The product of all the prime numbers

[BTGmoderatorDC]

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

How will I start solving it? What is the right solution to the problem?

OA C

[GMATGuruNY]

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

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

Since the answer choices are VERY far apart, we can BALLPARK.
For every value that we round UP, we should compensate by rounding another value DOWN.

2*3*5*7*11*13*17*19
210 * 10*15 * 15*20
200*150*300 = 9,000,000.

The closest power of 10 = 10,000,000 = 10�.

The correct answer is C.

[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

As Mitch has noted, 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.

So, here's another 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,

[Jeff@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

We need to determine the product of:

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:

5 x 19 is about 100 = 10^2

So, we are left with:

2 x 3 x 7 x 11 x 13 x 17

7 x 13 is about 100 = 10^2

So, we are left with:

2 x 3 x 11 x 17

2 x 3 x 17 is about 100 = 10^2

Finally, we have 11, which is about 10 = 10^1.

Thus, the product of all the prime numbers less than 20 is closest to 10^2 x 10^2 x 10^2 x 10^1 = 10^7.

Answer: C

[BTGmoderatorDC]

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

We need to determine the product of:

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:

5 x 19 is about 100 = 10^2

So, we are left with:

2 x 3 x 7 x 11 x 13 x 17

7 x 13 is about 100 = 10^2

So, we are left with:

2 x 3 x 11 x 17

2 x 3 x 17 is about 100 = 10^2

Finally, we have 11, which is about 10 = 10^1.

Thus, the product of all the prime numbers less than 20 is closest to 10^2 x 10^2 x 10^2 x 10^1 = 10^7.

Answer: C

Thanks a lot!

[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