chứng minh rằng 11^n+2 + 12^2n+1 chia hết cho 133 10/11/2021 Bởi Eden chứng minh rằng 11^n+2 + 12^2n+1 chia hết cho 133
Đáp án: Giải thích các bước giải: ${{11}^{n+2}}+{{12}^{2n+1}}\,\,\,\vdots \,\,\,133$ $\bullet \,\,\,\,\,{{11}^{n+2}}+{{12}^{2n+1}}$ $={{11}^{n}}{{.11}^{2}}+{{12}^{2n}}.12$ $={{11}^{n}}.121+{{\left( {{12}^{2}} \right)}^{n}}.12$ $={{11}^{n}}.121+{{144}^{n}}.12$ $={{11}^{n}}\left( 133-12 \right)+{{144}^{n}}.12$ $={{11}^{n}}.133-{{11}^{n}}.12+{{144}^{n}}.12$ $={{11}^{n}}.133+12\left( {{144}^{n}}-{{11}^{n}} \right)$ $\bullet \,\,\,$Ta có công thức ${{a}^{n}}-{{b}^{n}}\,\,\,\vdots \,\,\,\left( a-b \right)$, áp dụng công thức đó để giải bài toán này: $\bullet \,\,\,$Ta thấy: ${{144}^{n}}-{{11}^{n}}\,\,\,\vdots \,\,\,\left( 144-11 \right)$ $\to {{144}^{n}}-{{11}^{n}}\,\,\,\vdots \,\,\,133$ $\to 12\left( {{144}^{n}}-{{11}^{n}} \right)\,\,\,\vdots \,\,\,133$ Mà ${{11}^{n}}.133\,\,\,\vdots \,\,\,133$ $\to {{11}^{n}}.133+12\left( {{144}^{n}}-{{11}^{n}} \right)\,\,\,\vdots \,\,\,133$ $\to {{11}^{n+2}}+{{12}^{2n+1}}\,\,\,\vdots \,\,\,133$ Bình luận
`11^{n+2}` + `12^{2n+1}` = `11^n` . 121 + `144^n` . 12 Có : 144≡11(mod 133) ⇒ `144^n`≡ `11^n`(mod 133) ⇒`144^n` . 121≡`11^n` . 121(mod 133) ⇒`144^n`.121 + `144^12`.12≡`11^n` . 121 +`144^n` .12(mod 133) ⇒ `11^n` . 121+`144^n` . 12 ≡ `144^n`(121+12)(mod133) ⇒`11^n` . 121+`144^n` . 12 ≡`144^n` . 133(mod 133) Ta có 133 chia hết cho 133 ⇒ `144^n` . 133 chia hết cho 133 ⇒`11^n` . 121+`144^n` . 12 ≡ 0(mod 133) ⇒ `11^{n+2}` + `12^{2n+1}`≡0(mod 133) ⇒ `11^{n+2}` + `12^{2n+1}` chia hết cho 133 (đpc/m) Bình luận
Đáp án:
Giải thích các bước giải:
${{11}^{n+2}}+{{12}^{2n+1}}\,\,\,\vdots \,\,\,133$
$\bullet \,\,\,\,\,{{11}^{n+2}}+{{12}^{2n+1}}$
$={{11}^{n}}{{.11}^{2}}+{{12}^{2n}}.12$
$={{11}^{n}}.121+{{\left( {{12}^{2}} \right)}^{n}}.12$
$={{11}^{n}}.121+{{144}^{n}}.12$
$={{11}^{n}}\left( 133-12 \right)+{{144}^{n}}.12$
$={{11}^{n}}.133-{{11}^{n}}.12+{{144}^{n}}.12$
$={{11}^{n}}.133+12\left( {{144}^{n}}-{{11}^{n}} \right)$
$\bullet \,\,\,$Ta có công thức ${{a}^{n}}-{{b}^{n}}\,\,\,\vdots \,\,\,\left( a-b \right)$, áp dụng công thức đó để giải bài toán này:
$\bullet \,\,\,$Ta thấy:
${{144}^{n}}-{{11}^{n}}\,\,\,\vdots \,\,\,\left( 144-11 \right)$
$\to {{144}^{n}}-{{11}^{n}}\,\,\,\vdots \,\,\,133$
$\to 12\left( {{144}^{n}}-{{11}^{n}} \right)\,\,\,\vdots \,\,\,133$
Mà ${{11}^{n}}.133\,\,\,\vdots \,\,\,133$
$\to {{11}^{n}}.133+12\left( {{144}^{n}}-{{11}^{n}} \right)\,\,\,\vdots \,\,\,133$
$\to {{11}^{n+2}}+{{12}^{2n+1}}\,\,\,\vdots \,\,\,133$
`11^{n+2}` + `12^{2n+1}` = `11^n` . 121 + `144^n` . 12
Có : 144≡11(mod 133)
⇒ `144^n`≡ `11^n`(mod 133)
⇒`144^n` . 121≡`11^n` . 121(mod 133)
⇒`144^n`.121 + `144^12`.12≡`11^n` . 121 +`144^n` .12(mod 133)
⇒ `11^n` . 121+`144^n` . 12 ≡ `144^n`(121+12)(mod133)
⇒`11^n` . 121+`144^n` . 12 ≡`144^n` . 133(mod 133)
Ta có 133 chia hết cho 133
⇒ `144^n` . 133 chia hết cho 133
⇒`11^n` . 121+`144^n` . 12 ≡ 0(mod 133)
⇒ `11^{n+2}` + `12^{2n+1}`≡0(mod 133)
⇒ `11^{n+2}` + `12^{2n+1}` chia hết cho 133 (đpc/m)