cumulonimbus wrote:
VeritasPrepKarishma wrote:
Alterego wrote:
What is the remainder when \(43^{86}\) is divided by 5?
A) 0
B) 1
C) 2
D) 3
E) 4
Please provide a detail explanation on how you achieved the correct answer.
Thanks
First check out the post on binomial on this link:
http://www.veritasprep.com/blog/2011/05 ... ek-in-you/Now the question will take you 15 secs.
\(43^{86} = (40 + 3)^{86}\)
Since 40 is completely divisible by 5, you only have to think about \(3^{86}\)
\(3^{86} = 9^{43} = (10 - 1)^{43}\)
Again, 10 is completely divisible by 5 so we only need to worry about (-1). The remainder will be \((-1)^{43} = -1\) which means the remainder is 5 - 1 = 4
If you are uncomfortable with negative remainders, check this post:
http://www.veritasprep.com/blog/2011/05 ... emainders/Hi Karishma/Bunnel,
Please help with this problem:
Find the unit's digit of 47^(73^80)
Here 73^80 is number with unit's digit 1.
There fore a number ending in 7 when raised to the power a number ending in 1 - the units digit will vary
for example -
unit's digit of xy7^11 = 3,
unit's digit of xy7^21 = 7,
unit's digit of xy7^31 = 3,
however the explanation is as shown in the file attached.
I didnt get the part "If the remainder is 1, that means we will take the first term of the cycle of 7^x.
Another similar problem : 28^(43^20)
here according to me units digit will vary according to the power:
unit's digit of xy8^11 = 2,
unit's digit of xy8^21 = 8,
Please guide.
This question uses two related but different concepts - cyclicity and remainders
What is the last digit of \(47^5\)?
You know that the last digit is obtained by focusing on the last digit on the number i.e. 7.
7 has a cyclicity of 7, 9, 3, 1
47^1 = 47
47^2 = ...9
47^3 = ....3
47^4 = ....1
47^5 = ....7
47^6 = ....9
47^7 = .....3
47^8 = ....1
and so on
So the last digit depends on the remainder obtained when the power is divided by 4. If the power is a multiple of 4, the last digit will always be 1. If the power is 1 more than a multiple of 4 (remainder 1 when power is divided by 4), the last digit will be 7. If the power is 2 more than a multiple of 4, the last digit will be 9. If the power is 3 more than a multiple of 4, the last digit will be 3.
The concept remains the same here: \(47^x\) where \(x = 73^{80}\)
The last digit depends on whether x is divisible by 4 or not. If not, what is the remainder?
So basically we have to take two steps:
Step I: Find the remainder when x is divided by 4.
Step II: Find the corresponding last digit for that remainder.
Step I:
\(73^{80}/4\)
\((72+1)^{80}/4\)
(This concept is discussed above)
72 is divisible by 4.
So when \(73^{80}\) is divided by 4, remainder is 1.
So the power of 47 is 1 more than a multiple of 4.
Step II
Knowing the cyclicity of 7, this means the last digit will be 7.
_________________
Karishma
Veritas Prep GMAT Instructor
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