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PS question

by prernamalhotra » Wed May 14, 2014 1:52 am
Is there an easy way to solve the below kind of problems:


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


Thank you,
Prerna
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by GMATGuruNY » Wed May 14, 2014 2:12 am
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 sanju09 » Wed May 14, 2014 3:48 am
prernamalhotra wrote:Is there an easy way to solve the below kind of problems:


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


Thank you,
Prerna
Very a few numbers to deal with, 2, 3, 5, 7, 11, 13, 17, and 19 only

2, 5 make a perfect 10; 3, 7 make nearly double of 10; 11, 19 make nearly double of 10^2, and finally 13, 17 make somewhat more than double of 10^2. Our answer is slightly more than (2)(2)(2)(10)(10)(10^2)(10^2) which is very near to [spoiler]10^7.

Choose (C).[/spoiler]
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by Brent@GMATPrepNow » Wed May 14, 2014 6:47 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

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