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

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.

Regards,
Fabio.

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.