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Is there a quick and accurate way to solve this?

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Is there a quick and accurate way to solve this?

by Blues » Sun Aug 30, 2009 9:28 pm
Hello,

From OG 12 Diagnostic Question #15:
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

Answer: [spoiler]c - 10^7[/spoiler]
To get this, I just wrote out the primes under 20, counted them out, and selected that as my choice. I got it right, but it was more of an educated guess. I have a hard time believing this is the best way to go. Aside from actually multiplying this out, is there another way to determine the answer?

Thanks for you help.

Regards,
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by DanaJ » Sun Aug 30, 2009 10:21 pm
Actually, you have 8 primes under 20, so... I guess it was indeed a lucky one for you! What you can do is notice that 10 = 2*5 and approximate everything into multiples of 5 and 10:

2 - 2
3 - 5
5 - 5
7 - 5
11 - 10
13 - 10
17 - 20
19 - 20

As you can see, I tried to "balance out" the approximations: I approximated 3 to 5, by adding 2, but I approximated 7 to 5 as well, by subtracting 2. Same goes for (11, 13) and (17, 19): I subtracted 4 from the first group, but added the 4 to the second group.

Now it will be much, much easier to multiply stuff.
5^3 = 125, so the first 4 numbers will be 250.
10*10*20*20 = 40000.

250*40000 = 10 000 000 = 10^7.
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by Blues » Mon Aug 31, 2009 12:04 am
DanaJ wrote:Actually, you have 8 primes under 20, so... I guess it was indeed a lucky one for you!
Ha! I just went back to the list I made, and sure enough I somehow forgot the #2. I can't believe I did that!
DanaJ wrote:What you can do is notice that 10 = 2*5 and approximate everything into multiples of 5 and 10:

2 - 2
3 - 5
5 - 5
7 - 5
11 - 10
13 - 10
17 - 20
19 - 20

As you can see, I tried to "balance out" the approximations: I approximated 3 to 5, by adding 2, but I approximated 7 to 5 as well, by subtracting 2. Same goes for (11, 13) and (17, 19): I subtracted 4 from the first group, but added the 4 to the second group.

Now it will be much, much easier to multiply stuff.
5^3 = 125, so the first 4 numbers will be 250.
10*10*20*20 = 40000.

250*40000 = 10 000 000 = 10^7.
Hm...I've never been comfortable doing approximations for questions that don't specifically ask for it - I'm not really sure why. I guess it feels like an easy way to go down the wrong path. That said, it look like the best way to do this problem without a lot of slow, error prone number crunching.

Thanks!
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by heshamelaziry » Mon Aug 31, 2009 3:09 am
I don't understand Dana's explanation. Could you guys explain once more ??
sorry for the trouble
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by Blues » Mon Aug 31, 2009 3:51 am
heshamelaziry wrote:I don't understand Dana's explanation. Could you guys explain once more ??
sorry for the trouble
To start off, list all the primes from 0-20:

2 (the one I initially forgot :oops: ), 3, 5, 7, 11, 13, 17 and 19.

Now, approximate them to the nearest power of 5 or 10. The only expection is 2, which was left at 2 for simplicity sake, I assume.

2 becomes 2
3 becomes 5
5 becomes 5
7 becomes 5
11 becomes 10
13 becomes 10
17 becomes 20
19 becomes 20

2*5*5*5=250
10*10=100
20*20=400

250*100=25000
25000*400=10000000=10^7

It's a bit rough, since you're estimating numbers, but the original problem asks for an estimation. Dana picked numbers that are close, and make the math MUCH easier than actually multiplying all the prime numbers under 20.
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by DanaJ » Mon Aug 31, 2009 6:42 am
I did not approximate 2 because 10 is 2*5, so 2 will be useful in some later multiplications (i.e. when 3 becomes 5, for instance, you use 2*5). That's the only reason.
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by fueledGMAT » Wed Nov 16, 2011 11:24 am
DanaJ,

Even with these approximations, the math is still a bit time consuming. Are there any properties out there to determine an approximate product based on a set of numbers?
"Perfection is achieved, not when there is nothing more to add, but when there is nothing left to take away." ~ Antoine de Saint-Exupery
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by DanaJ » Thu Nov 17, 2011 11:34 am
Can't think of any - but trust me, approximating to 2, 5 and 10 (and multiples of 10 as necessary) is the best thing to do! It's much easier to calculate stuff then, because you just add zeroes to whatever you need.
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by Anurag@Gurome » Thu Nov 17, 2011 11:41 pm
DanaJ wrote:Actually, you have 8 primes under 20, so... I guess it was indeed a lucky one for you! What you can do is notice that 10 = 2*5 and approximate everything into multiples of 5 and 10:

2 - 2
3 - 5
5 - 5
7 - 5
11 - 10
13 - 10
17 - 20
19 - 20

As you can see, I tried to "balance out" the approximations: I approximated 3 to 5, by adding 2, but I approximated 7 to 5 as well, by subtracting 2. Same goes for (11, 13) and (17, 19): I subtracted 4 from the first group, but added the 4 to the second group.

Now it will be much, much easier to multiply stuff.
5^3 = 125, so the first 4 numbers will be 250.
10*10*20*20 = 40000.

250*40000 = 10 000 000 = 10^7.

The primes under 20 are 2, 3, 5, 7, 11, 13, 17, and 19.
Try to group them into sets such that the product of numbers in any set is close to a power of 10.
Now, 2*3*5*7*11*13*17*19 can be written as (5*19)*(2*3*17)*(7*13*11) = 95*102*1001
This is approximately equal to 100*100*1000 = 10^7
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