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by pzazz12 » Fri Sep 24, 2010 5:24 pm
Two different prime numbers between 4 and 18 are chosen. When their sum is subtracted from their product, which of the following numbers could be obtained ?

A. 21
B. 60
C. 119
D. 180
E. 231

Help me to solve this......................
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by diebeatsthegmat » Fri Sep 24, 2010 5:49 pm
pzazz12 wrote:Two different prime numbers between 4 and 18 are chosen. When their sum is subtracted from their product, which of the following numbers could be obtained ?

A. 21
B. 60
C. 119
D. 180
E. 231

Help me to solve this......................
i didnt sleep last night so i am not sure if i could do this correctly with correct answer but i will try...
there are 4 prime numbers from 4 to 18 : 5.11.13.17
sum is subtrated from the product = the products of 2 prime number a*b - the added 2 prime number a+b
i find the answer is C
11*13=143-(11+13)=119
any other number multipled together then subtracted to their sums get no right answer from the above answer
correct answer is C
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by narik11 » Mon Sep 27, 2010 10:29 am
pzazz12 wrote:Two different prime numbers between 4 and 18 are chosen. When their sum is subtracted from their product, which of the following numbers could be obtained ?

A. 21
B. 60
C. 119
D. 180
E. 231

Help me to solve this......................
Multiplication of two odd numbers is odd and Sum of two odd numbers is even.
Therefore the difference should be odd

And u can eliminate Option B and D.
Option A can be eliminated by plugging in lowest prime values.


Hope this helps
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by Brian@VeritasPrep » Mon Sep 27, 2010 10:57 am
Nice work, everyone. Taking Narik's strategy a step further:

The product will be odd (odd * odd) and the sum will be even (odd + odd), so subtracting those two you'll end up with an odd number as the result.

How do we know to do this? Well, there's just too much calculation possible - when calculations seem to be involved and tedious, you should always think about using number properties as a way to shorten them. There are simply too many combinations of these numbers (5, 7; 5, 11; 7, 11; 13, 17; 11, 17; etc.) to try double-digit, odd-number multiplication and do the subtraction. If there's a way to simplify that process - and knowing that you have some odd and some even answer choices is another nice clue - you should give that a shot first.


Now we know that B and D are out, leaving A, C, and E. What to do? I see a few clues:

-Those choices are all pretty nicely spread out - 21, 119, and 231. It's very, very possible that at least one of the outer choices will not be possible.

-If we do want to plug in primes via trial-and-error, it makes sense to be methodical. Try to get 21 first, and if we can't then move on - if we start with smaller primes, they'll be easier to use and we may find patterns that will help.

If we do that, we'll find that the SMALLEST pairing of 5 and 7 yields:

5*7 - (5+7) = 35 - 12 = 23

23 is too large to be 21, and that's the smallest we can get, so it's eliminated.

Now...we know that our remaining choices are 119 and 231, and we've already recognized that there's a limited range to the set of potential values here. Trial-and-error dictates that we need to try bigger numbers anyway, so let's try the biggest to see if we can either eliminate 231 or recognize how close we are to that range (for example, we might notice that 13 * 17 gets us very close to 231, but reducing one of the variables would do the trick...). So, let's try 13 and 17:

13*17 - (13+17) = 221 - 30, and we know that even at the biggest this number can be, we won't reach 231. Therefore E is eliminated and the correct answer must be C - it's the only plausible choice left.



When problems seem to involve too many calculations, too extensive of calculations, or both, try to use number properties and/or test the range of the answer choices to find a more efficient way of answering the questions. Even if you can only eliminate a few or limit the range somehow, you'll still drastically reduce your workload and the time it takes to complete these.
Brian Galvin
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Chief Academic Officer
Veritas Prep

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by diebeatsthegmat » Mon Sep 27, 2010 5:41 pm
Brian@VeritasPrep wrote:Nice work, everyone. Taking Narik's strategy a step further:

The product will be odd (odd * odd) and the sum will be even (odd + odd), so subtracting those two you'll end up with an odd number as the result.

How do we know to do this? Well, there's just too much calculation possible - when calculations seem to be involved and tedious, you should always think about using number properties as a way to shorten them. There are simply too many combinations of these numbers (5, 7; 5, 11; 7, 11; 13, 17; 11, 17; etc.) to try double-digit, odd-number multiplication and do the subtraction. If there's a way to simplify that process - and knowing that you have some odd and some even answer choices is another nice clue - you should give that a shot first.


Now we know that B and D are out, leaving A, C, and E. What to do? I see a few clues:

-Those choices are all pretty nicely spread out - 21, 119, and 231. It's very, very possible that at least one of the outer choices will not be possible.

-If we do want to plug in primes via trial-and-error, it makes sense to be methodical. Try to get 21 first, and if we can't then move on - if we start with smaller primes, they'll be easier to use and we may find patterns that will help.

If we do that, we'll find that the SMALLEST pairing of 5 and 7 yields:

5*7 - (5+7) = 35 - 12 = 23

23 is too large to be 21, and that's the smallest we can get, so it's eliminated.

Now...we know that our remaining choices are 119 and 231, and we've already recognized that there's a limited range to the set of potential values here. Trial-and-error dictates that we need to try bigger numbers anyway, so let's try the biggest to see if we can either eliminate 231 or recognize how close we are to that range (for example, we might notice that 13 * 17 gets us very close to 231, but reducing one of the variables would do the trick...). So, let's try 13 and 17:

13*17 - (13+17) = 221 - 30, and we know that even at the biggest this number can be, we won't reach 231. Therefore E is eliminated and the correct answer must be C - it's the only plausible choice left.



When problems seem to involve too many calculations, too extensive of calculations, or both, try to use number properties and/or test the range of the answer choices to find a more efficient way of answering the questions. Even if you can only eliminate a few or limit the range somehow, you'll still drastically reduce your workload and the time it takes to complete these.
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this is a nice solution. thanks
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by GMATGuruNY » Mon Sep 27, 2010 8:22 pm
pzazz12 wrote:Two different prime numbers between 4 and 18 are chosen. When their sum is subtracted from their product, which of the following numbers could be obtained ?

A. 21
B. 60
C. 119
D. 180
E. 231

Help me to solve this......................
The prime numbers between 4 and 18 are: 5, 7, 11, 13, 17
product = odd*odd = odd
sum = odd+odd = even
product - sum = odd-even = odd

The correct answer must be odd. Eliminate B and D.

Largest product is 13*17 = 221. Eliminate E.

Let's try 119.
Since 119 = product - sum, we need two prime numbers whose product is just a little more than 119.
11*13 = 143
11+13 = 24
143-24 = 119. Success!

The correct answer is C.
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