A sequence of positive integers a_n is defined by a_n = 2^n + 5^n for all integers n \ge 1… — GMAT

A sequence of positive integers $a_n$ is defined by $a_n = 2^n + 5^n$ for all integers $n \ge 1$ . What is the remainder when $a_{100}$ is divided by $7$ ?

Answer: 4

Answer

The remainder when

a_{100}

is divided by

7

4

By reducing

5 \equiv -2 \pmod 7

, we observe that

5^{100} \equiv (-2)^{100} = 2^{100} \pmod 7

because the exponent

100

is even. The powers of

2 \pmod 7

follow a 3-step cycle (

2, 4, 1

). Dividing the exponent

100

3

leaves a remainder of

1

, meaning

2^{100} \equiv 2^1 = 2 \pmod 7

. Adding the remainders for both terms yields

2 + 2 = 4

Step-by-Step Solution

Use modular arithmetic to simplify the base

5 \pmod 7

5 \equiv -2 \pmod 7

, which implies

5^{100} \equiv (-2)^{100} = 2^{100} \pmod 7

because the exponent

100

is even.

Converting

5

-2

allows both terms to be expressed using powers of

2

Determine the remainder cycle of powers of

2

when divided by

7

2^1 \equiv 2 \pmod 7

2^2 \equiv 4 \pmod 7

, and

2^3 = 8 \equiv 1 \pmod 7

. The pattern repeats every

3

powers.

Finding the period of cyclicity simplifies evaluating large powers.

Divide the exponent

100

by the cycle length

3

100 = 3 \times 33 + 1

, leaving a remainder of

1

. Thus,

2^{100} \equiv 2^1 = 2 \pmod 7

The remainder of the exponent modulo the cycle length determines the equivalent reduced power.

Combine the remainders for

2^{100}

and

5^{100}

a_{100} = 2^{100} + 5^{100} \equiv 2 + 2 = 4 \pmod 7

Adding the individual modular values gives the overall remainder.

Key Concept

Modular arithmetic cyclicity and negative congruences

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