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Product of all prime numbers less than 20

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Product of all prime numbers less than 20

by 4meonly » Tue Aug 19, 2008 4:01 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

OG11 offers to multiply 2*3*5*7*11*13*17*19 to get 9699690.

Can anybody find easier way?
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by pepeprepa » Tue Aug 19, 2008 6:28 am
Personnally, I did the way of OG
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Re: Product of all prime numbers less than 20

by sudhir3127 » Tue Aug 19, 2008 8:45 am
4meonly 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

OG11 offers to multiply 2*3*5*7*11*13*17*19 to get 9699690.

Can anybody find easier way?
iam not very sure of any easy method .. but this how i did ...

2*3*5*7*11*13*17*19

7*11*13 2*5 17*19 * 3

1001 * 10 * (18-1))(18+1) 3

approximately 10^4 *(324-1) *3

10^4 * 323*3

9690000

nearest power 10 is 10^7
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by preetha_85 » Tue Aug 19, 2008 11:10 am
Hi...

Just look at all the nos.. 2,3,5,7,11,13,17,19....
Group them such that they are almost equal as in....
2*3*5*7= 210,
11*19=209,
17*13=221
Since its all approximately 200-> the ans will be 8000000 + ((10+19+21)*200

This ans will certainly be less than 10^7 nd greater than 10^6...
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by 4meonly » Thu Aug 21, 2008 3:52 am
Guys, thanx!

For me these approaches are better that OG offers :D
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by sk818020 » Mon May 03, 2010 6:55 pm
I just ran across this problem during my prep and got the right answer, but I wanted to make sure the logic I used wouldn't lead me astray if used in the future on a similar problem.

As noted already the primes are; 2, 3, 5, 7, 11, 13, 17, 19.

I noted 5*2=10 and 3*7=21=(10+10+1)

Thus I simplified the equation:

2*3*5*7*11*13*17*19 to;

(10)(10+10+1)(10+1)(10+3)(10+7)(10+9)

I counted up the tens and saw that there were 7, so I picked C, which was 10^7.

Is this not a correct way to think about the problem?
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by nakul_anand » Tue May 04, 2010 2:52 am
Product of the prime numbers

2*3*5*7*11*13*17*19

=(2*5)*(3*7)*(11*19)*(13*17)

=(10)*(21)*(209)*(221)

=(10)*(10*2.1)*(10*10*2.09)*(10*10*2.21)

=10^6* (2.1*2.09*2.21)

Now, 10^6*(2.1*2.09*2.21) is definitely greater than 10^6 and definitely less than 10^7, but is closer to 10^7.

Hence Ans: C
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by kevincanspain » Tue May 04, 2010 2:59 am
4meonly 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

OG11 offers to multiply 2*3*5*7*11*13*17*19 to get 9699690.

Can anybody find easier way?
You could also round as you multiply

30 * 7 approx 200
200 x 11 approx 2000
2000 * 13 approx 30000
30000 * 17 approx 500, 000
500,000 x 19 approx 10,000,000

If you are quick with numbers, you could regroup

(19 x 5) x (17 x 2 x 3) x (7 x 13 x 11) approx 100 x 100 x 1000
Kevin Armstrong
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by oliviamika » Tue Nov 02, 2010 7:15 pm
sk818020 wrote:I just ran across this problem during my prep and got the right answer, but I wanted to make sure the logic I used wouldn't lead me astray if used in the future on a similar problem.

As noted already the primes are; 2, 3, 5, 7, 11, 13, 17, 19.

I noted 5*2=10 and 3*7=21=(10+10+1)

Thus I simplified the equation:

2*3*5*7*11*13*17*19 to;

(10)(10+10+1)(10+1)(10+3)(10+7)(10+9)

I counted up the tens and saw that there were 7, so I picked C, which was 10^7.

Is this not a correct way to think about the problem?
I prefer ur ans~thank you~
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by rajatdelhi » Tue Nov 02, 2010 8:37 pm
The approach I took was to approximate the number

2*3*5*7*11*13*17*19
->210*11*13*17*19
->200*10*10*20*20
->2*2*2*10^6
->8*10^6
->10*10^6= 10^7
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by Rezinka » Wed Nov 03, 2010 8:54 pm
I had seen it done this way somewhere which I found very easy :
We have;
2 * 3 * 5 * 7 * 11 * 13 * 17 * 19
= (2 * 5) * (3 * 7) * 11 * 13 * 17 * 19
= 10 * 21 * 11 * 13 * 17 * 19
approximating ;
10 * 20 * 10 * 10 * 20 * 20
=8 * 10^6
approximating :
10 * 10^6
= 10^7

Hope it helps..!!
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by heather208 » Tue Apr 26, 2011 5:57 pm
I'm not sure if this is easier, but it was for me:

I just regrouped the numbers.. 2, 3, 5, 7, 11, 13, 17, 19. The goal is to rewrite expression in terms of 10^1 :

2 x 19 = 2(10+9) = 20+18 = 38, which is approximately 4 x 10^1
3 x 17 = 3(10+7) = 30+21 = 51, which is approximately 5 x 10^1
5 x 13 = 5(10+3) = 50+15 = 65, which is approximately 7 x 10^1
7 x 11 = 7(10+1) = 70+7 = 77, which is approximately 8 x 10^1

Now, it is easy to multiply single digits and exponents with the same base of 10.

10^1 x 10^1 x 10^1 x 10^1 = 10^4
4 x 5 x 7 x 8 = approximately 1000, which can be expressed as 10^3

Multiply:
10^4 x 10^3 = 10^7 --> Final Answer
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by itheenigma » Fri Sep 02, 2011 11:26 pm
I'm afraid this method is incorrect. You can't count the number of 10s in the equation and add it up. In this particular problem, you were just lucky.
Consider another scenario where the 7 is replaced by another 5 in the sequence.
Then you would have -
2*3*5*5*11*13*17*19 -->
(10)(10+5)(10+1)(10+3)(10+7)(10+9)
Counting the 10s will give you 10^6.

But the actual answer for 2*3*5*5*11*13*17*19 = 6,928,350
again closer to 10^7

The only way to simplify this problem is by approximation.
Please see Rezinka's solution.

Cheers!
sk818020 wrote:I just ran across this problem during my prep and got the right answer, but I wanted to make sure the logic I used wouldn't lead me astray if used in the future on a similar problem.

As noted already the primes are; 2, 3, 5, 7, 11, 13, 17, 19.

I noted 5*2=10 and 3*7=21=(10+10+1)

Thus I simplified the equation:

2*3*5*7*11*13*17*19 to;

(10)(10+10+1)(10+1)(10+3)(10+7)(10+9)

I counted up the tens and saw that there were 7, so I picked C, which was 10^7.

Is this not a correct way to think about the problem?
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by AbhiJ » Mon Sep 05, 2011 2:38 am
2*3*5*7*11*13*17*19

(2*5)(17*3*19)(7*11*13)
10*(51*19)(91*11)
10*10^3*10^3
10^7
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by Amzad Sadik » Fri Aug 11, 2017 11:08 pm
How did you find 8 ? I know question is silly but please tell me. Thanks.
Rezinka wrote:I had seen it done this way somewhere which I found very easy :
We have;
2 * 3 * 5 * 7 * 11 * 13 * 17 * 19
= (2 * 5) * (3 * 7) * 11 * 13 * 17 * 19
= 10 * 21 * 11 * 13 * 17 * 19
approximating ;
10 * 20 * 10 * 10 * 20 * 20
=8 * 10^6
approximating :
10 * 10^6
= 10^7

Hope it helps..!!
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