MathRevolution wrote:

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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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