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If a and b are positive integers, what is the remainder when
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26 Feb 2012, 15:30
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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 Ok  this is how I am trying to solve this.
Statement 1
a = 1. Does not tell anything about b therefore is insufficient on its own to answer the question.
Statement 2
b = 2
2a + 1 + b becomes
2a (even) + 1 (odd) + b (even) = ODD. So the exponent to 4 is ODD. So I understand that if we put 3, 5 etc I get the remainder 4, but why can't I put exponent as 1 as 1 is ODD too. Can you please help?
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Re: Remainder when divided by 10
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26 Feb 2012, 15:58
If a and b are positive integers, what is the remainder when \(4^{2a+1+b}\) is divided by 10?This is a classic "C trap" question: "C trap" is a problem which is VERY OBVIOUSLY sufficient if both statements are taken together. When you see such question you should be extremely cautious when choosing C for an answer. Back to the question: 4 in positive integer power can have only 2 last digits: 4, when the power is odd or 6 when the power is even. Hence, to get the remainder of 4^x/10 we should know whether the power is odd or even: if it's odd the remainder will be 4 and if it's even the remainder will be 6. (1) a = 1 > \(4^{2a+1+b}=4^{3+b}\) depending on b the power can be even or odd. Not sufficient. (2) b = 2 > \(4^{2a+1+b}=4^{2a+3}=4^{even+odd}=4^{odd}\) > the remainder upon division of \(4^{odd}\) by 10 is 4. Sufficient. Answer: B. enigma123 wrote: 2a (even) + 1 (odd) + b (even) = ODD. So the exponent to 4 is ODD. So I understand that if we put 3, 5 etc I get the remainder 4, but why can't I put exponent as 1 as 1 is ODD too. Can you please help? The power of 4 is \(2a+3\) and since \(a\) is a positive integer then the lowest value of \(2a+3\) is 5, for \(a=1\). Next, even if the power were 1 then 4^1=4 and the remainder upon division of 4 by 10 would still be 4. Hope it's clear.
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Re: Remainder when divided by 10
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26 Feb 2012, 15:39
powers of 4 go like this:
The unit place is: 1 = 4 2 = 6 3 = 4 4 = 6
So all even exponents have 6 in unit place, and all off exponents have 4 in unit place. To solve the problem we need to find whether the 2a + 1 + b is even or odd
As a is +ve integer, 2a is always even. 2a + 1 will be odd. Now to determine whether (2a + 1 + b) is even or odd, we need to know only b.
Therefore, the answer is B




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Re: If a and b are positive integers, what is the remainder when
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11 Aug 2013, 08:20
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
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03 Sep 2015, 03:48
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
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27 Jan 2017, 09:38
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
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20 Jun 2017, 19:55
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
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Re: If a and b are positive integers, what is the remainder when
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27 Aug 2018, 11:23
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
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27 Aug 2018, 12:53
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 Attachment:
1535403274897.jpg [ 31.4 KiB  Viewed 1329 times ]
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Re: If a and b are positive integers, what is the remainder when
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28 Aug 2018, 03:07
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
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28 Aug 2018, 03:34




Re: If a and b are positive integers, what is the remainder when
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28 Aug 2018, 03:34






