When positive intiger n is divided by 3, the remainder is 2 and when positive intiger t is divided by 5, the remainder is 3. What is the remainder when the product of nt is divided by 15?
(1) n-2 is divisible by 5
(2) t is divisble by 3
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- Borcho
answer
Answer is C.
- StaceyKoprince
- ManhattanGMAT Staff
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When told that the divisor is 3 and the remainder is 2, we can write that as n = 3x+2. By the same process, we can write t = 5y+3.
Statement 2 is easier, so let's start with that one. If t is div. by 3, then 5y is div by 3, which means y by itself is div by 3. That means y could be 3, 6, 9, etc. By itself, this tells me nothing about what n (or x) could be, so there's more than one possibility for the remainder I'm trying to find. For example, if y = 3, then t = 18. n could be 5, in which case nt = 90, giving me a remainder of zero when divided by 15. If n is 8 instead, then nt = 144 which will not give me a remainder of zero (note that I don't actually bother to find this remainder - I can just tell I won't have the same number, zero, and that's enough). Insufficient.
Statement 1 says n-2 is div by 5. If I take my equation n = 3x+2 and subtract 2 from each side, I get n-2 = 3x, which means that 3x is div. by 5, which means x by itself is div. by 5. So x could be 5, 10, 15, etc. Again, by itself, tells me nothing about t or y, so insufficient.
Together, I know some traits and narrowed possibilities for both y and x. Let's see if that gives me a definitive remainder. t could be 18, 33, 48, etc. n could be 17, 32, 47, etc. ***
Try the two smallest numbers: 17*18 = 306, which gives a remainder of 6 when divided by 15. Next, let's try 17*33 = 561, which gives a remainder of 6 when divided by 15.
Now, here's where the actual shortcut can come in. I notice the pattern for n and t: each one goes up by 15, every time (go look up at the possibilities for n and t). So I realize that I don't even have to try 17*18 etc - the remainder is going to remain constant as long as I'm dividing by 15 because each possible number is generated by adding 15 each time. Now I know the next time I do a problem like this, I can stop when I get to the *** asterisk signs I typed above, because I can see the relevant pattern there.
Statement 2 is easier, so let's start with that one. If t is div. by 3, then 5y is div by 3, which means y by itself is div by 3. That means y could be 3, 6, 9, etc. By itself, this tells me nothing about what n (or x) could be, so there's more than one possibility for the remainder I'm trying to find. For example, if y = 3, then t = 18. n could be 5, in which case nt = 90, giving me a remainder of zero when divided by 15. If n is 8 instead, then nt = 144 which will not give me a remainder of zero (note that I don't actually bother to find this remainder - I can just tell I won't have the same number, zero, and that's enough). Insufficient.
Statement 1 says n-2 is div by 5. If I take my equation n = 3x+2 and subtract 2 from each side, I get n-2 = 3x, which means that 3x is div. by 5, which means x by itself is div. by 5. So x could be 5, 10, 15, etc. Again, by itself, tells me nothing about t or y, so insufficient.
Together, I know some traits and narrowed possibilities for both y and x. Let's see if that gives me a definitive remainder. t could be 18, 33, 48, etc. n could be 17, 32, 47, etc. ***
Try the two smallest numbers: 17*18 = 306, which gives a remainder of 6 when divided by 15. Next, let's try 17*33 = 561, which gives a remainder of 6 when divided by 15.
Now, here's where the actual shortcut can come in. I notice the pattern for n and t: each one goes up by 15, every time (go look up at the possibilities for n and t). So I realize that I don't even have to try 17*18 etc - the remainder is going to remain constant as long as I'm dividing by 15 because each possible number is generated by adding 15 each time. Now I know the next time I do a problem like this, I can stop when I get to the *** asterisk signs I typed above, because I can see the relevant pattern there.
Stacey Koprince
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Director, Content & Curriculum
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- christiancryan
- Course Students
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- Joined: Thu Jul 31, 2003 10:44 am
This is a legitimately scary problem! I do think it can be done in 2 min (barely!), but you have to get facile with representing remainder and divisibility relationships in certain ways, introducing variable names as appropriate on the fly and also sometimes just using "int" to represent a random integer (that is, an integer that's not generally the same as anywhere else I write "int"). This notation is key, at least for me.
I'll stick with Stacey's variable notation:
Stem:
n = 3x + 2
t = 5y + 3 (x, y both ints)
nt = 15xy + 10y + 9x + 6 --> what's the rem after divide by 15?
Drop 15xy (that's automatically divisible by 15) -->
10y + 9x + 6 = 15*int + what remainder?
1) n-2 = 5*int (at this point, I don't introduce yet another variable)
n = 5*int + 2
Matching up to first equation, I get
3x = 5*int
So x = (5*int)/3 = 5*(int/3) = 5*another int (since x = integer too -- think about this; if that int/3 is NOT an integer, then neither will 5*int/3 be an integer...)
Now, we can stick x = 5*int into the question:
10y + 9*5*int + 6 = 15*int + what remainder?
Since we don't know anything about y, we don't have enough info.
2) t = 3*int = 5y + 3
5y = 3*int - 3 = 3*another int
By same reasoning, y =3*yet another int. (Another way to see this is to think of the prime factorizations of both sides of the equation 5y = 3*int. y must contribute the 3 that shows up on the right side.)
Subbing into the original question, we get
30*int + 9x + 6 = 15*int + what remainder? Also not enough.
Together, we have 30*int + 9*5*int + 6 = 15*int + what remainder?
The first two terms are divisible by 15, so the remainder will definitely be 6. Answer is C.
I hate multiplying two-digit numbers together, so even though this is long, I personally prefer it to testing numbers.
I'll stick with Stacey's variable notation:
Stem:
n = 3x + 2
t = 5y + 3 (x, y both ints)
nt = 15xy + 10y + 9x + 6 --> what's the rem after divide by 15?
Drop 15xy (that's automatically divisible by 15) -->
10y + 9x + 6 = 15*int + what remainder?
1) n-2 = 5*int (at this point, I don't introduce yet another variable)
n = 5*int + 2
Matching up to first equation, I get
3x = 5*int
So x = (5*int)/3 = 5*(int/3) = 5*another int (since x = integer too -- think about this; if that int/3 is NOT an integer, then neither will 5*int/3 be an integer...)
Now, we can stick x = 5*int into the question:
10y + 9*5*int + 6 = 15*int + what remainder?
Since we don't know anything about y, we don't have enough info.
2) t = 3*int = 5y + 3
5y = 3*int - 3 = 3*another int
By same reasoning, y =3*yet another int. (Another way to see this is to think of the prime factorizations of both sides of the equation 5y = 3*int. y must contribute the 3 that shows up on the right side.)
Subbing into the original question, we get
30*int + 9x + 6 = 15*int + what remainder? Also not enough.
Together, we have 30*int + 9*5*int + 6 = 15*int + what remainder?
The first two terms are divisible by 15, so the remainder will definitely be 6. Answer is C.
I hate multiplying two-digit numbers together, so even though this is long, I personally prefer it to testing numbers.
- rschunti
need more information
Hi,
I wanted to clear my concept and master on the steps needed to solve similar type of problems as mentioned above using general equation. Pls can you refer me to the maths books that explains this concept and have sufficient exercises with explanations. I appreciate any advice. Thanks
I wanted to clear my concept and master on the steps needed to solve similar type of problems as mentioned above using general equation. Pls can you refer me to the maths books that explains this concept and have sufficient exercises with explanations. I appreciate any advice. Thanks
- RonPurewal
- Students
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Re: need more information
rschunti Wrote:Hi,
I wanted to clear my concept and master on the steps needed to solve similar type of problems as mentioned above using general equation. Pls can you refer me to the maths books that explains this concept and have sufficient exercises with explanations. I appreciate any advice. Thanks
that's quite a request.
try searching online for 'elementary number theory', 'divisibility and primes', and the like, and check out what you see. the trick is to avoid sites that veer too far off into advanced stuff; because of the nature of the internet, most of the pages (and especially textbooks) you're going to find will be way, way too advanced for the gmat. that's not to say you couldn't comprehend them, but you'd be wasting your time with most of the material that's in them. (remember that the gmat doesn't test much beyond basic algebra; they just do so in really nasty, devious, unconventional ways)
you can try the PDF files on this page, which are intended for students preparing for the australian math olympiads:
http://www.maths.uwa.edu.au/~gregg/Academy/
in fact, if you're looking for the right combination of depth and accessibility, try looking for files dedicated to students taking the math olympiads in various countries (especially at the junior-high or early secondary level). those tests focus on divisibility and primes often, without getting into the theoretical depth that characterizes university courses in number theory (and doesn't characterize gmat problems).
good luck!
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