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Expert replies
by bblast » Tue May 17, 2011 11:00 am
The product of all 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

OA is C -i browsed 5-6 threads but still did not find a "best" solution to this.
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Source: — Problem Solving |

by Brent@GMATPrepNow » Tue May 17, 2011 11:49 am
bblast wrote:The product of all 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

OA is C -i browsed 5-6 threads but still did not find a "best" solution to this.
Here's one approach.

Since the numbers are very spread apart (each answer choice is 10 times greater than the next answer choice), we can be somewhat aggressive with our estimation.

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 just a bit larger that 10, so let's call it 10+

This means (10)(3)(7)(11)(13)(17)(19) = (10)(3)(7)(10+)(13)(17)(19)

Next, (7)(13)=91, which is a bit less than 100, so we get (10)(3)(7)(10+)(13)(17)(19) = (10)(3)(100-)(10+)(17)(19)

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

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by GMATGuruNY » Tue May 17, 2011 12:20 pm
bblast wrote:The product of all 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

OA is C -i browsed 5-6 threads but still did not find a "best" solution to this.
2*3*5 * 7 * 11*13* 17*19
≈ 30 * 7 * 10*15 * 15*20
≈ 200 * 150 * 300
≈ 9,000,000

The correct answer is [spoiler]C: 10� = 10,000,000[/spoiler].
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by abhi0697 » Wed May 18, 2011 4:51 am
=2*3*5*7*11*13*17*19
=10*21*11*13*17*19
=2.1*1.1*1.3*1.7*1.9*10^6
=9.75*10^6
~10^7

Hence, C
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