MathRevolution wrote:
[
Math Revolution GMAT math practice question]
What is the units digit of \((3^{101})(7^{103})\)?
\(A. 1\)
\(B. 3\)
\(C. 5\)
\(D. 7\)
\(E. 9\)
When an integer is raised to consecutive powers, the resulting units digits repeat in a CYCLE.
\(3^{101}\):
3¹ --> units digit of 3.
3² --> units digit of 9. (Since the product of the preceding units digit and 3 = 3*3 = 9.)
3³ --> units digit of 7. (Since the product of the preceding units digit and 3 = 9*3 = 27.)
3⁴ --> units digit of 1. (Since the product of the preceding units digit and 3 = 7*3 = 21.)
From here, the units digits will repeat in the same pattern: 3, 9, 7, 1.
The units digit repeat in a CYCLE OF 4.
Implication:
When an integer with a units digit of 3 is raised to a power that is a multiple of 4, the units digit will be 1.
Thus, \(3^{100}\)has a units digit of 1.
From here, the cycle of units digits will repeat: 3, 9, 7, 1...
Thus, \(3^{101}\) has a units digit of 3.
7¹⁰³:
7¹ --> units digit of 7.
7² --> units digit of 9. (Since the product of the preceding units digit and 7 = 7*7 = 49.)
7³ --> units digit of 3. (Since the product of the preceding units digit and 7 = 9*7 = 63.)
7⁴ --> units digit of 1. (Since the product of the preceding units digit and 7 = 3*7 = 21.)
From here, the units digits will repeat in the same pattern: 7, 9, 3, 1.
The units digit repeat in a CYCLE OF 4.
Implication:
When an integer with a units digit of 7 is raised to a power that is a multiple of 4, the units digit will be 1.
Thus, \(7^{100}\) has a units digit of 1.
From here, the cycle of units digits will repeat: 7, 9, 3, 1...
\(7^{101}\)--> units digit of 7.
\(7^{102}\) --> units digit of 9.
\(7^{103}\)--> units digit of 3.
Result:
\(3^{101}7^{103}\) = (integer with a units digit of 3)(integer with a units digit of 3) = integer with a units digit of 9.
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