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The product of all the prime numbers less than 20 is closest [#permalink]
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10 Jul 2017, 01:07
Multiply 2,5,3,7 and get 210 now rest numbers 11,13,17,19 when multiplied beginning with 210 derived would add a digit more to the number each time and finally we get a 7 digit number, hence which should be near to 10^7.
Please advise if this is feasible solution when calculating other such problems.



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Re: The product of all the prime numbers less than 20 is closest [#permalink]
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12 Jul 2017, 17:02
Bunuel wrote: The product of all the 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 We need to determine the product of: 2 x 3 x 5 x 7 x 11 x 13 x 17 x 19 Let’s group some of these numbers to get powers of 10: 5 x 19 is about 100 = 10^2 So, we are left with: 2 x 3 x 7 x 11 x 13 x 17 7 x 13 is about 100 = 10^2 So, we are left with: 2 x 3 x 11 x 17 2 x 3 x 17 is about 100 = 10^2 Finally, we have 11, which is about 10 = 10^1. Thus, the product of all the prime numbers less than 20 is closest to 10^2 x 10^2 x 10^2 x 10^1 = 10^7. Answer: C
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Re: The product of all the prime numbers less than 20 is closest [#permalink]
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14 Aug 2017, 14:48
stonecold wrote: High Quality Question from the Official Set. Its always good to remember the first 10 primes 2 3 5 7 11 13 17 19 23 29
Now the product of all primes <20 => 2*3*5*7*11*13*19 Product => 210*143*323 Approximating => Product =>200*150*300 => 9000000 which is close to 10^7 Hence C hi Everytime I attempted the problem I ended up with 10^6, and, very naturally, I also found some or many people here with such answer, everybody, however, is claiming the answer to be 10^7... I have seen you online now, so can you please shed some light on this issue ...? thanks in advance ..



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Re: The product of all the prime numbers less than 20 is closest [#permalink]
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14 Aug 2017, 15:06
ScottTargetTestPrep wrote: Bunuel wrote: The product of all the 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 We need to determine the product of: 2 x 3 x 5 x 7 x 11 x 13 x 17 x 19 Let’s group some of these numbers to get powers of 10: 5 x 19 is about 100 = 10^2 So, we are left with: 2 x 3 x 7 x 11 x 13 x 17 7 x 13 is about 100 = 10^2 So, we are left with: 2 x 3 x 11 x 17 2 x 3 x 17 is about 100 = 10^2 Finally, we have 11, which is about 10 = 10^1. Thus, the product of all the prime numbers less than 20 is closest to 10^2 x 10^2 x 10^2 x 10^1 = 10^7. Answer: C hi I am very much worried about the question, especially because, in many ways we can end up with 10^6 depending on the errors that can stem from the numbers we can approximate in many ways ..... can you please shed some more light on this issue, especially on the number picking process, for example, we have multiplied 5x19, some other people can multiply 5x2, and so on ... can you please provide some guiding principles to handle questions as such ... thank you in advance ....



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Re: The product of all the prime numbers less than 20 is closest [#permalink]
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15 Aug 2017, 11:49
Bunuel wrote: SOLUTION
The product of all the 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
We should find the approximate value of 2*3*5*7*11*13*17*19 to some power of 10.
# of different approximations are possible.
Approach #1:
2*5=10; 3*7=~20 (actually more than 20); 11*19=~200 (actually more than 200); 13*17=~200 (actually more than 200);
\(2*3*5*7*11*13*17*19\approx{10*20*200*200=8*10^6}\approx{10^7}\).
Answer: C.
Approach #2:
2*5=10 3*17=~50 (actually more than 50); 7*13=~100 (actually less than 100); 11*19=~200 (actually more than 200)
\(2*3*5*7*11*13*17*19\approx{10*50*100*200}=10^7\).
Answer: C. I think I have got this high quality official question... 2,3,5,7,11,13,17,19 2x5 =10 3x11 =30 7x13= 100 17x19 = 400 =10 x1.2 x 10^6 =1.2 x 10^7 cheers ...



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Re: The product of all the prime numbers less than 20 is closest [#permalink]
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04 Sep 2017, 00:17
It is 2*3*5*7*11*13*17*19 ~10* 21* 140 * 400 ~200 * 48000 ~96 00 00 0 ~ 10^ 7
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Re: The product of all the prime numbers less than 20 is closest [#permalink]
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23 Jan 2018, 17:41
Bunuel wrote: The product of all the 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 Diagnostic Test Question: 15 Page: 22 Difficulty: 650 It is a good idea to have the representation of numbers in the solution in terms of that in the problem. Primes below 20 are 2,3,5,7,11,13,17,19. So we have (108)(107)(105)(103)(10+1)(10+3)(10+7)(10+9) Rearranging we have (108)*(10+9) * (107)*(10+7) * (103)*(10+3) *(105)*((10+1) Approximately these 4 pairs are: 90*50*90*50= approximately 10^7
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Re: The product of all the prime numbers less than 20 is closest [#permalink]
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02 May 2018, 09:05
Bunuel wrote: The product of all the 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
Since the answer choices are very spread apart (each number 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 10First (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 Since 10,000,000 = 10^7, the best answer is C Cheers, Brent
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Re: The product of all the prime numbers less than 20 is closest
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