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04 Jan 2021, 01:15
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
42% (02:37) correct
58%
(02:02)
wrong
based on 102
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If x and y and z are positive integers, what is the remainder when \(3^{(4 + 4x)} + 9^y + 4^{(y+181)} + 6^x+ 7^{(4x+28)} + 383^{4z} * 47^{4z+1}\) is divided by 5?
(1) y is divisible by n!, where n is an integer greater than 2
(2) y is even number
Are You Up For the Challenge: 700 Level Questions
(1) y is divisible by n!, where n is an integer greater than 2
(2) y is even number
Are You Up For the Challenge: 700 Level Questions
04 Jan 2021, 03:00
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Bunuel
Since we are looking for remainder when divided by 5, the units digit will suffice to answer the question.
Now there are 3 variables with different bases as
a) x....\(3^{(4+4x)}=3^{4(1+x)}, \ \ 6^x, 7^{4x+28}=7^{4(x+7)}\).
In all of these, the units digit can be found as 3 and 7 to the power of a multiple of 4 will have same units digit as \(3^4\) and \(7^4\). Also irrespective of x, the units digit of base 6 will be 6.
b) z....
Again all the powers containing z {\(383^{4z} * 47^{4z+1}=(383*47)^{4z}*47^1=(1....1)^{4z}*47=1*47\)} are with 4z, which means that the units digit can be found irrespective of the value of z.
c) y.....
The bases where y is power are 9 and 4, which have a cyclicity of 2 for units digit.
For 9, the units digit cyclicity is 9,1,9,1,9....
For 4, the units digit cyclicity is 4,6,4,6,4....
So we don’t require to know values of x or z. But only whether y is even.
(1) y is divisible by n!, where n is an integer greater than 2
This means y=1*2*..n. Hence y is even
Sufficient
(2) y is even number
Sufficient
D
07 Jan 2021, 07:32
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Bunuel
To know the remainder when dividing by 5, we only need to know the last digit of the quotient.
A winning shortcut for this question is to remember is that any digit to the power of 4 has a confirmed last digit. Take \(3^{(4 + 4x)}\) as an example, it is a power of \(3^4 = 81\) so the last digit of that term is always 1. Another example is \(7^{4x + 28} = {(7^4)}^{x + 7} = {(...1)}^{x+7}\) must have a last digit of 1.
Now we can start chopping down the equation and focus on the parts we cannot confirm. Anything with a power that is a multiple of 4 is confirmed using the knowledge above, so taking all powers of 4 away we are left with:
\(9^y + 4^{y + 181}\)
The last term is gone as we can write that as \(383^{4z}*47^{4z} * 47\) which has a known last digit regardless of z. Also \(6^x\) always ends in 6 so we can erase that too.
Therefore we are interested in the cycles of 9 and 4, the powers of 9 follow the last digit cycle of 9, 1, 9 , 1 .... while the powers of 4 follow the last digit cycle of 4, 6, 4, 6 ... so we end up only needing to know whether y is odd or even.
Statement 1:
y is divisible by an even number so y is even. Sufficient.
Statement 2:
Sufficeint.
Ans: D
