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akurathi12
GMAT 1: 490 Q47 V13
Posts: 111
06 Jan 2019, 04:25
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
55%
(hard)
Question Stats:
64% (02:06) correct
36%
(02:13)
wrong
based on 215
sessions
History
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Time
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What is the remainder when a positive integer t is divided by 15?
(1) When t is divided by 45, the remainder is 13.
(2) When t is divided by 10, the remainder is 3 and when t is divided by 6, the remainder is 1.
(1) When t is divided by 45, the remainder is 13.
(2) When t is divided by 10, the remainder is 3 and when t is divided by 6, the remainder is 1.
Updated on: 06 Jan 2019, 07:09
Originally posted by Dare Devil on 06 Jan 2019, 07:08.
Last edited by Dare Devil on 06 Jan 2019, 07:09, edited 1 time in total.
Last edited by Dare Devil on 06 Jan 2019, 07:09, edited 1 time in total.
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What is the remainder when a positive integer t is divided by 15?
(1) When t is divided by 45, the remainder is 13.
t%45 = 13
But, 15*3 is 45, so all the numbers divisible by 45 are divisible by 15
After dividing by 45 reminder is 13 < 15, so even after dividing by 15 reminder would be 13
As reminder < quotient
Sufficient
(2) When t is divided by 10, the remainder is 3 and when t is divided by 6, the remainder is 1.
t%10 = 3
So, t can take values 13,23,33,43,53,63,73,.......
t%6 = 1
So, t can take values 7,13,19,25,...43....73.....
Common among two are 43,73.....
43%15 = 13
73%15 = 13
..........
Sufficient
Option D is correct
(1) When t is divided by 45, the remainder is 13.
t%45 = 13
But, 15*3 is 45, so all the numbers divisible by 45 are divisible by 15
After dividing by 45 reminder is 13 < 15, so even after dividing by 15 reminder would be 13
As reminder < quotient
Sufficient
(2) When t is divided by 10, the remainder is 3 and when t is divided by 6, the remainder is 1.
t%10 = 3
So, t can take values 13,23,33,43,53,63,73,.......
t%6 = 1
So, t can take values 7,13,19,25,...43....73.....
Common among two are 43,73.....
43%15 = 13
73%15 = 13
..........
Sufficient
Option D is correct
06 Jan 2019, 07:09
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akurathi12
Very beautiful problem, akurathi12 ... Congrats and kudos!
\(t \ge 1\,\,{\mathop{\rm int}} \,\,\left( * \right)\)
\(t = 15M + R\)
\(M\mathop \ge \limits^{\left( * \right)} 0\,\,{\mathop{\rm int}} \,\,,\,\,\,0 \le R\,\,{\mathop{\rm int}} \,\, \le 14\)
\(? = R\)
\(\left( 1 \right)\,\,t = 45K + 13 = 15\left( {3K} \right) + 13\,\,,\,\,K\mathop \ge \limits^{\left( * \right)} 0\,\,{\mathop{\rm int}} \,\,\,\, \Rightarrow \,\,\,\,\left\{ \matrix{\\
\,M = 3K \hfill \cr \\
\,R = 13\,\,\,\,\, \Rightarrow \,\,\,\,\,{\rm{SUFF}}. \hfill \cr} \right.\)
\(\left( 2 \right)\,\,\left\{ \matrix{\\
\,t = 10J + 3\,\,\,,\,\,\,J\mathop \ge \limits^{\left( * \right)} 0\,\,{\mathop{\rm int}} \,\,\,\,\,\mathop \Rightarrow \limits^{ \cdot \,\,3} \,\,\,\,\,\,3t = 30J + 9 \hfill \cr \\
t = 6L + 1\,\,\,,\,\,\,L\mathop \ge \limits^{\left( * \right)} 0\,\,{\mathop{\rm int}} \,\,\,\,\,\mathop \Rightarrow \limits^{ \cdot \,5} \,\,\,\,\,\,5t = 30L + 5 \hfill \cr} \right.\,\,\,\,\,\,\,\,\mathop \Rightarrow \limits^{\left( - \right)} \,\,\,\,\,2t = 30\left( {L - J} \right) - 4\)
\(2t = 30\left( {L - J} \right) - 4\,\,\,\,\, \Rightarrow \,\,\,\,\,t = 15\left( {L - J} \right) - 2 = 15\left( {L - J - 1} \right) + 15 - 2\,\,\,\,\, \Rightarrow \,\,\,\,\left\{ \matrix{\\
\,M = L - J - 1 \hfill \cr \\
\,R = 13\,\,\,\,\,\, \Rightarrow \,\,\,\,\,{\rm{SUFF}}. \hfill \cr} \right.\)
The correct answer is therefore (D).
This solution follows the notations and rationale taught in the GMATH method.
Regards,
Fabio.
