2021/015) Show that $2^{147}-1$ is divisible by 343

 We know that $343=7^3$

1st we know that $2^3-1 = 7$ or $2^3= 7 +1$

raise both sides to power 7 to get

$(2^3)^7= (7+1)^7 = \sum_{i=0}^7{7 \choose i}7^{i}1^{7-i}$

Except the 1st 2 terms each term is having a power of 7 greater than or equals to 2 so each term is divisible by $7^2=49$. the 1st term is 1 and the second term is $7 * 7$ or 49 so all except 1st term is divisible by 49 and 1st term is 1 so

$(2^3)^7= (1 + 49 q)$ for some q

or $2^{21}=  49q+ 1$ for some q

raise both sides to power 7 to get

$(2^{21})^7= (49q+1)^7 = \sum_{i=0}^7{7 \choose i}(49q){i}1^{7-i}$

 Except the 1st 2 terms each term is having a power of 49 greater than or equals to 2 so each term is divisible by $7^4$ and hence $7^3=343$ The 1st term is 1 and the second term is $7 * 49q$ or 343q so all except 1st term is divisible by 49 and 1st term is 1 so

$2^{147}$ leaves a remainder `1 when divided by 343 or $2^{147}-1$ is divisible by 343