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Zorluk: OrtaUnit Digit and Cyclicity

What is the unit digit of the numerical expression M=3105×782+896M = 3^{105} \times 7^{82} + 8^{96}?

Cevap: 3

Cevap

The unit digit of the given expression is 3.
The unit digit of an exponential term is determined by its base's cyclicity cycle and the exponent modulo 4. For 31053^{105}, 105(mod4)=1    31=3105 \pmod 4 = 1 \implies 3^1 = 3. For 7827^{82}, 82(mod4)=2    72=49    982 \pmod 4 = 2 \implies 7^2 = 49 \implies 9. The product of these terms has a unit digit of (3×9)(mod10)=7(3 \times 9) \pmod{10} = 7. For 8968^{96}, 96(mod4)=096 \pmod 4 = 0, which indicates the 4th position in the cyclicity pattern of 8 (8, 4, 2, 6), yielding a unit digit of 6. Finally, adding the unit digits gives (7+6)(mod10)=3(7 + 6) \pmod{10} = 3.

Adım Adım Çözüm

1
Determine the cyclicity and remainder for 31053^{105}
Unit digit is 3
The unit digits of powers of 3 repeat in a cycle of 4 (3, 9, 7, 1). Dividing the exponent by 4 gives 105(mod4)=1105 \pmod 4 = 1, corresponding to 31=33^1 = 3.
2
Determine the cyclicity and remainder for 7827^{82}
Unit digit is 9
The unit digits of powers of 7 repeat in a cycle of 4 (7, 9, 3, 1). Dividing the exponent by 4 gives 82(mod4)=282 \pmod 4 = 2, corresponding to 72=497^2 = 49 (unit digit 9).
3
Calculate the unit digit of the product 3105×7823^{105} \times 7^{82}
Unit digit is 7
Multiply the unit digits of the two factors: (3×9)=27(3 \times 9) = 27, which has a unit digit of 7.
4
Determine the cyclicity and remainder for 8968^{96}
Unit digit is 6
The unit digits of powers of 8 repeat in a cycle of 4 (8, 4, 2, 6). The exponent 96 is completely divisible by 4 (96(mod4)=096 \pmod 4 = 0). When remainder is 0, we use the 4th power in the cycle (84=40968^4 = 4096), which ends in 6.
5
Sum the unit digits of the two main terms
Unit digit is 3
Adding the unit digits gives (7+6)=13(7 + 6) = 13, which has a unit digit of 3.

Anahtar Kavram

Determining the unit digit of composite exponential expressions using power cyclicity.
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