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The shortest way to solve such a question..

Expert replies
by anurag_7 » Fri Jul 18, 2014 8:39 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
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Source: — Problem Solving |

by GMATinsight » Fri Jul 18, 2014 8:45 am
anurag_7 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
Required product = 2 x 3 x 5 x 7 x 11 x 13 x 17 x 19

19 x 5 = 95 is close to 100 i.e. 10^2
17 x 3 x 2 = 102 is close to 100 i.e. 10^2
7 x 11 x 13= 1001 is close to 1000 i.e. 10^3

Therefore the closest power is 10^2 x 10^2 x 10^3 = 10^7

Answer Option C
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by Brent@GMATPrepNow » Fri Jul 18, 2014 11:08 am
The product of all prime numbers less than 20 is closest to which of following powers of 10 ?
(A) 10^9
(B) 10^8
(C) 10^7
(D) 10^6
(E) 10^5
Here's another 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 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 Hanneson - Creator of GMATPrepNow.com
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by GMATGuruNY » Fri Jul 18, 2014 2:12 pm
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 = 10^7.

The correct answer is C.
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by mcdesty » Fri Jul 18, 2014 10:02 pm
Here is what it looked like on my scratch Paper.
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by GMATinsight » Fri Jul 18, 2014 10:24 pm
Since this question has used the application of divisibility of (7x11x13) therefore I would propose that it's good for GMAT aspirants to know the rule of divisibility of (7x11x13) as it may have uses in various other questions as well.

Another observation about the number divisible by 7x11x13 is

Any number of the type abcabc is always divisible by 1001 i.e. 7, 11 and 13
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by GMATinsight » Fri Jul 18, 2014 10:26 pm
Another question based on the same principle is as mentioned below:

a, b, and c are positive integers. If a, b, and c are assembled into the six-digit number abcabc, which one of the following must be a factor of abcabc?

(A) 16
(B) 13
(C) 5
(D) 3
(E) none of the above

Answer: Option B
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