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Zorluk: OrtaRemainders and Units Digit Cyclicity

What is the smallest positive integer mm such that the remainder when 6m6^m is divided by 1111 is equal to 88?

Cevap: 7

Cevap

The smallest positive integer mm is 7.
Evaluating powers of 6 modulo 11 sequentially gives: 6166^1 \equiv 6, 6236^2 \equiv 3, 6376^3 \equiv 7, 6496^4 \equiv 9, 65106^5 \equiv 10, 6656^6 \equiv 5, and 678(mod11)6^7 \equiv 8 \pmod{11}. Therefore, m=7m = 7 is the smallest positive integer satisfying the condition.

Adım Adım Çözüm

1
Calculate the remainder of powers of 6 when divided by 11 starting from m=1m = 1.
616(mod11)6^1 \equiv 6 \pmod{11}, 623(mod11)6^2 \equiv 3 \pmod{11}, 637(mod11)6^3 \equiv 7 \pmod{11}, 649(mod11)6^4 \equiv 9 \pmod{11}, 6510(mod11)6^5 \equiv 10 \pmod{11}, 665(mod11)6^6 \equiv 5 \pmod{11}, 678(mod11)6^7 \equiv 8 \pmod{11}.
Modular multiplication allows computing powers step-by-step by multiplying the previous remainder by 6 and finding the remainder modulo 11.
2
Identify the smallest exponent mm that yields a remainder of 8.
m=7m = 7
At m=7m = 7, 67=279,936=11×25,448+86^7 = 279,936 = 11 \times 25,448 + 8, which leaves a remainder of 8.

Anahtar Kavram

Modular Arithmetic and Cyclicity of Powers
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