更新Hexo版本（8.1.1）；数学公式使用 katex 渲染；

Author: simon7073

.gitignore

@@ -2,10 +2,11 @@
 .deploy_heroku/
 node_modules/
 public/
+themes/
+source/_drafts/
 db.json
 *.log
 .DS_Store
 Thumbs.db
 NOGIT
-themes/
 package-lock.json

_config.next.yml

@@ -496,13 +496,13 @@ math:
   every_page: false
 
   mathjax:
-    enable: true
+    enable: false
     tags: ams # none | ams (自动方程编号) | all
 
   katex:
-    enable: false
+    enable: true
     # See: https://github.com/KaTeX/KaTeX/tree/master/contrib/copy-tex
-    copy_tex: false
+    copy_tex: true
 
 # 在您的网站上轻松启用快速Ajax导航.
 # For more information: https://github.com/next-theme/pjax
@@ -893,3 +893,8 @@ canvas_ribbon:
 css: css
 js: js
 images: images
+
+# config of hexo-renderer-pandoc
+pandoc:
+  args:
+    - --mathjax

package.json

@@ -9,23 +9,24 @@
     "server": "hexo server"
   },
   "hexo": {
-    "version": "6.3.0"
+    "version": "8.1.1"
   },
   "dependencies": {
-    "hexo": "^6.3.0",
+    "hexo": "^8.1.1",
     "hexo-deployer-git": "^4.0.0",
     "hexo-generator-archive": "^2.0.0",
     "hexo-generator-category": "^2.0.0",
     "hexo-generator-feed": "^3.0.0",
     "hexo-generator-index2": "^0.2.0",
-    "hexo-generator-searchdb": "^1.4.1",
+    "hexo-generator-searchdb": "^1.5.0",
     "hexo-generator-tag": "^2.0.0",
     "hexo-renderer-ejs": "^2.0.0",
-    "hexo-renderer-pandoc": "^0.3.1",
-    "hexo-renderer-stylus": "^2.1.0",
+    "hexo-renderer-markdown-it-plus": "^1.0.6",
+    "hexo-renderer-pandoc": "^0.5.0",
+    "hexo-renderer-stylus": "^3.0.1",
     "hexo-server": "^3.0.0",
-    "hexo-theme-landscape": "^0.0.3",
-    "hexo-theme-next": "^8.14.2",
-    "jquery": "^3.6.3"
+    "hexo-theme-landscape": "^1.1.0",
+    "hexo-theme-next": "^8.26.0",
+    "jquery": "^3.7.1"
   }
 }

source/_posts/2021-9-3-快速判断7的倍数.md

@@ -3,7 +3,7 @@ title: 快速判断7的倍数
 date: 2021-09-03 16:42:58
 tags: 
 categories: [未分类 ]
-mathjax: true
+mathjax: false
 ---
 
 
@@ -15,56 +15,71 @@ mathjax: true
 **方法一**：掐掉个位再减去掐掉数字的两倍
 例： 6615
 
-$$\begin{aligned}
+```
+$$
+\begin{aligned}
 661 - 2 * 5 &= 651 \\
 65 - 2 * 1 &= 63 \\
 6 - 2 * 3 &= 0 \\
-\end{aligned}$$
+\end{aligned}
+$$
+```
 
 当然，很明显 `63=7*9` 。所以当最后的数字是 0 或者是 7 的倍数时，该数可以判定为 7 的倍数；否则不是 7 的倍数。
 
 **方法二**：掐掉后三位再减去掐掉的数字
 例： 2222640
 
-$$\begin{aligned}
+```
+$$
+\begin{aligned}
 2222 - 640 &= 1582 \\
 1 - 582 &= -581 \\
-\end{aligned}$$
+\end{aligned}
+$$
+```
 
 此时再用法一，判断 581 是否为 7 的倍数
-
-$$\begin{aligned}
+```
+$$
+\begin{aligned}
 58 - 2 * 1 &= 56 \\
 5 - 2 * 6 &= -7 \\
-\end{aligned}$$
+\end{aligned}
+$$
+```
 
 很明显， `56=7*8` ，所以 2222640 是 7 的倍数。
 
 
 ## 简要分析
 
 **方法一**：假设 $X=10k+a_1$ ， ($a_1$ 为个位数)，则
-
-$$\begin{equation}
+```
+$$
+\begin{equation}
 \begin{aligned}
 X-21a_1 
 &= (10k + a_1) - 21a_1 \\
 &= 10k - 20a_1 \\
 &= 10(k-2a_1) \\
 \end{aligned}
-\end{equation}$$
-
+\end{equation}
+$$
+```
 若 $X$ 可以被7整除，则必然有 $ X'=k-2a_1$ 可以被7整除
 
 **方法二**：假设 $X=1000k + a_3$， ($a_3$ 为后三位数)，则
-
-$$\begin{equation}
+```
+$$
+\begin{equation}
 \begin{aligned}
 X+k-k
 &= (1000k + a_3) +k -k \\
 &= 1001k + a_3 -k \\
 &= (7*11*13)·k - (k -a_3) \\
 \end{aligned}
-\end{equation}$$
-
+\end{equation}
+$$
+```
 若 $X$ 可以被7整除，则必然有 $ X'=k -a_3$ 可以被7整除