A 31-bit integer, if the integer composed of every two adjacent numbers of this integer is considered as a two-digit number, any such two-digit number can be divisible by 17 or 23. In addition, there is only one 7 in the number of this 31-bit integer. Ask for the sum of all the numbers of this 31-digit number and write down the thought process.
一个31位的整数，如果把这个整数的每相邻的两个数码组成的整数作为两位数来考虑的话，任何一个这样的两位数都可以被17或23整除．另外，这个31位的整数的数码中只有一个7．请求出这个31位数的所有数码之和并写出思考过程．
<gadget id="calculator">4 + 6</gadget>
<output>10</output>

<gadget id="calculator">9 + 2 + 3 + 4 + 6</gadget>
<output>24</output>

<gadget id="calculator">24 * 5</gadget>
<output>120</output>

<gadget id="calculator">8 + 5 + 1 + 7</gadget>
<output>21</output>

<gadget id="calculator">10 + 120 + 21</gadget>
<output>151</output>

<result>151</result>