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A well-known simple trick for obtaining the REMAINDER after division by 3 is just the digital root of the dividend after division by 3, the digital root itself further simply obtained by first casting out 9's, then every digit pair that add up to 9, etc.

In the same vein, is there a short-cut method to rapidly obtain the QUOTIENT after division by 3?

Blue
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Beedassy Lekraj
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    No, you simply have to divide by $3$ to find out the quotient. In very special cases, the quotient can be obvious, but in general, there is no shortcut. – Peter Jul 30 '19 at 06:53

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Not sure if we can call it shortcut,

in base $10,$ Using Why is $a^n - b^n$ divisible by $a-b$? $$\dfrac{\sum_{r=0}^n10^ra_r}3=\sum_{r=0}^n\dfrac{10^r-1}3\cdot a_r+\dfrac{\sum_{r=0}^na_r}3$$

$$=\dfrac{\sum_{r=0}^na_r}3+\sum_{r=0}^n\underbrace{33\cdot33}_{r\text{ digits }}\cdot a_r$$

lab bhattacharjee
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  • In other words, the required quotient which is integer part of N/3 is just (digital sum of N)/3 + weighted digital sum of N by repdigits 333... Right ? – Beedassy Lekraj Jul 30 '19 at 07:11
  • @BeedassyLekraj, Right you are – lab bhattacharjee Jul 30 '19 at 07:14
  • Great ! For instance, 2020 ÷ 3 gives quotient [(2 + 0 + 2 + 0)/3] + (2×333) + (2×3) and remainder (2 + 0 + 2 + 0) mod 3, or (673 × 3) + 1, which is indeed correct ! – Beedassy Lekraj Jul 30 '19 at 07:28
  • In essence, this method simplifies computing the quotient after division by 3 to merely computing the quotient of digital sum of dividend after division by 3, and in general may be further iterated as many steps as may be deemed necessary ! [N/3] = [(digital sum of N)/3] + 3×(weighted sum of digits by appropriate repunits). – Beedassy Lekraj Jul 30 '19 at 08:44
  • BTW, the weighted sum of digits of dividend by repunits is actually equivalent to the sum of all successive digital truncatations from the right of dividend without the last end digit. For N = 2020, thus [2020/3] = 3×(202 + 20 + 2) + [(2 + 0 + 2 +0)/3] = 3×224 + 1 =673. – Beedassy Lekraj Jul 30 '19 at 10:30