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; 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.

OA C

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by chetan.sharma » Fri Oct 05, 2018 5:09 am

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BTGmoderatorDC 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.

OA C

Source: GMAT Prep
So n=3x+2 and t=5y+3
nt=(3x+2)(5y+3)=15xy+10y+9x+6
Now 15x is div by 15, so we have to check for 10y+9x+6
1) n-2 is div by 5
So 3x is div by 5 and thus 9x will be div by 15
But we cannot say anything about 10y..
It can leave a remainder of 5, or 10 or 0
Insufficient
2) t is divisible by 3.
5y+3 is div by 3 MEANS y is div y 3
So 10y is div by 15 but nothing about 9x
Insufficient

Combined
15xy and 10y and 9x are divisible by 15, so remainder is 6
Sufficient

C

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by fskilnik@GMATH » Fri Oct 05, 2018 7:56 am

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BTGmoderatorDC 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.

Source: GMAT Prep
Obs.: don´t be "frightened" by the many-integer-representing-letters 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.

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Fabio.
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by swerve » Sat Oct 06, 2018 11:00 am

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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 the remainder is 6, the answer is C.

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BTGmoderatorDC wrote:
Fri Oct 05, 2018 1:04 am
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.

OA C

Solution:

We see that n could be integers such as 2, 5, 8, 11, etc. and t could be 3, 8, 13, 18, etc. In other words, n = 3m + 2 and t = 5s + 3 for some non-negative integers m and s.

Statement One Only:

n - 2 is divisible by 5.

This means n could be integers such as 2, 17, 32, etc. In other words, n = 15k + 2 for some non-negative integer k. Since t = 5s + 3, we have:

nt = (15k + 2)(5s + 3) = 75ks + 45k + 10s + 6

Since the first two terms are divisible by 15, the remainder when nt is divided by 15 is determined by the last two terms, i.e., 10s + 6. We see that if s = 0, the remainder is 6. However, if s = 1, the remainder is 1 (note: 10(1) + 6 = 16 and 16/15 = 1 R 1). Statement one alone is not sufficient.

Statement Two Only:

t is divisible by 3.

This means t could be integers such as 3, 18, 33, etc. In other words, t = 15r + 3 for some non-negative integer r. Since n = 3m + 2, we have:

nt = (3m + 2)(15s + 3) = 45ks + 30s + 9m + 6

Since the first two terms are divisible by 15, the remainder when nt is divided by 15 is determined by the last two terms, i.e., 9m + 6. We see that if m = 0, the remainder is 6. However, if m = 1, the remainder is 0 (note: 9(1) + 6 = 15 and 15/15 = 1 R 0). Statement two alone is not sufficient.

Statements One and Two Together:

Since n = 15k + 2 and t = 15r + 3, we have:

nt = (15k + 2)(15s + 3) = 225ks + 45k + 30s + 6

Since the first three terms are divisible by 15, the remainder when nt is divided by 15 is determined by the last term, i.e., 6. In other words, regardless of what the values of k and s are, the remainder will be always 6. The two statements together are sufficient.

Answer: C

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