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Zorluk: ZorRemainders and Units Digit Cyclicity

What is the remainder when the integer N=383+783N = 3^{83} + 7^{83} is divided by 100100?

Cevap: 70

Cevap

The remainder when N=383+783N = 3^{83} + 7^{83} is divided by 100100 is 70.
Finding the remainder when an integer is divided by 100100 is equivalent to finding its last two digits. Powers of 33 repeat their last two digits in cycles of 2020, and 320≡1(mod100)3^{20} \equiv 1 \pmod{100}. Thus 383≡33=27(mod100)3^{83} \equiv 3^3 = 27 \pmod{100}. Powers of 77 repeat their last two digits in cycles of 44, with 74=2401≡1(mod100)7^4 = 2401 \equiv 1 \pmod{100}. Thus 783≡73=343≡43(mod100)7^{83} \equiv 7^3 = 343 \equiv 43 \pmod{100}. Combining these gives 27+43=7027 + 43 = 70.

Adım Adım Çözüm

1
Find the remainder of 3833^{83} modulo 100100 by identifying the cyclicity pattern of powers of 33.
320≡1(mod100)3^{20} \equiv 1 \pmod{100}, which implies 383=(320)4⋅33≡14⋅27=27(mod100)3^{83} = (3^{20})^4 \cdot 3^3 \equiv 1^4 \cdot 27 = 27 \pmod{100}.
Since 320≡1(mod100)3^{20} \equiv 1 \pmod{100}, exponent 8383 can be reduced to 83(mod20)=383 \pmod{20} = 3.
2
Find the remainder of 7837^{83} modulo 100100 by identifying the cyclicity pattern of powers of 77.
74=2401≡1(mod100)7^4 = 2401 \equiv 1 \pmod{100}, which implies 783=(74)20⋅73≡120⋅343≡43(mod100)7^{83} = (7^4)^{20} \cdot 7^3 \equiv 1^{20} \cdot 343 \equiv 43 \pmod{100}.
Since 74≡1(mod100)7^4 \equiv 1 \pmod{100}, exponent 8383 can be reduced to 83(mod4)=383 \pmod{4} = 3.
3
Sum the modular remainders to obtain the final remainder of NN.
N=383+783≡27+43=70(mod100)N = 3^{83} + 7^{83} \equiv 27 + 43 = 70 \pmod{100}.
The remainder of a sum is equal to the sum of the individual remainders modulo 100100.

Anahtar Kavram

Modular Arithmetic and Cyclicity of Last Two Digits
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