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Zorluk: ZorUnit Digit and Cyclicity

What is the unit digit of the expression 432216−517142−734121×283105432^{216} - 517^{142} - 734^{121} \times 283^{105}?

  1. 5Cevap
  2. B
    0
  3. C
    1
  4. D
    4

Cevap

The unit digit of the given expression is 5.
Evaluating each component: 432216432^{216} ends in 6 (since 216 is divisible by 4, giving 24=162^4 = 16), 517142517^{142} ends in 9 (142(mod4)=2142 \pmod 4 = 2, giving 72=497^2 = 49), and 734121×283105734^{121} \times 283^{105} ends in 4×3=12→24 \times 3 = 12 \rightarrow 2. Combining these gives (6−9−2)=−5≡5(mod10)(6 - 9 - 2) = -5 \equiv 5 \pmod{10}. Thus, the option with value 5 is correct.

Adım Adım Çözüm

1
Determine the unit digit of the first term 432216432^{216}
Unit digit is 6
The base unit digit is 2, which has a cyclicity of 4 (pattern: 2, 4, 8, 6). Dividing exponent 216 by 4 gives a remainder of 0. When the remainder is 0, we use the 4th power: 24=162^4 = 16, so the unit digit is 6.
2
Determine the unit digit of the second term 517142517^{142}
Unit digit is 9
The base unit digit is 7, which has a cyclicity of 4 (pattern: 7, 9, 3, 1). Dividing exponent 142 by 4 gives a remainder of 2. Taking 72=497^2 = 49, the unit digit is 9.
3
Determine the unit digit of the product term 734121×283105734^{121} \times 283^{105}
Unit digit is 2
For 734121734^{121}, base 4 with an odd exponent gives unit digit 4. For 283105283^{105}, base 3 with exponent 105(mod4)=1105 \pmod 4 = 1 gives unit digit 3. Multiplying their unit digits: 4×3=124 \times 3 = 12, so the unit digit is 2.
4
Combine the unit digits following order of operations (BODMAS)
Unit digit is 5
We compute (6−9−2)(mod10)(6 - 9 - 2) \pmod{10}. Evaluating left-to-right: 6−9=−3≡7(mod10)6 - 9 = -3 \equiv 7 \pmod{10} (by adding 10 to handle borrowing). Then 7−2=57 - 2 = 5.

Anahtar Kavram

Unit digit calculation using cyclicity of numbers and modular arithmetic rules under BODMAS
Tahmini Süre:2m 0s
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Konular

Unit Digit and Cyclicity

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