can anyone solve this problem with a short trick?

You can use trail and error method.
The remainder is 1 when divided by 5, thus the number should be greater than multiple of 5 by 1.
5*1 = 5 + 1 = 6. On dividing with 5 gives remainder 1, but is perfectly divisible by 2. This does not satisfy the condition.
5*2 = 10 + 1 = 11. And then divide the number by 3 to see if you get the remainder as 2. This satisfies the condition.

When the numbers are bigger, this method might be time consuming.

is option A(0) is the right answer?

Let's try to solve n first. It is given that when n is divided by 3, we get 2 as remainder and also that n-2 must be divisible by 5. Start looking at the table of 3, we can see that n can be 17, since 3*5 = 15 and when remainder is added, we get 17, plus n-2 = 15 is divisible by 5.
For t, start with the table of 3, and we can see that when t=18, this is divisible by 3 and when it is divided be 5, you get a remainder of 3.
Hence, n=17 and t=18.
Thus, nt = 17*18 = 306.
On dividing it with 15, you should get 6 as remainder.
Please check if the options you've typed are correct.

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thanks! I got your point , but for T I solve it as
t =3 , 3/3----> fully divisible and 3/5---> remainder is 3

so 17*3=51
51/15 generates a remainder 6
once again thanks for your productive response

its easy to understand this rule, just one question, when we should apply this? because questions are really different every time, so how to recognize where this rule will be applicable?

Given that when n is divided by 3, the remainder is 2 and when n is divided by 5, the remainder is 1. And we need to find the least possible value of n

Theory: Dividend = Divisor*Quotient + Remainder

n is divided by 3, the remainder is 2

n -> Dividend
3 -> Divisor
a -> Quotient (Assume)
2 -> Remainders
=> n = 3*a + 2 = 3a + 2

n is divided by 5, the remainder is 1

=> n = 5b + 1 (Assume b is the quotient)

So, that value of n is possible which will satisfy both the conditions
=> n = 5b + 1 = 3a + 2
=> 5b = 3a + 2 - 1 = 3a + 1
=> b = \(\frac{3a+1}{5}\)
So, for b to be integer 3a + 1 should be a multiple of 5
a = 1, 3a + 1 = 3*1 + 1 = 4
a = 2, 3a + 1 = 3*2 + 1 = 7
a = 3, 3a + 1 = 3*3 + 1 = 10 => POSSIBLE

=> n = 3a + 2 = 3*3 + 2 = 11

So, Answer will be 11
Hope it helps!

Watch the following video to learn the Basics of Remainders