Question

The first 66 natural numbers are written on the paper. Find all numbers that can be erased so that the sum of other 65 numbers would be divisible by 7.​

Answer 1

$\text { Because sum }$

$\begin{array}{l}=\frac{n(n+1)}{2} \\=\frac{65 \times 66}{2} \\=65 \times 33 \\=2145\end{array}$

$\text { So when number is } 0 \quad 245 \div 7 \approx 306.43$

$\text { when number is } 1(2451) \div 7 \approx 306.29$

$\text { when murber } 1 s 2 \quad(24554 \div 7 \approx 306.14$

$\text { when iamber is 3 }(2495.3) \div 7=306$

$\text { therisfore calculation result: }$

$3,10,17,24,31,38,45,52,59$