Calculating large #s.

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likewhoa174: Newbie | Next Rank: 10 Posts; Posts: 1; Joined: Fri Nov 04, 2011 1:02 pm

by likewhoa174 » Mon Mar 25, 2013 12:18 pm

I'm just starting my GMAT studies. I have been far removed from math for quite some time now. So quantifying large numbers intimidates me. I came across the following problem:

All the prime numbers between 2 and 20 inclusive is closest to which of the following?

a)10^5
b)10^6
c)10^7
d)10^8

To get the answer you had to multiply 2*3*5*7*11*13*17*19=9,699,690

This would take me forever to calculate using pencil and paper. Can you anyone please share the fastest and most acurrate way to solve this??? Also, what can I do to get more comfortable with doing this??? Thank you in advance!

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Source: Beat The GMAT — Problem Solving | Mon Mar 25, 2013 12:18 pm

rintoo22: Senior | Next Rank: 100 Posts; Posts: 73; Joined: Sun May 06, 2012 2:19 am; Location: Cape Town; Thanked: 6 times

by rintoo22 » Mon Mar 25, 2013 12:42 pm

Hi,

The way I would solve is to multiply the numbers in pairs to arrive to nearest multiple of 10.
example
2*3*5*7*11*13*17*19

19*2 = 38 => 40 => 4*10
17*3 = 51 => 50 => 5*10
13*7 = 91 => 90 => 9*10
11*5 = 55 => => 5.5*10

10^4 *(4*5*9*5.5) => 10^4 * 9900 => 10^4 * 10^3 => 10^7

Well there could be easier ways though.

Thanks
Ritesh

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Brent@GMATPrepNow: GMAT Instructor; Posts: 16207; Joined: Mon Dec 08, 2008 6:26 pmFollowed by: 1268 members; Location: Vancouver, BC; Thanked: 5254 times; GMAT Score:770

by Brent@GMATPrepNow » Mon Mar 25, 2013 2:41 pm

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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Alina Ross: Junior | Next Rank: 30 Posts; Posts: 16; Joined: Wed Mar 13, 2013 10:17 pmFollowed by: 1 member; Thanked: 2 times

by Alina Ross » Tue Mar 26, 2013 4:01 am

You can combine it into:

2,(10-7), 5, (10-3), 11, (10+3),(10+7), 20 = 10*50*90*11*20 =approx. 10^7

After multiplying since the numbers are large, approximation is more accurate.

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GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members; Location: New York, NY; Thanked: 13060 times; GMAT Score:790

by GMATGuruNY » Tue Mar 26, 2013 5:26 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 = 10â�·.

The correct answer is C.

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