Math problem: need help

santakdalai90 wrote:This question is from official guide 12th edition.
Q> The product of all the prime numbers less than 20 is closest to which powers of 10?
(A) 10^9
(B) 10^8
(C) 10^7
(D) 10^6
(E) 10^5

Ans : C

The explanation shows we have to multiply all the primes but that takes too long. Isn't there more logical and short way?

Here's one approach.

Since the numbers are very spread apart (each answer choice 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 just a bit larger that 10, so let's call it 10+

This means (10)(3)(7)(11)(13)(17)(19) = (10)(3)(7)(10+)(13)(17)(19)

Next, (7)(13)=91, which is a bit less than 100, so we get (10)(3)(7)(10+)(13)(17)(19) = (10)(3)(100-)(10+)(17)(19)

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

santakdalai90 wrote:This question is from official guide 12th edition.
Q> The product of all the prime numbers less than 20 is closest to which powers of 10?
(A) 10^9
(B) 10^8
(C) 10^7
(D) 10^6
(E) 10^5

Ans : C

The explanation shows we have to multiply all the primes but that takes too long. Isn't there more logical and short way?

We can regroup the numbers, 2, 3, 5, 7, 11, 13, 17, and 19, to multiply them.
2 * 3 * 5 * 7 * 11 * 13 * 17 * 19
= (2 * 5) * (3 * 7) * 11 * 13 * 17 * 19
= 10 * 21 * 11 * 13 * 17 * 19

Now approximating, we get 10 * 20 * 10 * 10 * 20 * 20 = 8 * 10^6, which is [spoiler]10^7[/spoiler] appr.

The correct answer is C.