This post is a sequel to the post How Mr. Bidder calculated logarithms published a few days ago. As with that post, this post is based on an excerpt from The Great Mental Calculators by Steven B. Smith.

Smith’s book says Arthur Benjamin squares large numbers using the formula

n² = (n + a)(n − a) + a²

where a is chosen to make the multiplication easier, i.e. to make n + a or n − a a round number. The method is then applied recursively to compute a², and the process terminates when you get to a square you have memorized. There are nuances to using this method in practice, but that’s the core idea.

The Great Mental Calculators was written in 1983 when Benjamin was still a student. He is now a mathematics professor, working in combinatorics, and is also well known as a mathemagician.

Major system

Smith quotes Benjamin giving an example of how he would square 4273. Along the way he needs to remember 184 as an intermediate result. He says

The way I remember it is by converting 184 to the word ‘dover’ using the phonetic code.

I found this interesting because I had not heard of anyone using the Major system (“the phonetic code”) in real time. This system is commonly used to commit numbers to long-term memory, but you’d need to be very fluent in the system to encode and decode a number in the middle of a calculation.

Maybe a lot of mental calculators use the Major system, or some variation on it, during calculations. Most calculators are not as candid as Benjamin in explaining how they think.

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George Parker Bidder (1806–1878) was a calculating prodigy. One of his feats was mentally calculating logarithms to eight decimal places. This post will explain his approach.

I’ll use “log” when the base of the logarithm doesn’t matter, and add a subscript when it’s necessary to specify the base. Bidder was only concerned with logarithms base 10.

Smooth numbers

If you wanted to calculate logarithms, a fairly obvious strategy would be to memorize the logarithms of small prime numbers. Then you could calculate the logarithm of a large integer by adding the logarithms of its factors. And this was indeed Bidder’s approach. And since he was calculating logarithms base 10, so he could make any number into a integer by shifting the decimal and then subtracting off the number of places he moved the decimal after taking the log of the integer.

Numbers with only small prime factors are called “smooth.” The meaning of “small” depends on context, but we’ll say numbers are smooth if all the prime divisors are less than 100. Bidder knew the logs of all numbers less than 100 and the logs of some larger numbers.

Rough numbers

But what to do with numbers that have a large prime factor? In this case he used the equation

log(a + b) = log(a(1 + b/a)) = log(a) + log(1 + b/a).

So if a number n doesn’t factor into small primes, but it is close to a number a that does factor into small primes, you’re left with the problem of finding log(1 + b/a) where b = n − a and the fraction b/a is small.

At this point you may be thinking that now you could use the fact that

log(1 + x) ≈ x

for small x. However, Bidder was interested in logarithms base 10, and the equation above is only valid for natural logarithms, logarithms base e. It is true that

log_e(1 + x) ≈ x

but

log₁₀(1 + x) ≈ log₁₀(e) x = 0.43429448 x.

Bidder could have used 0.43429448 x as his approximation, but instead he apparently [1] used a similar approximation, namely

log(1 + b/a) ≈ log(1 + 10^−m) 10^m b/a

where b/a is between 10^−m−1 and 10^−m. This approximation is valid for logarithms with any base, though Bidder was interested in logs base 10 and had memorized log₁₀1.01, log₁₀1.001, log₁₀1.0001, etc. [2]

Finding nearby smooth numbers

To get eight significant figures, the fraction b/a must be on the order of 0.001 or less. But not every number n whose log Bidder might want to calculate is so close to a smooth number. In that case Bidder might multiply n by a constant k to find a number so that kn is close to a smooth number, take the log of kn, then subtract log k.

Smith says in [1] “For example, to obtain log 877, he would multiply 877 by 13, which gives 11,401, or 600 × 19 + 1.” Then he could calculate

log(877) = log(13*877) −- log(13) = log(11400) + log(1/11400) − log(13)

and use his algorithm for approximating log(1/11400).

I can’t imagine thinking “Hmm. 877 isn’t close enough to a smooth number, but if I multiply it by 13 it will be.” But apparently Bidder thought that way.

Related posts

Here is a simple way to approximate logarithms to a couple significant figures without having to memorize anything.

See also this post on mentally calculating other functions to similar accuracy.

Notes

[1] This equation comes from The Great Mental Calculators by Steven B. Smith. Bidders methods are not entirely known, and Smith prefaces the approximation above by saying “it appears that Bidder’s approximation was …”.

[2]

|-------+-------------|
|     x | log_10(1+x) |
|-------+-------------|
| 10^-2 |  0.00432137 |
| 10^-3 |  0.00043408 |
| 10^-4 |  0.00004342 |
| 10^-5 |  0.00000434 |
| 10^-6 |  0.00000043 |
| 10^-7 |  0.00000004 |
|------+--------------|

Mentally calculating the day of the week will be especially easy in 2023. The five-step process discussed here reduces to three steps in 2023.

One of the steps involves leap years, and 2023 is not a leap year. Another step involves calculating and adding in the “year share,” and the year share for 2023 is zero. So the five steps reduce to these three steps.

1. Start with a constant corresponding to the month.
2. Add the day of the month.
3. Take the remainder by 7.

The constants for each month are as follows.

    | January | 6 | February | 2 | March     | 2 |
    | April   | 5 | May      | 0 | June      | 3 |
    | July    | 5 | August   | 1 | September | 4 |
    | October | 6 | November | 2 | December  | 4 |

You have to come up with your own mnemonics for memorizing this table. As I commented here,

It’s typically much easier to memorize something than to come up with a mnemonic that other people would find acceptable. No matter how natural your mnemonics sound to you, they usually sound like nonsense to anyone else.

Examples

To find the day of the week for July 4, 2023 we start with 5 for July, add 4, and get 9, which leaves a remainder of 2 when divided by 7. Days of the week are numbered starting with Sunday as 0, so 2 corresponds to Tuesday.

Pearl Harbor Day, December 7, will be on a Thursday next year because 4 + 7 leaves a remainder of 4 when divided by 7.

Doomsday

An alternative to the method above is John Conway’s “Doomsday” rule. This rule observes that that every year, the dates 4/4, 6/6, 8/8, 10/10, 12/12, 5/9, 9/5, 7/11, and 11/7 fall on the same day of the week. These dates are easy to memorize because they break down in double pairs of even digits—4/4, 6/6, 8/8, 10/10, and 12/12— and symmetric pairs of odd digits—5/9 and 9/5, 7/11 and 11/7.

For 2023 the Doomsday dates all fall on Tuesday. You can bootstrap your way to calculating other days of the week starting from these landmarks.

You can add “March 0”, i.e. the last day of February, to the list. In 2023 this is February 28 though of course it is sometimes February 29. And when March 0 is February 28, you can add February 0, i.e. January 31, to the list of Doomsday dates.

The Doomsday rule is less work up front than the rule above, but it’s more work in the long run.

Beyond 2023

In general you have to compute the year share in order to find the day of the week of a date, though in 2023 the year share is zero. This post gives multiple ways to compute the year share in your head.

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A couple weeks ago I wrote a post about the year share component of calculating the day of the week. To calculate the day of the week, you need to add the day of the week, a constant for the month, and the year share. Calculating year share is not that hard, but it’s the hardest step to carry out in your head.

The year share of a date y is

⌊5y/4⌋ % 7

where y is the last two digits of a year. There’s no need to multiply by 5, since you could compute

(y + ⌊y/4⌋) % 7

but this still takes some effort if y is large.

The earlier post gave nine different ways to compute the year share, and the easiest in my opinion is method #6:

1. Round y down to the most recent leap year and remember how much you had to round down. Call that amount r.
2. Let t be the tens value and u the units value.
3. Return (2t − u/2 + r) % 7.

In symbols,

1. Let y‘ = 4 ⌊y/4⌋ and r = y − y‘.
2. Let t = ⌊y’/10⌋ and u = y‘ % 10.
3. Return (2t − u/2 + r) % 7.

This method has a longer description but is easier to carry out mentally as the following examples illustrate.

Examples

Let y = 97. Here are the calculations required by the direct method.

1. ⌊y/4⌋ = 24
2. 97 + 24 = 121
3. 121 % 7 = 2

Here are the calculations required by the second method.

1. y‘ = 96 and r = 1.
2. t = 9 and u = 6.
3. (2*9 − 3 + 1) % 7 = 2

The former requires adding two 2-digit numbers. The latter requires doubling a single digit number and subtracting another single digit number.

The former requires reducing a 3-digit number mod 7 and the latter requires reducing a 2-digit number mod 7. Furthermore, the 2-digit number is never more than 18.

Proof

To prove that the two methods are equivalent we have to prove

⌊5y/4⌋ ≡ 2t − u/2 + r mod 7

which is kinda messy. The way t, u, and r are defined prevent this from being a simple algebraic calculation.

We can verify that

⌊5y/4⌋ ≡ 2t − u/2 + r mod 7

for y = 0, 1, 2, … 19 by brute force then prove the rest by induction. We’ll show that if the congruence holds for y then it holds for y + 20.

Suppose you increase y by 20. Then ⌊5y/4⌋ increases by 25. The value of t increases by 2, and the values of u and r remain unchanged, so the right side increases by 4. Since 25 % 7 = 4, the congruence still holds.

This post will be about psychology as much as math, looking at a number of algorithms for mentally calculating the same function.

The most difficult part of mentally computing days of the week is computing

⌊5y/4⌋ % 7

where y is the last two digits of a year. This quantity is called the year share because it is the year’s contribution to the day-finding algorithm.

The most obvious way to compute this function is to take the whole number part of y/4, add it to y, and take the remainder when dividing by 7. In Python, this would be

    def method0(y): return (y + y//4) % 7

The only reason there’s more to say is that the ease of description is not the same as ease of execution. There are several other methods that some people find easier to use.

Outline

In [1] the author gives several methods for computing the year share. Each of these methods has some advantage over the most direct approach. They generally work with smaller numbers and take advantage of operations that are mentally trivial such as splitting a number into its tens and ones digits.

I’ll describe one of these methods in words, the so-called “odd + 11” method, then implement all the methods in Python. Each method looks more complicated than the direct method if you just glance at the code. But if you mentally execute the code you can see why each method could be easier for a human to carry out.

You can find algebraic proofs of why each method works in [1]. Here I’ll use brute force: the methods only have to work for integer values of y from 0 to 99, so I’ll let Python verify that the methods all produce the same result. This brute force approach uncovered an error in the the paper that I correct in the code below.

Odd + 11 method

Here’s the “odd + 11” method:

1. If y is odd, add 11 to y.
2. Divide y by 2.
3. If y is odd after the step above, add 11.
4. Set y to the remainder when dividing y by 7.
5. Return 7 − y.

This method requires adding no more than 11, and it only requires one mental “register.” That is, all operations are done in place. With the direct method, you have to hold y and ⌊y/4⌋ in your head.

Python implementations

Here are Python implementations of 10 methods of computing a value congruent to ⌊5y/4⌋ mod 7. That is, the methods may produce different values, but the values differ by multiples of 7.

Note that the last three methods only require working with single-digit numbers.

Method #6 may be the easiest method to use. YMMV. (Update: See this follow-on post about this method.)

    from math import floor
    
    def method0(y):
        return y + y//4
    
    def method1(y):
        if y % 2 == 1:
            y += 11
        y /= 2
        if y % 2 == 1:
            y += 11
        y %= 7
        return 7 - y
    
    def method2(y):
        parity1 = y % 2
        if parity1:
            y -= 3
        y /= 2
        parity2 = y % 2
        if parity1 != parity2:
            y -= 3
        return -y 
    
    def method3(y):
        t = y % 12
        return y // 12 + t + t//4
    
    def method4(y):
        r = y % 4
        y -= r
        return r - y//2
    
    def method5(y):
        # placeholder to maintain the numbering in the paper
        return method0(y)
    
    def method6(y):
        r = y % 4
        y -= r # i.e. latest leap year
        t = y // 10
        u = y % 10
        return 2*t - u//2 + r
    
    def method7(y):
        t = y // 10
        u = y % 10
        x = (2*t + u) // 4
        x += u
        # The paper says to return 2t - x but it should be the opposite.
        return x - 2*t
    
    def method8(y):
        t = y // 10
        u = y % 10
        p = t % 2
        return (2*t + 10*p + u + (2*p + u)//4) % 7
    
    def method9(y):
        t = y // 10
        u = y % 10    
        return u - t + floor(u/4 - t/2)
    
    # verify all resultss are equal mod 7
    for y in range(100):
        for m in range(10):
            f = eval("method" + str(m))
            assert ((method0(y) - f(y)) % 7 == 0)

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[1] S. Kamal Abdali. Finding the year’s share in day-of-week calculations. Recreational Mathematics Magazine. Number 6, December 2016.