The product of all the prime numbers less than 20 is closest to which

Question

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

Bunuel · Accepted Answer

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.

amitbharadwaj7 · Answer

I have approached in the following way:-

2 x 3 x 5 x 7 x 11 x 13 x 17 x 19 = 10 x 21x 11 x 13 x 17 x 19 = 10 x 10 x 2.1 x 10 x 1.1 x 10 x 1.3 x 10 x 1.7 x 10 x 1.9 = 10^6 x (2.1 x1.1 x 1.3 x 1.7 x 1.9)

The approximation of the (2.1 x 1.7 x 1.9 x 1.3 x 1.1) will be close to 10, so the answer is 10^7

monkeya41 · Answer

Similar to Bunuel's first approach:
I thought it was easiest to just do :
2*3*5*7 = 210
11 roughly 10
13 roughly 10
17 roughly 20
19 roughly 20
so 210*10*10*20*20 = 8*10^6 = 10^7

cyberjadugar · Answer

Hi,

Difficulty level: 650

Product of prime numbers less than 20 = 2*3*5*7*11*13*17*19
=(2*5)*3*(7*13)*(11*17)*19
=3*10*91*189*19
~(30)*(90*19)*190
~5700*1710
~5700*1700
~9690000  (=10^6)

Thus, Answer is (D)

Regards,

EvaJager · Answer

The product is  2*3*5*7*11*13*17*19 = 10 * 21 * 11*(13*17) * 19 \approx10*20*10*15^2*20=900*10^4\approx10^7 .

Therefore, answer C

cyberjadugar · Answer

Hi,

The magnitude of error depends on the approximation done in the question.

For example: 51*671=34221
50*671=33550 (error = 671)
51*670=34170 (error = 51)

Thus, the when the larger number is approximated the error is less.

Regards,

KarishmaB · Answer

Quickly approximate
2, 3, 5, 7, 11, 13, 17, 19
Make groups
2*5 = 10
3*17 = 50 (approximately)
7*13 = 100 (approximately)
11*19 = 200 (approximately)
So you make 7 zeroes (the 2 and the 5 also make a 0). When you multiply all these, the answer will be close to 10^7

ScottTargetTestPrep · Answer

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

SVaidyaraman · Answer

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 (10-8)(10-7)(10-5)(10-3)(10+1)(10+3)(10+7)(10+9)
Rearranging we have (10-8)*(10+9) * (10-7)*(10+7) * (10-3)*(10+3) *(10-5)*((10+1)

Approximately these 4 pairs are: 90*50*90*50= approximately 10^7

BrentGMATPrepNow · Answer

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 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

Since 10,000,000 = 10^7, the best answer is C

Cheers,
Brent

Apatel91 · Answer

In above yellow colored, I don't understand following...
while 13*7= 91 (you have taken 100 instead of 90 (higher side 10th rounding))
and for 3*17 = 51 (you have taken 50 instead of 60 (lower side 10th rounding))

What is logic of it ?

Bunuel · Answer

We want an accurate approximation. So, we should try the products mentioned there not to be far off the real results. A little bit more or a little bit less is ok. That's it.

jabhatta2 · Answer

avigutman  - how would you do this ?

i did it this way but not sure if i can do this in 2 mins

avigutman · Answer

jabhatta2  My mind immediately went to differences of squares (this will look like a lot of work but it can be done in well under a minute): 
(3*7) * (2*5) * (11*19) * (13*17) = (5-2)(5+2) * (10) * (15-4)(15+4) * (15-2)(15+2) = 
(5^2 - 2^2) * (10) * (15^2 - 4^2) * (15^2 - 2^2) = 21 * 10 * 209 * 221 = 
(separating the factors of 2 from the factors of 10)   * 10*6 ≈ 10^7

GMATCoachBen · Answer

Short video solution here (1:39) : https://www.youtube.com/watch?v=k9M9Hl19o8U The key is that we can round quite aggressively, since the answers are a full power of 10 apart. (For example, 10^7 is an enormous 1000% of 10^6 ) In this solution video, I knew that 13*17 would round to be close to 200, since 13^2 = 169. However, it may be easier to round down the 13 to 10, and round up the 17 to 20, to make it 10*20. 2023-12-17 18_06_49-(2) Problem Solving (OG 2020)_ _The product of all the prime numbers less than 2.png

DanTheGMATMan · Answer

No sweat when we round:
 https://www.youtube.com/watch?v=2d2hpUV_YcI

Dbrunik · Answer

Posted from my mobile device

I don’t understand how this question has such a low accuracy rate. I was able to solve it and felt fairly confident the entire time that I would be able to solve it. The only problem is that it took me just over three minutes to solve, which is clearly far too long. The official solution appears to just tell you to multiply it all out. Has anybody figured out a faster way to do this

GGGMAT2 · Answer

Bunuel This is how I approached this question. Please let me know what's wrong with my approach.

2*3*5*7*11*13*17*19= 2*5 * 3*7 * 10(approximated 11 to 10) * 13*17 * 20 (approximated 19 to 20)
= 10 * 21 * 10 * 20 * 13*17 = 10 * 20 (approximated 21 to 20) * 10 * 20 * 13*17 = 4 * 10000 * 13*17 = 10000 * (4*13 ~ 50) * 17
= 5 * 100000 * 17 = 85*100000.

I chose   10^6   as the answer.

Krunaal · Answer

You can write  85*100000 = 8500000 = 8.5*10^6

10^6 < 8.5*10^6 < 10^7  =>  1*10^6 < 8.5*10^6 < 10*10^6

We can see  8.5*10^6  is closer to  10^7 than  10^6

The product of all the prime numbers less than 20 is closest to which

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The product of all the prime numbers less than 20 is closest to which

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