Question

Difficulty: HardRemainders and Units Digit Cyclicity

For any positive integer nn, let SnS_n denote the units digit of the sum 2n+3n+4n+7n2^n + 3^n + 4^n + 7^n. What is the remainder when the sum T=n=1102SnT = \sum_{n=1}^{102} S_n is divided by 99?

Answer: 1

Answer

The remainder when the sum T is divided by 9 is 1.
Evaluating the units digits of each exponent term reveals that SnS_n follows a repeating 4-term cycle (6,8,2,4)(6, 8, 2, 4) with a sum of 20 per cycle. For 102 terms, there are 25 full cycles plus the first two terms (S1=6S_1=6 and S2=8S_2=8). The total sum is T=25×20+6+8=514T = 25 \times 20 + 6 + 8 = 514. Dividing 514 by 9 yields 514=9×57+1514 = 9 \times 57 + 1, so the remainder is 1.

Step-by-Step Solution

1
Find the cyclicity of the units digit of each term 2n,3n,4n,2^n, 3^n, 4^n, and 7n7^n.
Units digits repeat in patterns of length 4: for 2n2^n (2, 4, 8, 6), for 3n3^n (3, 9, 7, 1), for 4n4^n (4, 6, 4, 6), and for 7n7^n (7, 9, 3, 1).
Units digits of positive integer powers cycle with periodicities that divide 4.
2
Compute SnS_n for n=1,2,3,4n = 1, 2, 3, 4 and find the sum of one 4-term period.
S1=units(2+3+4+7=16)=6S_1 = \text{units}(2+3+4+7=16) = 6, S2=units(4+9+6+9=28)=8S_2 = \text{units}(4+9+6+9=28) = 8, S3=units(8+7+4+3=22)=2S_3 = \text{units}(8+7+4+3=22) = 2, S4=units(6+1+6+1=14)=4S_4 = \text{units}(6+1+6+1=14) = 4. Sum of one period = 6+8+2+4=206+8+2+4 = 20.
The sum of the units digits of individual terms determines the units digit of the total expression.
3
Calculate the total sum T=n=1102SnT = \sum_{n=1}^{102} S_n.
Since 102=25×4+2102 = 25 \times 4 + 2, the sequence consists of 25 complete cycles of 4 terms plus the first 2 terms (S1=6S_1=6 and S2=8S_2=8). Thus, T=25×20+6+8=500+14=514T = 25 \times 20 + 6 + 8 = 500 + 14 = 514.
Dividing the total number of terms by the period length gives the number of full cycles and remaining initial terms.
4
Compute the remainder when T=514T = 514 is divided by 9.
The sum of the digits of 514 is 5+1+4=105 + 1 + 4 = 10, and 101(mod9)10 \equiv 1 \pmod 9. Alternatively, 514=9×57+1514 = 9 \times 57 + 1. Thus, the remainder is 1.
A positive integer and the sum of its digits leave the same remainder when divided by 9.

Key Concept

Units digit cyclicity of exponential terms and modular arithmetic on sequence sums
Estimated Time:2m 0s
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