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Zorluk: OrtaUnit Digit and Cyclicity

For the composite numerical expression E=(412120+317201)−814102E = (412^{120} + 317^{201}) - 814^{102}, which of the following represents its unit digit?

  1. 7Cevap
  2. B
    2
  3. C
    3
  4. D
    9

Cevap

7
Evaluating each term by cyclicity yields unit digits of 6 for 412120412^{120}, 7 for 317201317^{201}, and 6 for 814102814^{102}. Combining these according to the operations gives (6+7)−6=7(6 + 7) - 6 = 7, resulting in a unit digit of 7.

Adım Adım Çözüm

1
Determine the unit digit of 412120412^{120}
Unit digit is 6
The unit digit of base 412 is 2, which has a cyclicity of 4 (2, 4, 8, 6). The exponent 120 is divisible by 4 (120(mod4)=0120 \pmod 4 = 0), so we take the 4th power in the cycle: 24=162^4 = 16, giving a unit digit of 6.
2
Determine the unit digit of 317201317^{201}
Unit digit is 7
The unit digit of base 317 is 7, which has a cyclicity of 4 (7, 9, 3, 1). The exponent remainder is 201(mod4)=1201 \pmod 4 = 1, corresponding to 71=77^1 = 7.
3
Determine the unit digit of 814102814^{102}
Unit digit is 6
The unit digit of base 814 is 4, which has a cyclicity of 2 (4 for odd powers, 6 for even powers). Since 102 is an even exponent, the unit digit is 6.
4
Combine the unit digits according to the expression E=(412120+317201)−814102E = (412^{120} + 317^{201}) - 814^{102}
Unit digit is 7
Adding the unit digits of the first two terms gives 6+7=136 + 7 = 13, which has a unit digit of 3. Subtracting 6 gives 3−6=−33 - 6 = -3, which in modulo 10 arithmetic yields 13−6=713 - 6 = 7.

Anahtar Kavram

Cyclicity of numbers and modular arithmetic for unit digit determination
Tahmini Süre:1m 30s
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Konular

Unit Digit and Cyclicity

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