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Zorluk: Çok zorRemainders and Units Digit Cyclicity

Let S=k=182(1)kk7kS = \sum_{k=1}^{82} (-1)^k \cdot k \cdot 7^k. What is the remainder when S-S is divided by 1010?

  1. A
    1
  2. B
    3
  3. C
    7
  4. 9Cevap
  5. E
    0

Cevap

The remainder when S-S is divided by 1010 is 99.
Grouping the 82 terms into 20 blocks of 4 shows that each block contributes 6(mod10)6 \pmod{10}, summing to 1200(mod10)120 \equiv 0 \pmod{10}. The 81st and 82nd terms contribute 33 and 88 respectively, giving S111(mod10)S \equiv 11 \equiv 1 \pmod{10}. Therefore, S19(mod10)-S \equiv -1 \equiv 9 \pmod{10}.

Adım Adım Çözüm

1
Analyze the units digit cyclicity of 7k(mod10)7^k \pmod{10}.
The sequence of units digits for 7k7^k repeats every 4 powers: 7177^1 \equiv 7, 7297^2 \equiv 9, 7337^3 \equiv 3, 741(mod10)7^4 \equiv 1 \pmod{10}.
Units digits of powers of 7 follow a repeating cycle of length 4.
2
Evaluate the terms modulo 10 for a general block of 4 consecutive integers k=4m+1,4m+2,4m+3,4m+4k = 4m+1, 4m+2, 4m+3, 4m+4.
The four terms evaluate to: (1)4m+1(4m+1)74m+1(4m+1)(7)2m+3(mod10)(-1)^{4m+1}(4m+1)7^{4m+1} \equiv -(4m+1)(7) \equiv 2m+3 \pmod{10}; (1)4m+2(4m+2)74m+2(4m+2)(9)6m+8(mod10)(-1)^{4m+2}(4m+2)7^{4m+2} \equiv (4m+2)(9) \equiv 6m+8 \pmod{10}; (1)4m+3(4m+3)74m+3(4m+3)(3)8m+1(mod10)(-1)^{4m+3}(4m+3)7^{4m+3} \equiv -(4m+3)(3) \equiv 8m+1 \pmod{10}; and (1)4m+4(4m+4)74m+4(4m+4)(1)4m+4(mod10)(-1)^{4m+4}(4m+4)7^{4m+4} \equiv (4m+4)(1) \equiv 4m+4 \pmod{10}.
Modular arithmetic simplifies each term according to its position in the 4-term cycle.
3
Sum the four terms in a single block of 4.
(2m+3) + (6m+8) + (8m+1) + (4m+4) = 20m + 16 \equiv 6 \pmod{10}.
Every complete block of 4 consecutive terms contributes exactly 6 modulo 10, independent of mm.
4
Calculate the total sum S(mod10)S \pmod{10} for k=1k=1 to 8282.
82 terms contain 20 full blocks of 4 terms (k=1k=1 to 8080) plus 2 remaining terms (k=81k=81 and k=82k=82). The 20 blocks contribute 20×6=1200(mod10)20 \times 6 = 120 \equiv 0 \pmod{10}. For m=20m=20, k=81k=81 gives 2(20)+3=433(mod10)2(20)+3 = 43 \equiv 3 \pmod{10}, and k=82k=82 gives 6(20)+8=1288(mod10)6(20)+8 = 128 \equiv 8 \pmod{10}. Thus S0+3+8=111(mod10)S \equiv 0 + 3 + 8 = 11 \equiv 1 \pmod{10}.
Summing the contributions of all complete blocks and the remaining boundary terms gives S(mod10)S \pmod{10}.
5
Determine the positive remainder of S-S when divided by 10.
Since S1(mod10)S \equiv 1 \pmod{10}, S1101=9(mod10)-S \equiv -1 \equiv 10 - 1 = 9 \pmod{10}.
By definition of integer division, the remainder rr must satisfy 0r<100 \le r < 10.

Anahtar Kavram

Units Digit Cyclicity and Modular Arithmetic with Negative Integer Remainders
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