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When positive integer n is divided by 3, the remainder is 2 [#permalink]

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16 Mar 2009, 23:28

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When positive integer n is divided by 3, the remainder is 2; and when positive integer t is divided by 5, the remainder is 3. What is the remainder when the product nt is divided by 15?

n = 17 ( the last digit of n had to be either 2 or 7 since n-2 mod 5 =0, any number divisible by 5 has to have last digit as 0 or 5 therefore n-2=last digit 0 or 5, n= last digit 2 or 7 ) t= 18 ( t is divisible by 3, the 1st no. that is divisible by 3 and yields 3 as remainder is 18 )

I tried to proceed in this way: n = 3a+2 t = 5b+3 So nt = 15ab+9a+10b+6 .... no use

I stop here and deduct: whether nt divisible by 15 depends on "9a+10b+6" 1. n-2 (=3a) divisible by 5 so a must be divisible by 5, or a =5k , not suff 2. t-3 (=5b) divisible by 3 so b must be divisible by 3, or b=3r, not suff

1 and 2 combined: 9a+10b+6 =45k +30r+6 divided by 15, the remainder must be 6, suff

Picking number may be confusing and intimidating. so solving equation would be best.
_________________

When positive integer n is divided by 3, the remainder is 2; and when positive integer t is divided by 5, the remainder is 3. What is the remainder when the product nt is divided by 15?

1) n-2 is divisible by 5

2) t is divisible by 3

What is the OA. Somehow, I got E.

From the question, n = 3a + 2, t = 5b + 3. Hence, nt = 15ab + 9a + 10b + 6

From stmt1: n = 5x + 2 = 3a + 2 or, 3a = 5x. Hence, nt = 15ab + 15x + 10b + 6....insufficient.

Also, including stmt2: t=3y = 5b+3 Hence, nt = 15ab + 15x + 6y And, when nt is divided by 15, remainder will have different values....hence, E.

Please see above for the solutions - the OA is C, and the remainder is 6. Great little problem!

I looked at option 2 first. The question can be re-written as (NT/5*3). according to option 2, t is divisible by 3. But we can not tell if N will be divisible by 5 or not......insuff.

Lets analyse option 1. NT can be written as (N-2+2)T which is (N-2)T + 2T. So NT/15 = ((N-2)T + 2T)/15 or (N-2)T/5*3 + 2T/5*3 There is no way you can determine the divisibilty of T.......insuff.

If u combine 1 & 2 u know that T is divisible by 3 but there will be a 2/5 factor remaining which indicates that NT is not completely divisible by 15. Thats what we are looking for.......... Correct me if i went wrong...

n when divided by 3 leaves a remainder 2 ==> n-2 is divisible by 3. t when divided by 5 leaves a remainder 3 ==> t-3 is divisible by 5.

Question being asked is what is the remainder when nt is divided by 15.

Clue 1 ==> it is given n-2 is divisible by 5 and from given information, we know n-2 is also divisible by 2 ==> n-2 is divisible by 15. But nothing is mentioned abt t. Hence insufficient clue.

Clue 2 ==> it is given t is divisible by 3, we can derive that t-3 is also divisible by 3. From the given information, we know t-3 is divisble by 5 ==> t-3 is divisible by 3 and 5 ==> t-3 is divisible by 15. Nothing is mentioned abt n. Hence insufficient clue.

Combine both the clues

n-2 = 15a ==> n = 15a + 2 t-3 = 15b ==> t = 15b + 3

what's up guys? i too originally missed this question during a practice exam. However, after reviewing it without time constraints, I realized that even when you don't understand a DS problem, always employ logic if you have to guess.

from the stem, given:

n= 3x+2 t= 5y+3

the best approach i discovered is simply to algebraically determine what n and t are in order to determine what the remainder of nt is

(i) given (n-2)/5 which can be expressed as n-2 = 5x -now solve for n using the equation for n as given in the stem and substitute for x using (i)

n-2= 5x, x= (n-2)/5, now plugging into n as given in stem: n = 3((n-2)/5)+2 this solves for n = 2. However, since not given anything for t, can eliminate A&D

(ii) now, in this statement, t is divisible by 3 can be written as t = 3y

-now plugging in this equation into the equation given in the stem ( t = 5y+3) and substituting y = t/3 as given in (ii): t = 5(t/3) +3.......solving for this equation yields t=-9/2. However, this statement gives nothing about n as required to solve for the remainder of nt/15. Hence, we can eliminate B.

now combined: n=2 and t= -9/2.........nt = -9

no need to solve for remainder! However, if you chose to calculate, you would soon realize that the remainder is 0. Try it and see for yourself.....

According to sojafon: nt = 15ab + 9a + 10b + 6. However, ab+9a+10b is not equal to X from a distributive property standpoint as assumed by 15X+6. Therefore the remainder is not 6 as assumed.

This is the only true and proven method I can see to be employed in order to solve this problem. However, if you were to employ a systematic approach, you will see that (i) gives nothing about t, which means eliminate AD. And (ii) says nothing about n, eliminate B. This leaves you with a 50/50 guess for C or E.

According to sojafon: nt = 15ab + 9a + 10b + 6. However, ab+9a+10b is not equal to X from a distributive property standpoint as assumed by 15X+6. Therefore the remainder is not 6 as assumed. OA is C.

Incorrect. 15 wasn't factored out of 15ab alone, but 9a and 10b as well. I showed that 15ab, 9a and 10b are all divisible by 15 when you combine (1) and (2) and rewrote 15ab + 9a + 10b + 6 as 15x+6 to make the answer more obvious.

nt = 15ab + 9a + 10b + 6, nt = 15x + 6

x is just a variable that represents the sum/value that's left after factoring 15 out of 15ab + 9a + 10b. This is essentially the same as nt = 15p + 15q + 15r + 6 nt =15(p + q + r) + 6, Let x be the sum of p, q and r. x = p + q + r nt = 15x + 6.

I used the same method as priyankur_saha. I think the answer is C.

Also if using numbers , N=17 and T =18 remainder is 6. The only problems with plugging in numbers is that you are not sure if the number you picked represent all cases. So although the first method is longer, I would prefer it over the number plugging way. However, on the test, if you don't have the time to invest, use 17 and 18.
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Yeah. Solving these type of question by algebra is better option than plugging the number as sometimes we forget to put the the different kind of numbers.

As you can see, the remainder is different values, but if you remember from the original problem statement. You can see that when t is divided by 5, the remainder must be 3:

6*1/15 --> R=6 because 3*1/5 has R=3 6*6/15 --> R=6 because 3*6/5 has R=3

I think plugging numbers is the easiest way to do this.

This question is about pattern recognition. By listing out the possible solutions for each, you can see a definite pattern.

Possible Solutions for n given stem- 2, 5,8,11,14,17,20,23,26,29,32,35,38,41... Possible solutions for t given stem- 3,8,13,18,23,28,33,38,43,48,53,58...

I. Using n-2 is divisble by 5, you get n=17,32,47..., notice the difference is 15, meaning each remainder will be the same when multiplying (Note that it is important that they specify integers) However this gives us no indication as to the remainder of t - Insuff

II Using t is divisible by 3, you get t=3,18,33,48... again, the difference is 15, meaning each remainder will be the same. No indication of n- insuff

You know that each remainder will be the same (in this case 6), C is suff since each individual number when divided by 15 will give the same remainder.

Thoughts?
_________________

"Any school that meets you and still lets you in is not a good enough school to go to" - my mom upon hearing i got in Thanks mom.

It seems that the consensus is for C being the answer. Despite this I keep getting E, and would appreciate if anyone can point out the pitfall in my reasoning:

hence \(nt/15 = (3a+2)*(5b+3)/15=(15ab+9a+10b+6)/15=ab+3a/5+2b/3+2/5\) ab is always an integer, so for the purpose of the question can be neglected

I)\(n=5c+2=3a+2 a=5/3c\) \(nt/15=3/5*5/3c+2/3b+2/5=c+2/3b+2/5\) c is always an integer and can be neglected, but nothing can be said of \(2/3b+2/5\), hence NS

I+II)\(nt/15=c+2/5d\) c is always an integer and can be neglected. \(nt/15=2/5d\) or \(nt=int+6*d\) we cannot find a single value for the remainder, hence E

When positive integer n is divided by 3, the remainder is 2; and when positive integer t is divided by 5, the remainder is 3. What is the remainder when the product nt is divided by 15?

From the stem: \(n=3p+2\) and \(t=5q+3\). \(nt=15pq+9p+10q+6\), we should find the remainder when this expression is divided by 15.

(1) n-2 is divisible by 5 --> \(n-2=5m\) --> \(n=5m+2=3p+2\) --> \(5m=3p\), \(15m=9p\) --> \(nt=15pq+9p+10q+6=15pq+15m+10q+6\). Clearly \(15pq\) and \(15m\) are divisible by 15, so remainder by dividing these components will be 0. But we still know nothing about \(10q+6\). Not sufficient.

(2) t is divisible by 3 --> means that \(5q+3\) is divisible by 3 --> 5q is divisible by 3 or q is divisible by 3 --> \(5q=5*3z=15z\) --> \(10q=30z\) --> \(nt=15pq+9p+10q+6=15pq+9p+30z+6\). \(15pq\) and \(30z\) are divisible by 15. Know nothing about \(9p+6\). Not sufficient.

(1)+(2) \(9p=15m\) and \(10q=30z\) --> \(nt=15pq+9p+10q+6=15pq+15m+30z+6\). Remainder when this expression is divided by 15 is 6. Sufficient.

Answer: C.

OR:

From the stem: \(n=3p+2\) and \(t=5q+3\).

(1) n-2 is divisible by 5 --> \(n-2=5m\) --> \(n=5m+2\) and \(n=3p+2\) --> general formula for \(n\) would be \(n=15k+2\) (about deriving general formula for such problems at: good-problem-90442.html#p723049 and manhattan-remainder-problem-93752.html#p721341) --> \(nt=(15k+2)(5q+3)=15*5kq+15*3k+10q+6\) --> first two terms are divisible by 15 (\(15*5kq+15*3k\)) but we don't know about the last two terms (\(10q+6\)). Not sufficient.

(2) t is divisible by 3 --> \(t=3r\) and \(t=5q+3\) --> general formula for \(t\) would be \(t=15x+3\) --> \(nt=(3p+2)(15x+3)=15*3px+9p+15*2x+6\). Not sufficient.

(1)+(2) \(nt=(15k+2)(15x+3)=15*15kx+15*3k+15*2x+6\) this expression divided by 15 yields remainder of 6. Sufficient.

Here's more of a plain English explanation, for those who like that sort of thing.

An important principle here is that you can multiply remainders. For instance, 10/7 = 1 r2 and 9/7 = 1 r3. 10*9/7 = 12 r6. See? Remainder 2 * remainder 3 = remainder 6.

Notice that if the resulting remainder is greater than the divisor, it wraps around again. 10/4=2 r 2 and 15/4=3 r 3. 10*15/4 should be remainder 6, but since 4 goes into 6, there is only 2 left. The actual result is 17 r2.

So, back to our problem. Using the prompt and statement 1, we know that both n/3 and n/5 have remainders of 2. We also know that t/5 has a remainder of 3.

So, if we just wanted to know the remainder when nt is divided by 5, we could multiply our remainders: 2 for n and 3 for t = 6. Since 5 goes into 6, we would be left with a remainder of 1.

However, since we’re dealing with 15, it’s more complicated. We know the remainder when n is divided by 15. Since n must be 2 more than a multiple of 3 and 2 more than a multiple of 5, it will also be 2 more than a multiple of 15: 17, 32, 47 . . .

We don’t know about t, though, unless we bring in statement 2. Once we do, we know that it is not only 3 more than a multiple of 5, but also an exact multiple of 3.

Out of our original list (3, 8, 13, 18, 23, 28, 33 . . .basically, every number that ends in 8 or 3), this leaves 3, 18, 33, 48 . . . i.e., 3 more than a multiple of 15.

Now we can multiply our remainders. n/15 has a remainder of 2 and t/15 has a remainder of 3, so nt/15 has a remainder of 6.
_________________

Dmitry Farber | Manhattan GMAT Instructor | New York

Here's more of a plain English explanation, for those who like that sort of thing.

An important principle here is that you can multiply remainders. For instance, 10/7 = 1 r2 and 9/7 = 1 r3. 10*9/7 = 12 r6. See? Remainder 2 * remainder 3 = remainder 6.

Notice that if the resulting remainder is greater than the dividend, it wraps around again. 10/4=2 r 2 and 15/4=3 r 3. 10*15/4 should be remainder 6, but since 4 goes into 6, there is only 2 left. The actual result is 17 r2.

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