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When positive integer n is divided by 3, the remainder is 2
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01 Nov 2009, 13:06
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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? (1) n2 is divisible by 5. (2) t is divisible by 3.
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Re: GMAT Prep 2 remainder
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01 Nov 2009, 14:07
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) n2 is divisible by 5. (2) t is divisible by 3. 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) \(n2=5m\) > \(n=5m+2=3p+2\) > \(5m=3p\), \(15m=9p\) > \(nt=15pq+ 9p+10q+6=15pq+ 15m+10q+6\). 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) n2 is divisible by 5 > \(n2=5m\) > \(n=5m+2\) and \(n=3p+2\) > general formula for \(n\) would be \(n=15k+2\) (about deriving general formula for such problems at: goodproblem90442.html#p723049 and manhattanremainderproblem93752.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. Answer: C.
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Re: GMAT Prep 2 remainder
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24 May 2010, 17:52
If the explanation above is not helpful, you may find a step by step video solution of this question useful. On GMATFix site, this is GMATPrep question 1045
Best of luck, Patrick



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Re: DS problem : remainders
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28 Nov 2010, 18:36
hogwarts wrote: Saw this question on a GMATPrep test, and I can't figure out how to get to the correct answer. Can anyone help? Thanks! 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) n2 is divisible by 5. (2) t is divisible by 3. The correct answer is (C)  both statements together are sufficient, but neither statement alone is sufficient. Can anybody out there help explain to me how to get to this answer though? Thanks! This is how I would approach this question. When positive integer n is divided by 3, the remainder is 2; I say n = 3a + 2 ( a is a non negative integer) and when positive integer t is divided by 5, the remainder is 3. So t = 5b + 3 (b is a non negative integer.) What is the remainder when the product nt is divided by 15?So nt = (3a + 2)(5b + 3) = 15ab + 9a + 10b + 6 15ab is divisible by 15. But I don't know anything about ( 9a + 10b + 6) yet. Stmnt 1: n2 is divisible by 5. From above, n  2 is just 3a. If n  2 is divisible by 5, then 'a' must be divisible by 5. So I get that 9a is divisible by 15. I still don't know anything about b. If b = 1, remainder of nt is 1. If b = 2, remainder of nt is 11 and so on... Not sufficient. Stmnt 2: t is divisible by 3. If t is divisible by 3, then (5b + 3) is divisible by 3. Therefore, b must be divisible by 3. (If this is unclear, think: 15 + 3 will be divisible by 3 but 20 + 3 will not be. If the second term is 3, the first term must also be divisible by 3 to make the whole expression divisible by 3). So 10b is divisible by 15 but we do not know anything about a. If a = 1, remainder of nt is 0, if a = 2, remainder of nt is 9. Not sufficient. Using both statements together, we know 9a and 10b are divisible by 15. So remainder must be 6. Sufficient. Answer (C).
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Re: GMAT Prep 2 remainder
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16 Mar 2011, 00:55
Bunuel wrote: 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) n2 is divisible by 5. (2) t is divisible by 3.
Answer: C. I have another approach to this ds, plz correct me if i'm wrong. n=3x + 2 t = 5y + 3 Clearly we cannot solve the problem with either n or t. We need both information concerning n and t because we need to figure out the remaining of n*t. => left with C or E (1): n2 is divisible by 5 & n=3x + 2 => x is multiple of 5. (2): t is divisible by 3. & t = 5y + 3 => y is multiple of 3 (1)& (2) => n*t = (3x+2) (5y+3) = (3x*5y) + (9x) + (10y) + 6 we know that: x is multiple of 5, y is multiple of 3 so: (3x*5y) + (9x) + (10y) + 6 will have remaining of 6 because: each (3x*5y); (9x); (10y) is all multiple of 15.
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Re: GMAT Prep 2 remainder
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17 Mar 2011, 00:34
I tried plugging numbers: n = 3k + 2 = 2,5,8,11,14,17 t = 5m + 3 = 3,8,13,18,23,28 (1) n  2 = 5l => n = 5l + 2 = 2,7,12,17 So n = 15k + 2 = 2,17,32,47 But n* t = 6 (2*3), 16(2*8) so n/15 can have rem 1 or 6 Hence (1) is not enough (2) t = 3p = 3,6,12,15,18,21 So t = 15q + 3 = 3,18,33,48 But n*t = 6, 15, so rem can be 6, 0 etc. Combining (1) and (2), it can be seen that nt = 15 * an integer + 6, so remainder is 6, answer is C.
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Re: GMAT Prep 2 remainder
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02 Oct 2011, 17:42
I did a similar thing with plugging numbers.
First you have to see that (1) and (2) alone are not sufficient alone before it really works in a time effective manner though.
(1) n = 17,32,47,etc, t still has so many values and remainder can differ (17*3 and 17*8 for example) (2) t = 18,33,48,etc, same as above but with n
(1)+(2)
(15p+2)(15p+3) will always have r6 when divided by 15
breaking it out in factored form like that is helpful for me to see it very clearly.



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Re: DS problem : remainders
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19 Oct 2013, 08:00
VeritasPrepKarishma wrote: hogwarts wrote: Saw this question on a GMATPrep test, and I can't figure out how to get to the correct answer. Can anyone help? Thanks! 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) n2 is divisible by 5. (2) t is divisible by 3. The correct answer is (C)  both statements together are sufficient, but neither statement alone is sufficient. Can anybody out there help explain to me how to get to this answer though? Thanks! This is how I would approach this question. When positive integer n is divided by 3, the remainder is 2; I say n = 3a + 2 ( a is a non negative integer) and when positive integer t is divided by 5, the remainder is 3. So t = 5b + 3 (b is a non negative integer.) What is the remainder when the product nt is divided by 15?So nt = (3a + 2)(5b + 3) = 15ab + 9a + 10b + 6 15ab is divisible by 15. But I don't know anything about ( 9a + 10b + 6) yet. Stmnt 1: n2 is divisible by 5. From above, n  2 is just 3a. If n  2 is divisible by 5, then 'a' must be divisible by 5. So I get that 9a is divisible by 15. I still don't know anything about b. If b = 1, remainder of nt is 1. If b = 2, remainder of nt is 11 and so on... Not sufficient. Stmnt 2: t is divisible by 3. If t is divisible by 3, then (5b + 3) is divisible by 3. Therefore, b must be divisible by 3. (If this is unclear, think: 15 + 3 will be divisible by 3 but 20 + 3 will not be. If the second term is 3, the first term must also be divisible by 3 to make the whole expression divisible by 3). So 10b is divisible by 15 but we do not know anything about a. If a = 1, remainder of nt is 0, if a = 2, remainder of nt is 9. Not sufficient. Using both statements together, we know 9a and 10b are divisible by 15. So remainder must be 6. Sufficient. Answer (C). so from statement 1 we got 10b+6 if b=1 we get 16 then rem =1 if b=2 we got 106 so remainder =1 if b=3 we get 1006 so remainder =1 ..............................so i think a is sufficient .............what am i doing wrong



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Re: DS problem : remainders
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20 Oct 2013, 05:38
tyagigar wrote: VeritasPrepKarishma wrote: hogwarts wrote: Saw this question on a GMATPrep test, and I can't figure out how to get to the correct answer. Can anyone help? Thanks! 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) n2 is divisible by 5. (2) t is divisible by 3. The correct answer is (C)  both statements together are sufficient, but neither statement alone is sufficient. Can anybody out there help explain to me how to get to this answer though? Thanks! This is how I would approach this question. When positive integer n is divided by 3, the remainder is 2; I say n = 3a + 2 ( a is a non negative integer) and when positive integer t is divided by 5, the remainder is 3. So t = 5b + 3 (b is a non negative integer.) What is the remainder when the product nt is divided by 15?So nt = (3a + 2)(5b + 3) = 15ab + 9a + 10b + 6 15ab is divisible by 15. But I don't know anything about ( 9a + 10b + 6) yet. Stmnt 1: n2 is divisible by 5. From above, n  2 is just 3a. If n  2 is divisible by 5, then 'a' must be divisible by 5. So I get that 9a is divisible by 15. I still don't know anything about b. If b = 1, remainder of nt is 1. If b = 2, remainder of nt is 11 and so on... Not sufficient. Stmnt 2: t is divisible by 3. If t is divisible by 3, then (5b + 3) is divisible by 3. Therefore, b must be divisible by 3. (If this is unclear, think: 15 + 3 will be divisible by 3 but 20 + 3 will not be. If the second term is 3, the first term must also be divisible by 3 to make the whole expression divisible by 3). So 10b is divisible by 15 but we do not know anything about a. If a = 1, remainder of nt is 0, if a = 2, remainder of nt is 9. Not sufficient. Using both statements together, we know 9a and 10b are divisible by 15. So remainder must be 6. Sufficient. Answer (C). so from statement 1 we got 10b+6 if b=1 we get 16 then rem =1 if b=2 we got 106 so remainder =1 if b=3 we get 1006 so remainder =1 ..............................so i think a is sufficient .............what am i doing wrong 10b above means 10*b, 10 multiplied by b. If b=2, then 10b+6=10*2+6=26 not 106; If b=2, then 10b+6=10*3+6=36 not 1006. Does this make sense?
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Re: When positive integer n is divided by 3, the remainder is 2
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11 Nov 2015, 20:57
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 = 3a + 2, t = 5b + 3. Q: (3a + 2)(5b + 3)/15 = c + r/15. What is the value of r?
r is knowable only if it can be determined that each of 3a and 5b are divisible by 15, or in other words that a is divisible by 5 and b is divisible by 3, in which case the remainder would be the only remaining element that is clearly not divisible by 15.
nt = (3a +2)(5b +3) = 3a x 5b + 3a x3 + 10b + 6
if the first three elements of this equation are divisible by 15 the remainder would be 6. If divisibility of any of the variablecontaining elements of this equation cannot be determined then the remainder cannot be determined.
St1: n2 is divisible by 5.
n = 3a + 2 3a = n2
if n2 divisible by 5, then 3a is divisible by 5 (at least one factor of a = 5), and therefore 3a is divisible by 15.
However, divisibility of 5b by 15 cannot be determined, and therefore the value of r cannot be determined.
INSUFF.
(2) t is divisible by 3.
t = 5b + 3 5b + 3 is divisible by 3, therefore 5b is divisible by 3(at least one factor of b is 3) and 5b is divisible by 15.
However, divisibility of 3a by 15 cannot be determined, and therefore the value of r cannot be determined.
INSUFF.
Combined: 3a, and 5b are both divisible by 15, and therefore, all variablecontaining elements of the equation nt = (3a +2)(5b +3) = 3a x 5b + 3a x3 + 10b + 6 are divisible by 15 and the remainder is 6.
SUFF.
Answer is C.



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Re: When positive integer n is divided by 3, the remainder is 2
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28 Feb 2016, 10:09
Hi Bunuel, thanks for the solution. I tried to solve statement two and I below is what I am getting. What am i doing wrong? T = 5q+3 (given) T = 3k (statement 2) So 5q + 3 = 3k > 10q + 6 = 6k and this does not assures that 10q+6 is a multiple of 15. What am i missing in solving it this way? Thanks in advance! Bunuel wrote: 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) n2 is divisible by 5. (2) t is divisible by 3. 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) \(n2=5m\) > \(n=5m+2=3p+2\) > \(5m=3p\), \(15m=9p\) > \(nt=15pq+ 9p+10q+6=15pq+ 15m+10q+6\). 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) n2 is divisible by 5 > \(n2=5m\) > \(n=5m+2\) and \(n=3p+2\) > general formula for \(n\) would be \(n=15k+2\) (about deriving general formula for such problems at: goodproblem90442.html#p723049 and manhattanremainderproblem93752.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. Answer: C.



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Re: When positive integer n is divided by 3, the remainder is 2
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05 May 2016, 20:29
kt00381n wrote: 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) n2 is divisible by 5. (2) t is divisible by 3. As per the stated, n = 3p + 2 => n2 = 3p i.e. n2 is a multiple of 3. t = 5q + 3 => t3 = 5q i.e. t3 is a multiple of 5. Statement 1 says n2 is a multiple of 5. Hence, n2 is divisible by both 3 and 5. So, it should be divisible by 15. => n2 = 15m => n = 15m +2.... (1) But stat 1 does not say anything about t2. Hence insufficient. (or so u can deduce) Statement 2 says t2 is a multiple of 3. Hence, t3 is divisible by both 3 and 5. So, it should be divisible by 15. => t3 = 15m => t = 15s +3.... (2) But stat 2 does not say anything about n. Hence insufficient. On combining these two statements, nt= 15m*15s + 15m*3+15s*2+3*2 So nt on dividing by 15 gives 6 as the remainer. (All other terms are divisible by 15)



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Re: When positive integer n is divided by 3, the remainder is 2
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10 Jun 2016, 12:09
kt00381n wrote: 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) n2 is divisible by 5. (2) t is divisible by 3. n=3a+2 AND t=5b+3nt = 15ab + 10b + 9a + 6  divide it by 15 Nt/15 = ab + (2/3)b + (3/5)a + 6/15. We see if we know both b and a are divisible by 3 and 5 respectively then we remainder is 6 else remainder would be something else depending on the value of a and b.Stmt1:As we know n=3a+2 then n2 = 3a. Stmt1 says n2 is divisible by 5 that is3a is divisible by 5. Hence a is divisible by 5. BUT what about b? INSUFF. Stmt2:As we know t=5b+3 Stmt2 says t is divisible by 3 that is b is divisible by 3. BUT what about a? INSUFF. Combining stmt1 and stmt2 we know a and b are divisible by 5 and 3 respectively. This is what we required to know as mentioned above. Sufficient. Answer is C.
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Re: When positive integer n is divided by 3, the remainder is 2
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26 Dec 2016, 07:32
VeritasPrepKarishma wrote: hogwarts wrote: Saw this question on a GMATPrep test, and I can't figure out how to get to the correct answer. Can anyone help? Thanks! 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) n2 is divisible by 5. (2) t is divisible by 3. The correct answer is (C)  both statements together are sufficient, but neither statement alone is sufficient. Can anybody out there help explain to me how to get to this answer though? Thanks! This is how I would approach this question. When positive integer n is divided by 3, the remainder is 2; I say n = 3a + 2 ( a is a non negative integer) and when positive integer t is divided by 5, the remainder is 3. So t = 5b + 3 (b is a non negative integer.) What is the remainder when the product nt is divided by 15?So nt = (3a + 2)(5b + 3) = 15ab + 9a + 10b + 6 15ab is divisible by 15. But I don't know anything about ( 9a + 10b + 6) yet. Stmnt 1: n2 is divisible by 5. From above, n  2 is just 3a. If n  2 is divisible by 5, then 'a' must be divisible by 5. So I get that 9a is divisible by 15. I still don't know anything about b. If b = 1, remainder of nt is 1. If b = 2, remainder of nt is 11 and so on... Not sufficient. Stmnt 2: t is divisible by 3. If t is divisible by 3, then (5b + 3) is divisible by 3. Therefore, b must be divisible by 3. (If this is unclear, think: 15 + 3 will be divisible by 3 but 20 + 3 will not be. If the second term is 3, the first term must also be divisible by 3 to make the whole expression divisible by 3). So 10b is divisible by 15 but we do not know anything about a. If a = 1, remainder of nt is 0, if a = 2, remainder of nt is 9. Not sufficient. Using both statements together, we know 9a and 10b are divisible by 15. So remainder must be 6. Sufficient. Answer (C). Very straight forward, crisp and compact solution to this problem.
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Re: When positive integer n is divided by 3, the remainder is 2
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15 Jan 2017, 01:38
Hi everyone,
Thanks for all the explanations, it´s always extremely helpful.
I got the right answer but it took me 4min 44 sec. Any suggestions to get get the time down.
Thanks



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Re: When positive integer n is divided by 3, the remainder is 2
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15 Jan 2017, 01:45
lauramo wrote: Hi everyone,
Thanks for all the explanations, it´s always extremely helpful.
I got the right answer but it took me 4min 44 sec. Any suggestions to get get the time down.
Thanks HOW TO IMPROVE YOUR QUANT SCORE?How to Go From a 48 to 51 in GMAT QuantImproving from 3035 to 4044Improving from 44 to 50Other discussions dedicated to this issue:HOW TO SPEED UP?Want to speed up? Check this: Timing Strategies on the GMATHow to Improve Your Timing on the GMATOther discussions dedicated to this issue:
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Re: When positive integer n is divided by 3, the remainder is 2
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05 Oct 2018, 09:06
kt00381n wrote: 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) n2 is divisible by 5. (2) t is divisible by 3.
(The solution presented below is VERY similar to Karishma´s approach.) Obs.: don´t be "frightened" by the manyintegerrepresentingletters below. The MEANING they provide is simple and far more important than the "heavy notation"! \(n,t\,\, \ge \,\,1\,\,{\rm{ints}}\,\,\,\left( * \right)\) \(n = 3M + 2\,\,\,\,\left( {{\mathop{\rm int}} \,\,M \ge 0\,\,{\rm{by}}\,\,\left( * \right)} \right)\) \(t = 5K + 3\,\,\,\,\left( {{\mathop{\rm int}} \,\,K \ge 0\,\,{\rm{by}}\,\,\left( * \right)} \right)\) \(nt = \underbrace {15MK}_{{\text{multiple}}\,{\text{of}}\,\,15} + 9M + 10K + 6\,\,\,\,\,\, \Rightarrow \,\,\,\,\boxed{\,\,?\,\,\,\,:\,\,\,\,{\text{remainder}}\,\,{\text{by}}\,\,15\,\,\,}\) \(\left( 1 \right)\,\,n  2 = 5J\,\,\,\left( {{\mathop{\rm int}} \,\,J \ge 0\,\,{\rm{by}}\,\,\left( * \right)} \right)\,\,\,\,\,\left\{ \matrix{ \,{\rm{Take}}\,\,\left( {n,t} \right) = \left( {2,3} \right)\,\,\,\, \Rightarrow \,\,\,\,\left( {M,K} \right) = \left( {0,0} \right)\,\,\,\, \Rightarrow \,\,\,{\rm{?}}\,\,{\rm{ = }}\,\,{\rm{6}}\,\, \hfill \cr \,{\rm{Take}}\,\,\left( {n,t} \right) = \left( {2,8} \right)\,\,\,\, \Rightarrow \,\,\,\,\left( {M,K} \right) = \left( {0,1} \right)\,\,\,\, \Rightarrow \,\,\,{\rm{?}}\,\,{\rm{ = }}\,\,{\rm{1}}\,\, \hfill \cr} \right.\) \(\left( 2 \right)\,\,t = 3G\,\,\,\left( {{\mathop{\rm int}} \,\,G \ge 1\,\,{\rm{by}}\,\,\left( * \right)} \right)\,\,\,\,\,\left\{ \matrix{ \,\left( {{\rm{Re}}} \right){\rm{Take}}\,\,\left( {n,t} \right) = \left( {2,3} \right)\,\,\,\, \Rightarrow \,\,\,\,\left( {M,K} \right) = \left( {0,0} \right)\,\,\,\, \Rightarrow \,\,\,{\rm{?}}\,\,{\rm{ = }}\,\,{\rm{6}}\,\, \hfill \cr \,{\rm{Take}}\,\,\left( {n,t} \right) = \left( {5,3} \right)\,\,\,\, \Rightarrow \,\,\,\,\left( {M,K} \right) = \left( {1,0} \right)\,\,\,\, \Rightarrow \,\,\,{\rm{?}}\,\,{\rm{ = }}\,\,{\rm{0}}\,\, \hfill \cr} \right.\) \(\left( {1 + 2} \right)\,\,\) \(\left\{ \matrix{ \,\,n  2 = 3M \hfill \cr \,\,n  2 = 5J \hfill \cr} \right.\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,n  2 = 15H\,\,\,\left( {{\mathop{\rm int}} \,\,H \ge 0\,\,{\rm{by}}\,\,\left( * \right)} \right)\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,n = 15H + 2\) \(\left\{ \matrix{ \,\,t  3 = 5K \hfill \cr \,\,t = 3G\,\,\,\, \Rightarrow \,\,\,t  3 = 3W \hfill \cr} \right.\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,t  3 = 15D\,\,\,\left( {{\mathop{\rm int}} \,\,D \ge 0\,\,{\rm{by}}\,\,\left( * \right)} \right)\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,t = 15D + 3\) \(\left\{ \matrix{ n = 15H + 2 \hfill \cr t = 15D + 3 \hfill \cr} \right.\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,?\,\,\,:\,\,\,\,\,\,nt = 15\,\underbrace {\left[ {15HD + 3H + 2D} \right]}_{{\mathop{\rm int}} } + 6\,\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\,? = 6\,\,\,\,\, \Rightarrow \,\,\,\,\,{\rm{SUFF}}.\) This solution follows the notations and rationale taught in the GMATH method. Regards, Fabio.
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Re: When positive integer n is divided by 3, the remainder is 2
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14 Mar 2019, 12:08
Hi, Here are my two cents for this question. First I would like to summarize the concept of General Numbers here. Concept of General number(A) if the reminders are same
Say n is divided by 3 reminder is 2 and when n is divided by 5 reminder is 2. What would be the general number n = 15M+2. where M>0 for M=1 , n=17, M=2 , n=32. How to derive general number Say n when divided by 5 gives remainder 2, we can represent this algebraically n= 5P+2 (I) Now if n is divided by 3 reminder is 2 which means \(\frac{5P+2 }{3}\) we can rewrite \(\frac{5P+2 }{3}\) as \(\frac{3P+2P+2 }{3}\)= \(\frac{3P}{3}\)+\(\frac{2P+2}{3}\) Now since weare given that when n is divided by 3 we get reaminder as 2 so this part \(\frac{2P+2}{3}\) yields a reminder 2. for what value of P will the reminder be 2 Say P = 3M then the part\(\frac{2P+2}{3}\) become \(\frac{2(3M)+2}{3}\) which gives us reminder as 2 So we can write equations (I) as n= 5(3M)+2 n=15M+2 (B) if the reminders are differentSay when n is divided by 11 the reminder is 7 and when n is divided by 13 reminder is 12 lets say n= 13T+12(II) when we divide this number by 11 we have reminder as 7 so which means \(\frac{13T+12 }{11}\) leaves us with reminder 7 we can rewrite \(\frac{13T+12 }{11}\) as \(\frac{11T+2T+11+1 }{7}\)= \(\frac{11(T+1)}{11}\)+\(\frac{(2T+1)}{11}\) Now since we are given that when n is divided by 11 we get remainder as 7 so this part \(\frac{(2T+1)}{11}\) yields a reminder 7. for what value of T will the reminder be 7 recall that if \(\frac{x}{y}\) reminder r is 0\(\leqslant\)r<y, Also if in case x<y then reminder is x Which means if 2T+1 = 7 then we can say that \(\frac{(2T+1)}{11}\) yields a reminder 7 So for what smallest positive value of T will 2T+1 = 7 Say T = 3 then the part\(\frac{2T+1}{11}\) become \(\frac{2(3)+1}{11}\) which gives us reminder as 7 So we can write equations (II) as n= 13(3)+12 n=51 So Smallest value of n which when divided by 11 leaves remainder 7 and when divided by 13 leaves remainder 12 is 51 From this we can say that the general number n= 11*13K+51 which when divided by 11 leaves reminder 7 and when divided by 13 leaves remainder 12 Now getting back to question. n = 3K+2 and t= 5R+3. Now the product of nt would be = 15KR+9K+10R+6. We need to find the remainder when nt is divided by 15. So here if know when n is divided by 5 and when t is divided by 3 what would be the reminder then we could certainly say what would the reminder if nt is divided by 15 Statement 1 Tells us that when n2 is divided by 5 reminder is 0 so n2= 5Q or n=5Q+2 and we have from stem that n=3K+2, which tells us that n is in the form of 15V+2 But without any information about T we cannot determine the remainder of "nt" So Insufficient Statement 2 Says that t is divisible by 3 so t= 3F and from Stem we have t= 5R+3., so general form of t=15W+3, which when divided by 3 leaves reminder 0 and when divided by 5 leaves reminder 3 Now in absence of any information about 'n' we can tell what would be the remainder when "nt" is divided by 15 hence insufficient. Now combining statements 1 and 2 we have n=15V+2 t=15W+3, we can infer from that when "nt" is divided by 15 we will have some reminder . Hence Both Statements are necessary for us to answer the Question. Probus
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Probus
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Re: When positive integer n is divided by 3, the remainder is 2
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14 Mar 2019, 12:08






