Um número de três dígitos é divisível por 7 se e somente se a soma do dobro do 1º dígito com o número formado pelos últimos dois dígitos der um resultado divisível por 7. Então, por exemplo, 938 é divisível por 7 porque 2 × 9 + 38 = 56 = 7 × 8.
O truque funciona porque a operação equivale a subtrair um múltiplo de 98 (e portanto de 7) do número original. Assim, se o resultado é divisível por 7, então o número original também é divisível por 7.
O truque pode ser estendido a números de qualquer comprimento: 229187 → 2 × 2291 + 87 = 4669 → 2 × 46 + 69 = 161 → 2 × 1 + 61 = 63 = 7 × 9.
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A three-digit number is evenly divisible by 7 if and only if twice its first digit added to the number formed by its two last digits gives a result that’s divisible by 7. So, for example, 938 is divisible by 7 because 2 × 9 + 38 = 56 = 7 × 8.
The trick works because the operation is equivalent to subtracting a multiple of 98 (and therefore of 7) from the original number. This way, if the result is divisible by 7, then the original number is also divisible by 7.
The trick can be extended to numbers of any length: 229187 → 2 × 2291 + 87 = 4669 → 2 × 46 + 69 = 161 → 2 × 1 + 61 = 63 = 7 × 9.
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(Fonte: Futility Closet, citando J. Kashangaki, “A Test for Divisibility by Seven,” Mathematical Gazette 80:487 [March 1996], 226.)
PCFilho
Nice.
ResponderExcluirThere are other methods for testing for divisibility by 7, of course, but this is probably the simplest one in the decimal (base 10) number system. However, there is one that is related to the test for divisibility by 9 and it involves changing the number base from base 10 to another number base. Here's the procedure, using the number 315 to illustrate:
First we need to express the number in the binary (base 2) number system:
1. Divide the number by 2 and set aside the remainder.
2. Repeat step 1 with the new number until the new number is 0.
3. Arrange the remainders in the reverse order from the order in which they were obtained.
So, with 315, here's what the successive divisions by 2 yield:
157 remainder 1
78 remainder 1
39 remainder 0
19 remainder 1
9 remainder 1
4 remainder 1
2 remainder 0
1 remainder 0
0 remainder 1
So, arranging these remainders in the reverse order, we get 100111011. This is what 315 looks like in base 2.
Next, we express the number in the octal (base 8) number system:
1. Partition the number into groups of three digits each, starting with the ones digit and proceeding to the left.
2. For each three digit group, replace it with a number from 0 to 7 in this way: 000=0, 001=1, 010=2, 011=3, 100=4, 101=5, 110=6, 111=7.
Here's what this looks like with 315, which in base 2 is 100111011:
Partitioning into groups of three digits each yields 100 111 011.
We replace 100 with 4, 111 with 7, and 011 with 3.
Therefore, 315 in base 8 is 473.
Alternatively, one can get to this point without converting to base 2 first by successively dividing by 8 and setting aside each remainder, but division by 2 is easier than division by 8, so that's why that step exists.
Now, we simply add the digits of the base-8 representation. If this sum is divisible by 7, the number is divisible by 7.
With 315, which in base 8 is 473, adding the digits of 473 gives 4 + 7 + 3 = 14, which is divisible by 7, so 315 is divisible by 7.
I just thought I should share that.
Very nice method, too, Jake!!
ExcluirThank you for sharing. :)