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Is there an easier way to solve Q 15 diagnostic test OG'16?

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Is there an easier way to solve Q 15 diagnostic test OG'16?

by lotrgandalf » Sat Jul 01, 2017 2:11 am
My strategy:
2x3x5x7 would be around point something 10^2.
11x13 would also be around point something 10^2
And similarly 17x19 too would be around point 10^2
So, answer would be around 10^6
But I got confused since the number before point can also multiply & be around 10^1 or 10^2.

Any trick or tips?
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by Brent@GMATPrepNow » Sat Jul 01, 2017 4:18 am
The product of all prime numbers less than 20 is closest to which of following powers of 10 ?
(A) 10^9
(B) 10^8
(C) 10^7
(D) 10^6
(E) 10^5
Here's another approach:

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 [approximately]

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

Cheers,
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by Brent@GMATPrepNow » Sat Jul 01, 2017 4:24 am
lotrgandalf wrote:My strategy:
2x3x5x7 would be around point something 10^2.
11x13 would also be around point something 10^2
And similarly 17x19 too would be around point 10^2
So, answer would be around 10^6
But I got confused since the number before point can also multiply & be around 10^1 or 10^2.

Any trick or tips?
Each of your above estimations are too low, and you end up missing out on a power of 10
2x3x5x7 would be around point something 10^2
3x5x7 = 105, so 2x3x5x7 = 210
11x13 would also be around point something 10^2
11x13 = 143
17x19 too would be around point 10^2
17x19 =323

So, you've rounded 210, 143 and 323 down to 100, which has caused you to get a product that's too small

When estimating, it's a good idea to round some calculations up and some down so that each offsets the other.

Cheers,
Brent
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by GMATGuruNY » Sat Jul 01, 2017 4:33 am
The product of all the prime numbers less than 20 is closest to which of the following powers of 10?

A.10^5
B.10^9
C.10^7
D.10^6
E.10^8
Since the answer choices are VERY far apart, we can BALLPARK.
For every value that we round UP, we should compensate by rounding another value DOWN.

2*3*5*7*11*13*17*19
210 * 10*15 * 15*20
200*150*300 = 9,000,000.

The closest power of 10 = 10,000,000 = 10^7.

The correct answer is C.
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by Matt@VeritasPrep » Wed Jul 05, 2017 1:22 am
Another quick and dirty approximation:

2 * 19 ≈ 40
3 * 17 ≈ 50
5 * 13 ≈ 60
7 * 11 ≈ 80

From there, we've got

40 * 50 * 60 * 80 =>

4 * 5 * 6 * 8 * 10 * 10 * 10 * 10

4 * 5 = 20 and 6 * 8 ≈ 50, so we've got

20 * 50 * 10�

2 * 10 * 5 * 10 * 10�

or 10�.

More important here is the clue that rough approximation is appropriate: the answers are far apart! Since they differ by powers of 10, we can get away with fudging the math pretty abominably.
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by elias.latour.apex » Fri Jul 07, 2017 7:09 am
There is a trick for doing fast math when one is multiplying numbers that are close together. For example, if someone is multiplying 11 x 13 you can simplify the math by using a reference number.

In this case, we will use the reference number of 10, because it's an easy number to work with.

(10) 11 x 13 =

We can see that eleven is higher than 10, so we can write down +1 under the 11
We can see that thirteen is higher than 10, so we can write down +3 under the 13.
Then we either add the +1 to 13 to get 14
or the +3 to 11 to get 14, and multiply that number by 10, yielding 140.
Then we multiply 1 x 3 to get 3. The final answer is 143.

This method has the advantage that it provides a pretty close estimation in the first phase, and a second phase that provides complete accuracy.

Similarly the 17x19 could be done base 20.

(20) 17 x 19 =
The 17 is -3 from 20.
The 19 is -1 from 20.
So the answer will be around (17-1)20 or (19-3)20. In either case it will be around 16x20 or 320. The exact answer is 323 because we add (-1)(-3) to 320 to get the final answer.

So fast estimation should get us:

2x5 = 10
3x7 = 21
11 x 13 ≈ 140
17 x 19 ≈ 320

And go from there.
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by Ian Stewart » Fri Jul 07, 2017 3:43 pm
I'd just use quick estimates here - the way I prefer to do this question is:

2*5 = 10
3*7 is roughly 20
11*19 is roughly 10*20
13*17 is roughly 10*20

so the product is roughly 10*20*10*20*10*20 = 8 * 10^6, which is closest to 10^7.

I'd never even contemplate doing any precise calculation here, since the answers are wildly far apart and the question only asks for an estimate, but if we're discussing shortcuts to calculate awkward products quickly and accurately, one way is to do as follows: for 13*17, say, take the median of the two numbers, 15, and rewrite each number by adding to or subtracting from 15:

13*17 = (15 - 2) (15 + 2)

Now we just have a difference of squares, easy if you know that 15^2 is 225:

13*17 = (15 - 2) (15 + 2) = 15^2 - 2^2 = 225 - 4 = 221

Similarly

11*19 = (15 - 4)(15 + 4) = 15^2 - 4^2 = 225 - 16 = 209

That only works out cleanly if you know the square of the median of your two numbers, but it can make certain difficult-looking products (87*93 say, which is just 90^2 - 3^2 = 8091) easy to work out.
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by Jay@ManhattanReview » Sun Jul 09, 2017 6:28 am
lotrgandalf wrote:My strategy:
2x3x5x7 would be around point something 10^2.
11x13 would also be around point something 10^2
And similarly 17x19 too would be around point 10^2
So, answer would be around 10^6
But I got confused since the number before point can also multiply & be around 10^1 or 10^2.

Any trick or tips?
Well, the problem is your estimation of the three products.

1. 2x3x5x7 = 210 = ~2*10^2
2. 11x13 = ~143 = ~1.5*10^
3. 17x19 = 323 = ~3*10^2

=> 2x3x5x7x11x13x17x19 =~2*10^2*~1.5x10^2*~3*10^2 = ~9*10^6 =~10^7.

The correct answer: C

Hope this helps!

-Jay

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by Scott@TargetTestPrep » Wed Jul 12, 2017 4:19 pm
The product of all prime numbers less than 20 is closest to which of 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

We are left with:

2 x 3 x 7 x 11 x 13 x 17

7 x 13 is about 100 = 10^2

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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