- 2 It Turns Out That An Integer Written As Usual In Base 10 Is Divisible By 3 If And Only If Its Sum Of Digits Is Divi 1 (35.83 KiB) Viewed 38 times
2: It turns out that an integer (written as usual in base-10) is divisible by 3 if and only if its sum of digits is divi
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2: It turns out that an integer (written as usual in base-10) is divisible by 3 if and only if its sum of digits is divi
2: It turns out that an integer (written as usual in base-10) is divisible by 3 if and only if its sum of digits is divisible by 3 . (Also, you can use this recursively: For instance, 59388 is divisible by 3 because 5+9+3+8+8=33 and 3+3=6.) Please prove that this trick works. Hint: Note that 10=1mod3 and observe that any integer written in base-10 is simply a sum of powers of 10 . The proof is then short, and just a matter of setting notation correctly.
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