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28 Jan 2020, 19:56
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Difficulty:
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For all positive integers \(A\) and \(B\), If \(N = 84^A*59^B\), What is the unit digit of N ?
(1) \(A + B = 10\)
(2) \(B = 4\)
Posted from my mobile device
(1) \(A + B = 10\)
(2) \(B = 4\)
Posted from my mobile device
28 Jan 2020, 21:17
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Hi,
I have taken the following approach to solve this question. Please correct me if I am wrong.
To calculate the unit's digit of N, we will have to find out A & B.
First let us check the cyclicity of 4 and 9,
cyclicity of 4 is 2
4^1=4
4^2=16
4^3=64
4^4=256
and so on --
so we can see a pattern that if 4 ^ odd then, unit's digit is 4 and if 4 ^ even then, unit's digit is 6
cyclicity of 9 is 2
9^1=9
9^2=81
9^3=729
9^4=---1
and so on --
so we can see a pattern that if 9 ^ odd then, unit's digit is 9 and if 9 ^ even then, unit's digit is 1
Statement 1:-
A + B = 10
Let us see the available options for this equation,
A + B = 10
1 + 9 => Both A and B odd => 4^odd * 9^odd => 4*9 => unit's digit = 6
2 + 8 => Both A and B even => 4^even * 9^even => 6 * 1 => unit's digit = 6
3 + 7 => Both A and B odd => 4^odd * 9^odd => 4*9 => unit's digit = 6
4 + 6 => Both A and B even => 4^even * 9^even => 6 * 1 => unit's digit = 6
5 + 5 => Both A and B odd => 4^odd * 9^odd => 4 * 9 => unit's digit = 6
In any case, we get the same unit's digit. So we get a definite answer, so we can keep options A and D and eliminate all other options.
Statement 2:-
In this statement, we only know that B=4 but we don't have any information about A ,
Now let us assume A is odd => 4^odd * 9^4=>4 * 1 => Unit's digit of N => 4
Now let us assume A is even => 4^even * 9^4=> 6 * 1 => Unit's digit of N => 6
So in this statement we are not getting a definite answer, so this statement is insufficient.
So we can eliminate option D.
The correct answer is A.
I have taken the following approach to solve this question. Please correct me if I am wrong.
To calculate the unit's digit of N, we will have to find out A & B.
First let us check the cyclicity of 4 and 9,
cyclicity of 4 is 2
4^1=4
4^2=16
4^3=64
4^4=256
and so on --
so we can see a pattern that if 4 ^ odd then, unit's digit is 4 and if 4 ^ even then, unit's digit is 6
cyclicity of 9 is 2
9^1=9
9^2=81
9^3=729
9^4=---1
and so on --
so we can see a pattern that if 9 ^ odd then, unit's digit is 9 and if 9 ^ even then, unit's digit is 1
Statement 1:-
A + B = 10
Let us see the available options for this equation,
A + B = 10
1 + 9 => Both A and B odd => 4^odd * 9^odd => 4*9 => unit's digit = 6
2 + 8 => Both A and B even => 4^even * 9^even => 6 * 1 => unit's digit = 6
3 + 7 => Both A and B odd => 4^odd * 9^odd => 4*9 => unit's digit = 6
4 + 6 => Both A and B even => 4^even * 9^even => 6 * 1 => unit's digit = 6
5 + 5 => Both A and B odd => 4^odd * 9^odd => 4 * 9 => unit's digit = 6
In any case, we get the same unit's digit. So we get a definite answer, so we can keep options A and D and eliminate all other options.
Statement 2:-
In this statement, we only know that B=4 but we don't have any information about A ,
Now let us assume A is odd => 4^odd * 9^4=>4 * 1 => Unit's digit of N => 4
Now let us assume A is even => 4^even * 9^4=> 6 * 1 => Unit's digit of N => 6
So in this statement we are not getting a definite answer, so this statement is insufficient.
So we can eliminate option D.
The correct answer is A.
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