First of all we see that 1==1, 10==3, 100==2, 1000==6, 10000==4, and 100000==5 in mod 7, and the cycle repeats from here. Thus, if we want to check the divisibility of (A1 A2 A3 . . . . . . .An), an 'n' digit number, we can simply add the numbers from the right, multiplying them with 1, 3, 2, 6, 4, 5, and so on. For example, if we want to check whether 125783 is divisible by 7, we will find the sum 3(1)+8(3)+7(2)+5(6)+2(4)+1(5)=84, which is divisible by 7, thus 125783 is divisible by 7.

I also found out another divisibility test for 7, in which we subtract twice the units digit of a number from the remaining number till we get a single digit, and if this number is divisible, the whole number is divisible. This can be proved by the fact that 10(a)+b==10(a-2b) in mod 7, where 'b' is the units digit, and 'a' the others.

For divisibility with 7 do the following : Suppose you have a no. abcde. Double the last digit i.e. 2e and subtract it from abcd (abcd - 2e), if this difference is divisible by 7, the no. is also. Keep on repeating the process with d,ç and b. You will get the answer.

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TopNewestFirst of all we see that 1==1, 10==3, 100==2, 1000==6, 10000==4, and 100000==5 in mod 7, and the cycle repeats from here. Thus, if we want to check the divisibility of (A1 A2 A3 . . . . . . .An), an 'n' digit number, we can simply add the numbers from the right, multiplying them with 1, 3, 2, 6, 4, 5, and so on. For example, if we want to check whether 125783 is divisible by 7, we will find the sum 3(1)+8(3)+7(2)+5(6)+2(4)+1(5)=84, which is divisible by 7, thus 125783 is divisible by 7.

I also found out another divisibility test for 7, in which we subtract twice the units digit of a number from the remaining number till we get a single digit, and if this number is divisible, the whole number is divisible. This can be proved by the fact that 10(a)+b==10(a-2b) in mod 7, where 'b' is the units digit, and 'a' the others.

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Nice, can you prove for \(7^n\)? BTW, may you use LaTeX?

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\[\overline { { x }_{ n }{ x }_{ n-1 }{ x }_{ n-2 }\dots { x }_{ 1 } } \equiv 0\quad \left( \text{mod 7} \right) \\ 10\overline { { x }_{ n }{ x }_{ n-1 }{ x }_{ n-2 }\dots { x }_{ 2 } } +{ x }_{ 1 }\equiv 0\quad \left( \text{mod 7} \right) \\ 50\overline { { x }_{ n }{ x }_{ n-1 }{ x }_{ n-2 }\dots { x }_{ 2 } } +5{ x }_{ 1 }\equiv 0\quad \left( \text{mod 7} \right) \\ \overline { { x }_{ n }{ x }_{ n-1 }{ x }_{ n-2 }\dots { x }_{ 2 } } +5{ x }_{ 1 }\equiv 0\quad \left( \text{mod 7} \right) \]

I haven't found out the test for \(7^n\) yet but it can probably be found out in a similar manner.

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For divisibility with 7 do the following : Suppose you have a no. abcde. Double the last digit i.e. 2e and subtract it from abcd (abcd - 2e), if this difference is divisible by 7, the no. is also. Keep on repeating the process with d,ç and b. You will get the answer.

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But can you prove it?

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