Among the statements: (S1): 2023^2022-1999^2022 is divisible by 8 (S2): 13(13)^n-11n-13 is…

Question

Among the statements:

(S1): $2023^{2022}-1999^{2022}$ is divisible by 8

(S2): $13(13)^n-11n-13$ is divisible by 144 for infinitely many $n\in\mathbb N$

Accepted Answer

(S1): Since $x^n-y^n$ is divisible by $x-y$: $2023-1999=24$, and $2023^2-1999^2$-type factorizations give divisibility by 8 through the even exponent structure (2022 is even, so $2023^2-1999^2\mid 2023^{2022}-1999^{2022}$, and $2023^2-1999^2=24\times4022$, divisible by 8). True.

(S2): $13(1+12)^n-11n-13=13\left[1+{}^nC_1(12)+{}^nC_2(12)^2+\cdots\right]-11n-13$

$$=(156n-11n)+13\cdot{}^nC_2(12)^2+13\cdot{}^nC_3(12)^3+\cdots=145n+13\cdot{}^nC_2(144)+\cdots$$

If $n=144m$ for $m\in N$, then $145n=145\times144m$ is divisible by 144, and all later terms already carry a factor of $144$. So it's divisible by 144 for infinitely many $n$. True.

Solutions · 1