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Re: If a and b are positive integers, what is the remainder when [#permalink]
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enigma123 wrote:
If a and b are positive integers, what is the remainder when \(4^{2a+1+b}\) is divided by 10?

(1) a = 1
(2) b = 2


We need to determine the remainder when 4^(2a+1+b) is divided by 10. Since the remainder when an integer is divided by 10 is always equal to the units digit of that integer, we need to determine the units digit of 4^(2a+1+b).

We can also simplify 4^(2a+1+b):

4^(2a+1+b) = (4^2a)(4^1)(4^b)

Let’s now evaluate the pattern of the units digits of 4^n for positive integer values of n. That is, let’s look at the pattern of the units digits of powers of 4. When writing out the pattern, notice that we are ONLY concerned with the units digit of 4 raised to each power.

4^1 = 4

4^2 = 6

4^3 = 4

4^4 = 6

The pattern of the units digit of powers of 4 repeats every 2 exponents. The pattern is 4–6. In this pattern, all exponents that are odd will produce a 4 as its units digit and all exponents that are even will produce a 6 as its units digit. Thus:

4^1 has a units digit of 4 and 4^2a has a units digit of 6.

Thus, to determine the units digit of (4^2a)(4^1)(4^b), we only need the value of b.

Statement One Alone:

a = 1

Since we do not have the value of b, statement one alone is not sufficient to answer the question.

Statement Two Alone:

b = 2

Since we have the value of b, statement two is sufficient to determine the remainder when (4^2a)(4^1)(4^b) is divided by 10.

Answer: B
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Re: If a and b are positive integers, what is the remainder when [#permalink]
REM(4^(2a+1+b))/10

Means we have to find last digit of the expression.So rephrasing the question

What is the last digit of 4^(2a+1+b)

(1).
a=1

Break the expression as 4^2a * 4^1 * 4^b.

b is unknown hence INSUFFICIENT

(2).

b=2

4^2a * 4^1 * 4^b.

If you can observe the expression 4^2a, you will see that this will always give last digit as '6' you can try out numbers if you want.

So knowing the expression and value of b last digit can be calculated and hence the remainder can also be calculated.

Hence (B) it is !!
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If a and b are positive integers, what is the remainder when [#permalink]
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Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and equations ensures a solution.


If a and b are positive integers, what is the remainder when 4 2a+1+b is divided by 10?

(1) a = 1
(2) b = 2


Transforming the original condition and the question, 4^(2a+1+b)=(4^2a)(4^(1+b))=(16^a)(4^(1+b))=(.......6)(4^(1+b)), because the first digit is always 6 when it's multiplied by 6. Since b is all we need to know, the answer is B.
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Re: If a and b are positive integers, what is the remainder when [#permalink]
Imo C
Got it wrong on my first attempt .
We require only Statement 2 4^3+2a on division by 10 will give us a remainder of 4 , as the power will always be odd.
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If a and b are positive integers, what is the remainder when [#permalink]
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If a and b are positive integers, what is the remainder when \(4^{2a+1+b}\) is divided by 10?

(1) a=1

(2) b=2
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Re: If a and b are positive integers, what is the remainder when [#permalink]
carcass wrote:
If a and b are positive integers, what is the remainder when \(4^{2a+1+b}\) is divided by 10?

(1) a=1

(2) b=2



we are looking for remainder of \(4^{2a+1+b}\) when divided by 10.

Statement 1: NOT sufficient. why?

a = 1.

2a = all time even. we don't know anything about b whether it's even or odd. Surely b has great impact on this question and remainder will differ based on the value of b.

Statement 2: b=2

the problem we have on statement 1 is solved here. a has no impact as it's a part of 2a, all time even.

Say , a = 1 b = 2.

\(4^{ 2*1 + 1 + 2}\)

\(4^4\)

\(\frac{64}{10}\)

= 6*10 + 4

Remainder is 4. and this time no change in remainder . In all cases , i hope, it's true.

The best answer is B.
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Re: If a and b are positive integers, what is the remainder when [#permalink]
carcass wrote:
If a and b are positive integers, what is the remainder when \(4^{2a+1+b}\) is divided by 10?

(1) a=1

(2) b=2
Answer is B
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Re: If a and b are positive integers, what is the remainder when [#permalink]
The remainder of any integer when divided by 10 is the integer's Unite Digit.
The unit digit of 4 to the power of n depends of ODD or Even nature of n
if n=ODD, then unite digit will be 4
if n=EVEN, then unite digit will be 6

Here we need to find whether 2a+b+1 is ODD or EVEN
Because 2a is Even and 1 is ODD then sum of them will be ODD, therefore we need to find whether b is ODD or EVEN

Statement 2 is sufficient
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Re: If a and b are positive integers, what is the remainder when [#permalink]
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carcass wrote:
If a and b are positive integers, what is the remainder when \(4^{2a+1+b}\) is divided by 10?

(1) a=1

(2) b=2

________________________
Merging topics.
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Re: If a and b are positive integers, what is the remainder when [#permalink]
enigma123 wrote:
If a and b are positive integers, what is the remainder when \(4^{2a+1+b}\) is divided by 10?

(1) a = 1
(2) b = 2


cycles of 4^1,2,3,4={4,16,64,256..}
4^n/10 has remainder 4 or 6

2a=even, 1=odd, b=odd/even
2a+1=even+odd=odd; and,
odd+b=odd+odd=even; or,
odd+b=odd+even=odd

ans (B)
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Re: If a and b are positive integers, what is the remainder when [#permalink]
Cyclicity of 4 = {4, 6}

(1) a = 1

Gives us no information on B; insufficient.

(2) b = 2

We know that 2a + 1 will always be odd. Odd + 2 = Odd. This statement tells us that the units digit will be 4.
4 / 10 = remainder 4.

Statement 2 is sufficient. Answer is B.
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Re: If a and b are positive integers, what is the remainder when [#permalink]
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