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

This solution is only detailed for better understanding. Hope it helps....

n = 3a + 2
t = 5b + 3
What is the remainder when nt/15?

1) n-2 is divisible by 5
2) t is divisible by 3

1) n - 2 is divisible by 5.

n = 3a + 2
n - 2 = 3a
3a is divisible by 5 and therefore by 15

nt = 3a(5b + 3) + 2(5b + 3)
=15ab + 9a + 10b + 6
15ab and 9a are divisible by 15,
10b: no info from (1)
INSUFFICIENT

2) t = 5b + 3,
so 5b is divisible by 3 and therefore, by 15.

nt = 15ab + 9a + 10b + 6
10b is divisible by 15
15ab and 9a: no info from (2)
INSUFFICIENT

When you combine (1) and (2)...
15ab, 9a, and 10b are divisible by 15.
Therefore, nt = 15ab + 9a + 10b + 6
=15(x) + 6

C.

Here is one other way of solving the problem.

It is given in the question stem -

n when divided by 3 leaves a remainder 2 ==> n-2 is divisible by 3.
t when divided by 5 leaves a remainder 3 ==> t-3 is divisible by 5.

Question being asked is what is the remainder when nt is divided by 15.

Clue 1 ==> it is given n-2 is divisible by 5 and from given information, we know n-2 is also divisible by 2 ==> n-2 is divisible by 15. But nothing is mentioned abt t. Hence insufficient clue.

Clue 2 ==> it is given t is divisible by 3, we can derive that t-3 is also divisible by 3. From the given information, we know t-3 is divisble by 5 ==> t-3 is divisible by 3 and 5 ==> t-3 is divisible by 15. Nothing is mentioned abt n. Hence insufficient clue.

Combine both the clues

n-2 = 15a ==> n = 15a + 2
t-3 = 15b ==> t = 15b + 3

nt = (15a + 2)(15b + 3) ==> 15.15.a.b + 15.3.a + 15.2.b + 6

Clearly first 3 terms are divisible by 15. So the remainder should be 6.

Hence both the clues are sufficient to say the remainder of nt.

I think plugging numbers is the easiest way to do this.

This question is about pattern recognition.
By listing out the possible solutions for each, you can see a definite pattern.

Possible Solutions for n given stem- 2, 5,8,11,14,17,20,23,26,29,32,35,38,41...
Possible solutions for t given stem- 3,8,13,18,23,28,33,38,43,48,53,58...

I. Using n-2 is divisble by 5, you get n=17,32,47..., notice the difference is 15, meaning each remainder will be the same when multiplying (Note that it is important that they specify integers)
However this gives us no indication as to the remainder of t - Insuff

II Using t is divisible by 3, you get t=3,18,33,48... again, the difference is 15, meaning each remainder will be the same.
No indication of n- insuff

You know that each remainder will be the same (in this case 6), C is suff since each individual number when divided by 15 will give the same remainder.

Thoughts?

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.

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.

Here's more of a plain English explanation, for those who like that sort of thing.

An important principle here is that you can multiply remainders. For instance, 10/7 = 1 r2 and 9/7 = 1 r3. 10*9/7 = 12 r6. See? Remainder 2 * remainder 3 = remainder 6.

Notice that if the resulting remainder is greater than the divisor, it wraps around again. 10/4=2 r 2 and 15/4=3 r 3. 10*15/4 should be remainder 6, but since 4 goes into 6, there is only 2 left. The actual result is 17 r2.

So, back to our problem. Using the prompt and statement 1, we know that both n/3 and n/5 have remainders of 2. We also know that t/5 has a remainder of 3.

So, if we just wanted to know the remainder when nt is divided by 5, we could multiply our remainders: 2 for n and 3 for t = 6. Since 5 goes into 6, we would be left with a remainder of 1.

However, since we’re dealing with 15, it’s more complicated. We know the remainder when n is divided by 15. Since n must be 2 more than a multiple of 3 and 2 more than a multiple of 5, it will also be 2 more than a multiple of 15:
17, 32, 47 . . .

We don’t know about t, though, unless we bring in statement 2. Once we do, we know that it is not only 3 more than a multiple of 5, but also an exact multiple of 3.

Out of our original list (3, 8, 13, 18, 23, 28, 33 . . .basically, every number that ends in 8 or 3), this leaves 3, 18, 33, 48 . . . i.e., 3 more than a multiple of 15.

Now we can multiply our remainders. n/15 has a remainder of 2 and t/15 has a remainder of 3, so nt/15 has a remainder of 6.

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?

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?

n = 3a + 2, t = 5b + 3. Q: (3a + 2)(5b + 3)/15 = c + r/15. What is the value of r?

r is knowable only if it can be determined that each of 3a and 5b are divisible by 15, or in other words that a is divisible by 5 and b is divisible by 3, in which case the remainder would be the only remaining element that is clearly not divisible by 15.

nt = (3a +2)(5b +3) = 3a x 5b + 3a x3 + 10b + 6

if the first three elements of this equation are divisible by 15 the remainder would be 6. If divisibility of any of the variable-containing elements of this equation cannot be determined then the remainder cannot be determined.



St1: n-2 is divisible by 5.

n = 3a + 2
3a = n-2

if n-2 divisible by 5, then 3a is divisible by 5 (at least one factor of a = 5), and therefore 3a is divisible by 15.

However, divisibility of 5b by 15 cannot be determined, and therefore the value of r cannot be determined.

INSUFF.

(2) t is divisible by 3.

t = 5b + 3
5b + 3 is divisible by 3, therefore 5b is divisible by 3(at least one factor of b is 3) and 5b is divisible by 15.

However, divisibility of 3a by 15 cannot be determined, and therefore the value of r cannot be determined.

INSUFF.


Combined: 3a, and 5b are both divisible by 15, and therefore, all variable-containing elements of the equation nt = (3a +2)(5b +3) = 3a x 5b + 3a x3 + 10b + 6 are divisible by 15 and the remainder is 6.

SUFF.

Answer is C.

n=3a+2 AND t=5b+3

nt = 15ab + 10b + 9a + 6 ---- divide it by 15

Nt/15 = ab + (2/3)b + (3/5)a + 6/15. We see if we know both b and a are divisible by 3 and 5 respectively then we remainder is 6 else remainder would be something else depending on the value of a and b.

Stmt-1:
As we know n=3a+2 then n-2 = 3a.

Stmt-1 says n-2 is divisible by 5 that is3a is divisible by 5. Hence a is divisible by 5. BUT what about b? INSUFF.

Stmt-2:

As we know t=5b+3

Stmt-2 says t is divisible by 3 that is b is divisible by 3. BUT what about a? INSUFF.

Combining stmt-1 and stmt-2 we know a and b are divisible by 5 and 3 respectively. This is what we required to know as mentioned above. Sufficient.

Answer is C.

Hi everyone,

Thanks for all the explanations, it´s always extremely helpful.

I got the right answer but it took me 4min 44 sec. Any suggestions to get get the time down.

Thanks

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(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.

Hi,

Here are my two cents for this question.

First I would like to summarize the concept of General Numbers here.

Concept of General number

(A) if the reminders are same

Say n is divided by 3 reminder is 2 and when n is divided by 5 reminder is 2.

What would be the general number n = 15M+2. where M>0 for M=1 , n=17, M=2 , n=32.

How to derive general number

Say n when divided by 5 gives remainder 2, we can represent this algebraically n= 5P+2 -------(I)

Now if n is divided by 3 reminder is 2 which means \(\frac{5P+2 }{3}\)

we can rewrite \(\frac{5P+2 }{3}\) as \(\frac{3P+2P+2 }{3}\)= \(\frac{3P}{3}\)+\(\frac{2P+2}{3}\)

Now since weare given that when n is divided by 3 we get reaminder as 2 so this part \(\frac{2P+2}{3}\) yields a reminder 2.

for what value of P will the reminder be 2 Say P = 3M then the part\(\frac{2P+2}{3}\) become \(\frac{2(3M)+2}{3}\) which gives us reminder as 2

So we can write equations (I) as
n= 5(3M)+2
n=15M+2

(B) if the reminders are different

Say when n is divided by 11 the reminder is 7 and when n is divided by 13 reminder is 12

lets say n= 13T+12-----------(II)

when we divide this number by 11 we have reminder as 7

so which means \(\frac{13T+12 }{11}\) leaves us with reminder 7

we can rewrite \(\frac{13T+12 }{11}\) as \(\frac{11T+2T+11+1 }{7}\)= \(\frac{11(T+1)}{11}\)+\(\frac{(2T+1)}{11}\)

Now since we are given that when n is divided by 11 we get remainder as 7 so this part \(\frac{(2T+1)}{11}\) yields a reminder 7.

for what value of T will the reminder be 7

recall that if \(\frac{x}{y}\) reminder r is 0\(\leqslant\)r<y,
Also if in case x<y then reminder is x

Which means if 2T+1 = 7 then we can say that \(\frac{(2T+1)}{11}\) yields a reminder 7
So for what smallest positive value of T will 2T+1 = 7 Say T = 3 then the part\(\frac{2T+1}{11}\) become \(\frac{2(3)+1}{11}\) which gives us reminder as 7

So we can write equations (II) as
n= 13(3)+12
n=51
So Smallest value of n which when divided by 11 leaves remainder 7 and when divided by 13 leaves remainder 12 is 51

From this we can say that the general number n= 11*13K+51 which when divided by 11 leaves reminder 7 and when divided by 13 leaves remainder 12


Now getting back to question.
n = 3K+2 and t= 5R+3. Now the product of nt would be = 15KR+9K+10R+6. We need to find the remainder when nt is divided by 15.

So here if know when n is divided by 5 and when t is divided by 3 what would be the reminder then we could certainly say what would the reminder if nt is divided by 15

Statement 1 Tells us that when n-2 is divided by 5 reminder is 0
so n-2= 5Q or n=5Q+2 and we have from stem that n=3K+2,
which tells us that n is in the form of 15V+2
But without any information about T we cannot determine the remainder of "nt"
So Insufficient

Statement 2 Says that t is divisible by 3 so t= 3F and from Stem we have t= 5R+3., so general form of t=15W+3, which when divided by 3 leaves reminder 0 and when divided by 5 leaves reminder 3
Now in absence of any information about 'n' we can tell what would be the remainder when "nt" is divided by 15

hence insufficient.

Now combining statements 1 and 2 we have
n=15V+2
t=15W+3,

we can infer from that when "nt" is divided by 15 we will have some reminder .

Hence Both Statements are necessary for us to answer the Question.


Probus

I tried this way:
Upfront from the question stem:

According to the question stem,

n can be = 2, 5, 8, 11, 14, 17,...etc.

And

t can be = 3, 8, 13, 18, .... etc.

__

Statement 1:
Find the values of n (from above list) which satisfy this condition.

So n can be 2, 17, ... etc.
We dont know anything about t.
INSUFFICIENT.
__

Statement 2:
Same task. find the values of t (from the above list) which satisfy this condition.

So t can be 3, 18, etc.
We don't know anything about n.
INSUFFICIENT.

_

Together,
We know,
n = 2, 17 ...etc
t = 3, 18 ...etc.

So nt = 6, 36, 51 .... etc.
All gives remainder 6 if divided by 15.
This a reliable pattern. Reminder questions follows pattern.

Sufficient.

Answer is (C).
_____

Posted from my mobile device

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

This one is not supposed to be solved i two minutes, right?
even if I have a general idea of to solve this type of qns, it still takes 3-4 minutes, isn't it?
please share your timing on this one