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Zorluk: OrtaRemainders and Units Digit Cyclicity

For any positive integer nn, what is the units digit of the expression S=74n+1+92n+34n+3S = 7^{4n+1} + 9^{2n} + 3^{4n+3}?

  1. A
    0
  2. B
    3
  3. 5Cevap
  4. D
    7
  5. E
    9

Cevap

The units digit of the expression is 5.
Each base has a predictable units digit cyclicity. The pattern for powers of 77 is 7,9,3,17, 9, 3, 1 (period 4), giving a units digit of 77 for exponent 4n+14n+1. The pattern for powers of 99 is 9,19, 1 (period 2), giving a units digit of 11 for even exponent 2n2n. The pattern for powers of 33 is 3,9,7,13, 9, 7, 1 (period 4), giving a units digit of 77 for exponent 4n+34n+3. Adding these units digits gives 7+1+7=157 + 1 + 7 = 15, whose units digit is 55.

Adım Adım Çözüm

1
Determine the units digit of 74n+17^{4n+1}
The units digit of powers of 77 follows a repeating pattern of length 4: 7,9,3,17, 9, 3, 1. Since the exponent 4n+14n+1 has a remainder of 11 when divided by 44, the units digit is 77.
Units digit cyclicity of base 77 repeats every 4 powers.
2
Determine the units digit of 92n9^{2n}
The units digit of powers of 99 follows a repeating pattern of length 2: 9,19, 1. For any positive integer nn, the exponent 2n2n is even, so the units digit is 11.
Even powers of 99 always end in 11.
3
Determine the units digit of 34n+33^{4n+3}
The units digit of powers of 33 follows a repeating pattern of length 4: 3,9,7,13, 9, 7, 1. Since the exponent 4n+34n+3 has a remainder of 33 when divided by 44, the units digit is 77.
Units digit cyclicity of base 33 repeats every 4 powers.
4
Sum the units digits and take the final units digit
7+1+7=157 + 1 + 7 = 15, which has a units digit of 55.
The units digit of a sum of numbers equals the units digit of the sum of their individual units digits.

Anahtar Kavram

Units digit cyclicity of integer powers
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