct answer here? Experts, can you show me how would you solve this PS question?

Thanks in advanced.

2 x 3 x 5 x 7 x 11 x 13 x 17 x 19

Let's group some of these numbers to get powers of 10:

2 x 5 = 10 = 10^1

7 x 11 x 13 = 1001 ≈ 10^3

Now we are left with 3, 17 and 19:

3 x 17 x 19 = 51 x 19 ≈ 50 x 20 = 1000 = 10^3

Thus, the product of all the prime numbers less than 20 is closest to 10^1 x 10^3 x 10^3 = 10^7.

Answer: C