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Zorluk: OrtaRemainders and Units Digit Cyclicity

When a positive integer nn is divided by 1212, the remainder is 77. What is the remainder when n2+5n+11n^2 + 5n + 11 is divided by 1212?

Cevap: 11

Cevap

11
Since nn leaves a remainder of 77 when divided by 1212, we can substitute n7(mod12)n \equiv 7 \pmod{12} directly into n2+5n+11n^2 + 5n + 11. Evaluating 72+5(7)+117^2 + 5(7) + 11 gives 49+35+11=9549 + 35 + 11 = 95. Dividing 9595 by 1212 gives 77 with a remainder of 1111.

Adım Adım Çözüm

1
Express nn using modular arithmetic
n7(mod12)n \equiv 7 \pmod{12}
A positive integer nn that leaves a remainder of 77 when divided by 1212 can be expressed as n=12k+7n = 12k + 7 for some non-negative integer kk.
2
Substitute n7(mod12)n \equiv 7 \pmod{12} into the target expression
n2+5n+1172+5(7)+11(mod12)n^2 + 5n + 11 \equiv 7^2 + 5(7) + 11 \pmod{12}
By the algebraic properties of remainders (modular arithmetic), substituting the remainder 77 for nn yields an equivalent remainder modulo 1212.
3
Evaluate the arithmetic sum
49+35+11=9549 + 35 + 11 = 95
Computing 72=497^2 = 49, 5×7=355 \times 7 = 35, and adding 1111 gives 9595.
4
Find the remainder of 9595 modulo 1212
95=12×7+11    9511(mod12)95 = 12 \times 7 + 11 \implies 95 \equiv 11 \pmod{12}
Dividing 9595 by 1212 yields a quotient of 77 and a remainder of 1111.

Anahtar Kavram

Modular Arithmetic and Polynomial Remainders
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