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Emma 27/02/2018 at 09:24
We have:
\(\overline{a_na_{n-1}...a_3a_2a_1}=\overline{a_na_{n-1}..a_4}.1000+\overline{a_3a_2a_1}\)
Since \(1000⋮4\) => \(\overline{a_na_{n-1}..a_4}.1000\) \(⋮\) \(4\).
\(\overline{a_na_{n-1}...a_3a_2a_1}\) \(⋮\) \(4\) iff \(\overline{a_3a_2a_1}\) \(⋮\) \(4\)
In other word: a number is divisibility by 4 if only if the number formed by its three right digits is divisibility by 4.
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¤« 28/02/2018 at 16:04
Another way:
We have: ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯xnxn−1...x3x2x1=100.¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯xnxn−1...x3+¯¯¯¯¯¯¯¯¯¯¯x2x1
Since 100⋮4
=>100.¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯xnxn−1...x3⋮4
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯xnxn−1...x3x2x1⋮4
=>¯¯¯¯¯¯¯¯¯¯¯x2x1⋮4
In other word: a number is divisibility by 4 when the last three digits on the right-hand is divisibility by 4
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FC Alan Walker 28/02/2018 at 08:38
Another way:
We have: \(\overline{x_nx_{n-1}...x_3x_2x_1}=100.\overline{x_nx_{n-1}...x_3}+\overline{x_2x_1}\)
Since \(100⋮4\)=>\(100.\overline{x_nx_{n-1}...x_3}⋮4\)
\(\overline{x_nx_{n-1}...x_3x_2x_1}⋮4\)=>\(\overline{x_2x_1}⋮4\)
In other word: a number is divisibility by 4 when the last three digits on the right-hand is divisibility by 4