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Zorluk: ZorRemainders and Units Digit Cyclicity

If nn is a positive integer such that the units digit of 7n7^n is 33, what is the remainder when 3n+2+8n+13^{n+2} + 8^{n+1} is divided by 55?

  1. A
    00
  2. B
    11
  3. C
    22
  4. D
    33
  5. 44Cevap

Cevap

The correct remainder is 44.
The cyclicity pattern of the units digit of 7n7^n repeats every 44 terms: 7,9,3,17, 9, 3, 1. For the units digit to be 33, nn must leave a remainder of 33 when divided by 44 (n≡3(mod4)n \equiv 3 \pmod 4). Using this form, n+2≡5≡1(mod4)n+2 \equiv 5 \equiv 1 \pmod 4, which means 3n+2≡31≡3(mod5)3^{n+2} \equiv 3^1 \equiv 3 \pmod 5. Similarly, n+1≡4≡0(mod4)n+1 \equiv 4 \equiv 0 \pmod 4, and since 8≡3(mod5)8 \equiv 3 \pmod 5, we have 8n+1≡30≡1(mod5)8^{n+1} \equiv 3^0 \equiv 1 \pmod 5. Adding these values gives 3+1=43 + 1 = 4, so the remainder when divided by 55 is 44.

Adım Adım Çözüm

1
Determine the remainder of nn when divided by 44 using the units digit cyclicity of 7n7^n.
n≡3(mod4)n \equiv 3 \pmod 4.
The units digits of powers of 77 follow a repeating pattern of period 44: 71→77^1 \rightarrow 7, 72→97^2 \rightarrow 9, 73→37^3 \rightarrow 3, 74→17^4 \rightarrow 1. Since the units digit of 7n7^n is 33, nn must be of the form 4k+34k + 3.
2
Evaluate 3n+2(mod5)3^{n+2} \pmod 5.
3n+2≡3(mod5)3^{n+2} \equiv 3 \pmod 5.
Substitute n=4k+3n = 4k + 3 into the exponent: n+2=4k+5=4(k+1)+1n + 2 = 4k + 5 = 4(k+1) + 1. The powers of 3(mod5)3 \pmod 5 repeat every 44 powers (31≡3,32≡4,33≡2,34≡13^1 \equiv 3, 3^2 \equiv 4, 3^3 \equiv 2, 3^4 \equiv 1). Thus, 34k+5≡31≡3(mod5)3^{4k+5} \equiv 3^1 \equiv 3 \pmod 5.
3
Evaluate 8n+1(mod5)8^{n+1} \pmod 5.
8n+1≡1(mod5)8^{n+1} \equiv 1 \pmod 5.
First simplify the base: 8≡3(mod5)8 \equiv 3 \pmod 5, so 8n+1≡3n+1(mod5)8^{n+1} \equiv 3^{n+1} \pmod 5. Substitute n=4k+3n = 4k + 3: n+1=4k+4=4(k+1)n + 1 = 4k + 4 = 4(k+1). Since the exponent is a multiple of 44, 34(k+1)≡34≡1(mod5)3^{4(k+1)} \equiv 3^4 \equiv 1 \pmod 5.
4
Sum the modular results to find the final remainder modulo 55.
(3+1)(mod5)=4(3 + 1) \pmod 5 = 4.
By properties of modular addition, (3n+2+8n+1)(mod5)≡(3+1)(mod5)=4(3^{n+2} + 8^{n+1}) \pmod 5 \equiv (3 + 1) \pmod 5 = 4.

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