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When positive integer n is divided by 3, the remainder is 2

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When positive integer n is divided by 3, the remainder is 2 [#permalink] New post 01 Nov 2009, 12:06
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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) n-2 is divisible by 5.
(2) t is divisible by 3.
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Re: GMAT Prep 2 remainder [#permalink] New post 01 Nov 2009, 13:07
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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) n-2 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) 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.

Answer: C.
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Re: GMAT Prep 2 remainder [#permalink] New post 24 May 2010, 16:52
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

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Re: DS problem : remainders [#permalink] New post 28 Nov 2010, 17:36
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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) n-2 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: n-2 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 [#permalink] New post 15 Mar 2011, 23:55
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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) n-2 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): n-2 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 [#permalink] New post 16 Mar 2011, 23:34
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 [#permalink] New post 02 Oct 2011, 16:42
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: When positive integer n is divided by 3, the remainder is 2 [#permalink] New post 07 Oct 2013, 20:15
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Re: DS problem : remainders [#permalink] New post 19 Oct 2013, 07:00
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) n-2 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: n-2 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 [#permalink] New post 20 Oct 2013, 04:38
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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) n-2 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: n-2 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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NEW TO MATH FORUM? PLEASE READ THIS: ALL YOU NEED FOR QUANT!!!

PLEASE READ AND FOLLOW: 11 Rules for Posting!!!

RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders; 11. GMAT Prep Software Analysis NEW!!!; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) NEW!!!; 12. Tricky questions from previous years. NEW!!!;

COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: DS problem : remainders   [#permalink] 20 Oct 2013, 04:38
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