chưng minh rằng:
a,( 222^333+333^222) chia hét cho 13
b, ( 36^36- 9^10) chia hết cho 45
CMR
(222^333 +333^222)chia hết cho 13
(36^36-9^10)chia hết cho 45
Ai nhanh mk like
Áp dụng hằng đẳng thức sau
an−1=(a−1).[an−1+an−2+...+1]=(a−1).pan−1=(a−1).[an−1+an−2+...+1]=(a−1).p (nn là 1 số nguyên dương)
an+1=(a+1).[an−1−an−2+..+1]=(a+1).qan+1=(a+1).[an−1−an−2+..+1]=(a+1).q (nn là 1 số nguyên dương lẻ)
Thay vào ta được như sau:
+) 222333−1=(222−1).p=13.17.p222333−1=(222−1).p=13.17.p
+) 333222+1=(3332)111+1=110889111+1=(110889+1).q=13.8530.q333222+1=(3332)111+1=110889111+1=(110889+1).q=13.8530.q
=>=> 222333+333222=222333−1+333222+1=13(17p+8530q)⋮13222333+333222=222333−1+333222+1=13(17p+8530q)⋮13
Vậy: 222333+333222⋮13222333+333222⋮13 (đpcm)(đpcm)
CMR
(222^333 +333^222)chia hết cho 13
(36^36-9^10)chia hết cho 45
Ai nhanh mk like
\(\left(222^{333}+333^{222}\right)⋮13\)
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333222+1=(3332)111+1=110889111+1=(110889+1).q=13.8530.q" role="presentation" style="border:0px; color:rgb(40, 40, 40); direction:ltr; display:inline-block; float:none; font-family:helvetica,arial,sans-serif; font-size:18.06px; line-height:0; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; padding:1px 0px; position:relative; white-space:nowrap; word-wrap:normal" class="MathJax_CHTML mjx-chtml">
=>" role="presentation" style="border:0px; color:rgb(40, 40, 40); direction:ltr; display:inline-block; float:none; font-family:helvetica,arial,sans-serif; font-size:18.06px; line-height:0; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; padding:1px 0px; position:relative; white-space:nowrap; word-wrap:normal" class="MathJax_CHTML mjx-chtml">
a) \(222^{333}+333^{222}\)
\(=\left(111.2\right)^{333}+\left(111.3\right)^{222}\)
\(=111^{333}.2^{333}+111^{222}.3^{222}\)
\(=111^{222}.\left(111^{111}.2^{333}+3^{222}\right)\)
\(=111^{222}.\left(111^{111}.2^{3.111}+3^{2.111}\right)\)
\(=111^{222}.\left[111^{111}.\left(2^3\right)^{111}+\left(3^2\right)^{111}\right]\)
\(=111^{222}.\left(111^{111}.8^{111}+9^{111}\right)\)
\(=111^{222}.\left[\left(111.8\right)^{111}+9^{111}\right]\)
\(=111^{222}.\left(888^{111}+9^{111}\right)\)
\(=111^{222}.\left(888+9\right)\left[888^{110}-888^{109}.9+.....-888.9^{109}+9^{110}\right]\)
\(=111^{222}.7992\left[888^{110}-888^{109}.9+.....-888.9^{109}+9^{110}\right]\)
\(=111^{222}.897\left[888^{110}-888^{109}.9+.....-888.9^{109}+9^{110}\right]\)
\(=111^{222}.13.69\left[888^{110}-888^{109}.9+.....-888.9^{109}+9^{110}\right]⋮13\)
Vậy \(222^{333}+333^{222}⋮13\left(dpcm\right)\)
Chứng minh: a,222^333+333^222 chia hết cho 13
b, 3^105+4^105 chai hết cho 13 nhưng ko chia hết cho 11
a)
Ta có: \(222^{333}=\left(222^3\right)^{111}\equiv1^{111}=1\left(mod13\right)\)
\(\Rightarrow222^{333}+333^{222}\equiv1+333^{222}=1+\left(333^2\right)^{111}\)
\(\equiv1+12^{111}\equiv1+12^{110}\cdot12\equiv1+\left(12^2\right)^{55}\cdot12\)
\(\equiv1+1\cdot12\equiv13\equiv0\left(mod13\right)\)
Vậy $222^{333}+333^{222}$ chia hết cho $13.$
b) Ta có:
\(3^{105}\equiv\left(3^3\right)^{35}\equiv1^{35}\equiv1\) (mod13)
\(\Rightarrow3^{105}+4^{105}\equiv1+4^{105}\equiv1+\left(4^3\right)^{35}\)
\(\equiv1+12^{35}\equiv1+\left(12^2\right)^{17}\cdot12\equiv1+1\cdot12\equiv13\equiv0\left(mod13\right)\)
Vậy $3^{105}+4^{105}$ chia hết cho $13.$
Lại có:
\(3^{105}\equiv\left(3^3\right)^{35}\equiv5^{35}\equiv\left(5^5\right)^7\equiv1\left(mod11\right)\)
\(4^{105}\equiv\left(4^3\right)^{35}\equiv9^{35}\equiv\left(9^5\right)^7\equiv1\left(mod11\right)\)
Từ đây:\(3^{105}+4^{105}\equiv1+1\equiv2\left(mod11\right)\)
Vậy $3^{105}+4^{105}$ không chia hết cho $11.$
P/s: Rất lâu rồi không giải, không chắc.
Chứng minh: a,222^333+333^222 chia hết cho 13
b, 3^105+4^105 chai hết cho 13 nhưng ko chia hết cho 11
a) So sánh 222333 và 333222
b) Tìm các chữ số x,y để 1x8y2 chia hết cho 36
chứng minh rằng :222 mũ 333 +333 mũ 222 chia hết cho 13
a,so sánh 222333 và 333222
b, tìm các chữ số x ,y biết 1x8y2 chia hết cho 36
c,tìm số tự nhiên biết 1960 và 2002 chia a cùng số dư là 28
Chứng minh:
222^333 + 333^222 chia hết cho 13
chứng minh 222333+333222 chia hết cho 13