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) n-2 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 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.
_________________
Fabio Skilnik ::
GMATH method creator (Math for the GMAT)
Our high-level "quant" preparation starts here:
https://gmath.net