If x and y are positive integers, what is the remainder when

IMO, it is C. You must know both x and y


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Look, you know that 3^4 has its unit digit as 1. Thus, 3^4x will always have unit digit of 1, no matter what the value of x is, obviously > 0 as is stated in the question. So, 3^(4x+1) will be 3. Thus, if you know the value of y, you know the units digit of the resulting number. That number when divided by 10 will have its unit digit as the remainder.

Hope this helps.

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hi viveknegi and balhhs0e,

this is due to cyclicity..
Remainder by 10 means we are looking for last/unit's digit

every digit has cyclicity when it comes to last/unit's digit...each digit surely repeats the units digit after every 4th power..

i) 1, 5 and 6 and 0 repeat with each power..
1^1 or 1^2 will always give 1 and similarly for 5,6,0

ii) 4,9 repeat after increase of two powers..
4^1 or 4^3 or 4^5 will all leave 4 as last digit and 4^2 or 4^4 will leave 6 as last digits
so cyclicity is 4,6,4,6..
for 9 it is 9,1,9,1,9...

iii) remaining 2,3,7,8 repeat after every 4th power
so 2^1 , 2^(1+4), 2^(1+4+4) will leave 2 in each case
cyclicity is 2,4,8,6,2,4,8,6..
for 3 it is 3,9,7,1,3,9,..
for 7 it is 7,9,3,1,7,9,3,1...
for 8 it is 8,4,2,6,8,4,2,6..

Back to question
so 3^(4x+1) will always be same as 3^1 or 3 when x is positive integer

that is why 3^(4x+1)+y will depend on y for units digit ..
B gives you value of y hence sufficient

=>

Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.
The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question.
\(3^1 ~ 3^5 ~ 3^9 ~ … ~ 3\) : Integers of the form \(3^{4x+1}\) always have the remainder of \(3\) when they are divided by \(10\).
\(3^2 ~ 3^6 ~ 3^{10} ~ … ~ 9\) : Integers of the form \(3^{4x+2}\) always have the remainder of \(9\) when they are divided by \(10\).
\(3^3 ~ 3^7 ~ 3^{11} ~ … ~ 7\) : Integers of the form \(3^{4x+3}\) always have the remainder of \(7\) when they are divided by \(10\).
\(3^4 ~ 3^8 ~ 3^{12} ~ … ~ 1\) : Integers of the form \(3^{4x}\) always have the remainder of \(1\) when they are divided by \(10\).
Therefore, the remainder when \(3^{4x + 1} + y\) is divided by \(10\) depends only on the value of \(y\).

Only condition 2) gives us a value for \(y\).
Therefore, B is the answer.
Answer: B

B is the correct answer.

we can not have a unique solution from A as we don't know the value of y.
but in option B, we have the value of y and value of x is not relevant as prefix 4 is there with 4x . since the cyclicity of 3 is 4, irrespective of value of x, we will always get the same unit value in the given equation.

hence option B is correct

If \(x\) and \(y\) are positive integers, what is the remainder when \(3^{4x+1}+y\) is divided by \(10\)?

Cyclicity of 3 = 3, 9, 7, 1

\(1) x=2\)

Not sufficient; we don't know the value of y.

\(2) y=3\)

4x + 1 is always 1 more than a multiple of 4; thus, the units digit of \(3^{4x+1}\) is 3.

3 + 3 = 6
6/10 = remainder 6. Sufficient.

Answer is B.