Bunuel wrote:
What is the unit's digit in the numerical value of \((2)^{37} * (3)^{47} * (7)^{79} * (13)^{26}\)?
A. 2
B. 4
C. 6
D. 8
E. 9
We can use the cyclicity (pattern) of the powers of the digits - 2, 3, and 7 to solve this question:
Units digit of \((2)^{37} * (3)^{47} * (7)^{79} * (13)^{26}\)
= Units digit of \((2)^{37} * (3)^{47} * (7)^{79} * (3)^{26}\)
= Units digit of \((2)^{37} * (3)^{47+26} * (7)^{79}\)
= Units digit of \((2)^{37} * (3)^{73} * (7)^{79}\)
For 2: The cycle for 2 is: 2, 4, 8, 6 (4 places)
The exponent of 2 is 37
Dividing 37 by 4, we have remainder 1
=> Units digit of \((2)^{37}\) = Units digit of \((2)^{1}\) = 2
For 3: The cycle for 3 is: 3, 9, 7, 1 (4 places)
The exponent of 3 is 73
Dividing 73 by 4, we have remainder 1
=> Units digit of \((3)^{73}\) = Units digit of \((3)^{1}\) = 3
For 7: The cycle for 7 is: 7, 9, 3, 1 (4 places)
The exponent of 7 is 79
Dividing 79 by 4, we have remainder 3
=> Units digit of \((7)^{79}\) = Units digit of \((7)^{3}\) = 3
Thus, required units digit of \((2)^{37} * (3)^{73} * (7)^{79}\)
= Units digit of \(2 * 3 * 3\)
= 8
Answer DAlternate approach: Units digit of \((2)^{37} * (3)^{47} * (7)^{79} * (13)^{26}\)
= Units digit of \((2)^{37} * (3)^{47} * (7)^{79} * (3)^{26}\)
= Units digit of \((2)^{37} * (3)^{47+26} * (7)^{79}\)
= Units digit of \((2)^{37} * (3)^{73} * (7)^{79}\)
= Units digit of \((2 * 3)^{37} * (3)^{73 - 37} * (7)^{79}\)
= Units digit of \((6)^{37} * (3)^{36} * (7)^{79}\)
= Units digit of \(6 * (3 * 7)^{36} * (7)^{79 - 36}\) ... (Since the units digit of 6 raised to any positive integer is always 6)
= Units digit of \(6 * (1)^{36} * (7)^{43}\)
= Units digit of \(6 * 1 * (7)^{43}\)
For 7: The cycle for 7 is: 7, 9, 3, 1 (4 places)
The exponent of 7 is 43
Dividing 79 by 4, we have remainder 3
=> Units digit of \((7)^{43}\) = Units digit of \((7)^{3}\) = 3
Thus, the required units digit = Units digit of \(6 * 1 * 3\)
= 8
Answer D _________________
Sujoy Kumar Datta |
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