For any positive integer n, let u(n) denote the units digit of the sum 7^n + 3^{n+1} + 2^{… — GMAT

For any positive integer $n$ , let $u(n)$ denote the units digit of the sum $7^n + 3^{n+1} + 2^{n+2}$ . What is the remainder when the sum $u(1) + u(2) + u(3) + \dots + u(100)$ is divided by 7?

6Answer
B
5
C
2
D
0
E
1

Answer

The remainder when the sum is divided by 7 is 6.

Evaluating the units digit

u(n)

for

n = 1, 2, 3, 4

gives

u(1) = 4

u(2) = 2

u(3) = 6

, and

u(4) = 0

. This pattern of 4 values repeats indefinitely. The sum of one full cycle of 4 terms is

4 + 2 + 6 + 0 = 12

. Since there are 100 terms in total, there are

100 / 4 = 25

full cycles. The total sum is

25 \times 12 = 300

. Dividing 300 by 7 gives a quotient of 42 with a remainder of 6. Thus, the correct choice is 6.

Step-by-Step Solution

Determine the units digit cyclicity for each power term.

The units digits of

7^n

follow the 4-term cycle [7, 9, 3, 1]. The units digits of

3^{n+1}

follow [9, 7, 1, 3]. The units digits of

2^{n+2}

follow [8, 4, 2, 6].

Units digits of positive integer powers repeat periodically with a cycle length of 4.

Calculate

u(n)

for the first 4 terms to identify the repeating pattern of

u(n)

For

n=1

7+9+8 = 24 \implies u(1)=4

. For

n=2

9+7+4 = 20 \implies u(2)=0

. For

n=3

3+1+2 = 6 \implies u(3)=6

. For

n=4

1+3+6 = 10 \implies u(4)=0

. The sequence of

u(n)

[4, 0, 6, 0]

Evaluating individual terms determines the fundamental period and sum per period of

u(n)

Sum the values of

u(n)

over one complete cycle of 4 terms.

Sum of one cycle =

4 + 0 + 6 + 0 = 10

Finding the sum of a single period simplifies finding the total sum over 100 terms.

Calculate the total sum for 100 terms and determine its remainder modulo 7.

Since 100 terms contain

100 / 4 = 25

complete cycles, the total sum

S = 25 \times 10 = 250

. Dividing 250 by 7 gives

250 = 7 \times 35 + 5

? Wait:

u(2) = (9+7+4)

units digit is 0. Let's verify:

7^2=49

(9),

3^3=27

(7),

2^4=16

(6).

9+7+6=22 \implies u(2)=2

. Let's re-verify:

u(1)=4, u(2)=2, u(3)=6, u(4)=0

. Sum per cycle =

4+2+6+0 = 12

. Total sum

S = 25 \times 12 = 300

300 = 7 \times 42 + 6

. Remainder is 6.

Dividing the total sum of 300 by 7 yields a quotient of 42 and a remainder of 6.

Key Concept

Units Digit Cyclicity and Modular Arithmetic Sums

Estimated Time:1m 30s

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