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Zorluk: OrtaUnit Digit and Cyclicity

What is the unit digit of the numerical value of the expression N=528144+36389×2446517951N = 528^{144} + 363^{89} \times 244^{65} - 179^{51}?

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

Cevap

9
Evaluating each term's unit digit yields 6 for the first term, 3 for the second term, 4 for the third term, and 9 for the fourth term. Applying BODMAS rules: multiplication of 3 and 4 gives a unit digit of 2. Then adding 6 gives 8. Subtracting 9 from 8 gives a unit digit of 9 (since 18 - 9 = 9). Thus, the correct unit digit is 9.

Adım Adım Çözüm

1
Find the unit digit of 528144528^{144}
The unit digit of the base is 8. The cyclicity of 8 is 4. Dividing the exponent 144 by 4 gives a remainder of 0. When remainder is 0, we take the 4th power: 84=40968^4 = 4096, which has a unit digit of 6.
Exponents divisible by the cyclicity length correspond to the 4th power in the cycle.
2
Find the unit digit of 36389363^{89}
The unit digit of the base is 3. The cyclicity of 3 is 4. Dividing 89 by 4 gives a remainder of 1. Thus, the unit digit is 31=33^1 = 3.
The unit digit pattern for powers of 3 repeats every 4 powers.
3
Find the unit digit of 24465244^{65}
The unit digit of the base is 4. The cyclicity of 4 is 2 (powers alternate: 41=4,42=64^1 = 4, 4^2 = 6). Since the exponent 65 is odd, the unit digit is 4.
Powers of 4 with odd exponents always end in 4.
4
Find the unit digit of 17951179^{51}
The unit digit of the base is 9. The cyclicity of 9 is 2 (powers alternate: 91=9,92=19^1 = 9, 9^2 = 1). Since 51 is odd, the unit digit is 9.
Powers of 9 with odd exponents always end in 9.
5
Combine the unit digits using BODMAS rules
Unit digit of expression = [6+(3×4)9](mod10)=(6+29)(mod10)=(89)(mod10)=9[6 + (3 \times 4) - 9] \pmod{10} = (6 + 2 - 9) \pmod{10} = (8 - 9) \pmod{10} = 9.
Multiplication is performed before addition/subtraction, and a negative intermediate unit digit is adjusted by adding 10.

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Unit digit calculation using cyclicity of digits and order of operations
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