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

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 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

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,

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
_________________

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,

2*3*5*7*11*13*17*19 = 210*(10+1)(10+3)(10+7)(10+9)
210 ~ 10*10*2
So the original equation will be equivalent to 10*10*2*(10+1)(10+3)(10+7)(10+9)
The highest order of the equation will be 10^6, and other part of expression will not make it closer to 10^7.
I have solved it mostly on hunch, no sure how correct I am.

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

Product of Prime no between 1 & 20 =

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

210 * 143 * 323

\((10^2+10) * (10^1+43) * (10^3+23)\)

\((10^3+4300+10^2+430) * (10^3+23)\)

\(5830 * (10^3+23)\)

5830000 + 134090

= 6974090 this is near to 10^7

Answer = C

Here's my solution
2*3*5*7*11*13*17*19

1) 2*3*5*19 ~ 30*20
2) 11*13 ~ 140
3) 7*17 ~ 120

SO we have 600*140*120 = 10^4*168*6 (168*6 ~ 1000) --> ~ 10^4*10^3 ~10^7

Method 2

2*3*5*7*11*13*17*19 = 2*5*11*19*(10-7)(10+7)(10-3)(10+3) ~ 2*10*10*10*50*80=100*10*10*10*10*8 ~\(10^7\) Answer (D)

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
_________________

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
_________________

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
_________________

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

Prime numbers from 1-10= 2*3*5*7=210=200 roughly,

Prime numbers from 11-20 roughly:

11=10

13=10

17=20

19=20

Now,

200*10*10*20*20

=(2*10^2)*(10)*(10)*(2*10)*(2*10)

=2*2*2*10^6

=8*10^6

=10^7 approximately

Answer is C

Posted from my mobile device

Bunuel

I did something similar to approach # 1 you mentioned and got ~ 8 x 10^6


Q1) If the question was made harder to give an estimation like ~ 5 x 10^6... Could i still pick (C) for this or (D) be closer in that case

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 ?

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.
_________________

Since the answer choices were so far apart, I just estimated this way:

2*5=10
3*7=~20
11, 13, 17, 19 =~10*10*20*20

10^3*20^3=8*10^6 so closest to 10^7

My estimation was:

7*11*13 = 1000

*19 = 20 000

*5 = 100 000

*17 = 1 700 000

*6 = 10 200 000

The exact product is 9 699 690