Problematics | The key to a calendar trick

A 20th-century trick in which a magician derives the sum of secretly chosen dates. How does he do it?

During the middle of the 20th century, a time when mathematical and card tricks thrived, Martin Gardner described scores of them in his writings. In some cases, he explained why and how a trick worked. In other cases, he described how but not why, which is good for a column like ours. You should enjoy working out the mechanics of the trick described below.

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Gardner credits Walter Gibson, a New York writer, with creating this trick, which uses a calendar. Gibson’s handling was slightly altered by the New York broker and puzzler Royal Heath, and Gardner’s own description is based on Heath’s. My description, in turn, is based on Gardner’s writings, embellished with two illustrative examples of my own.

#Puzzle 220.1

Puzzle 220.1

A magician presents the calendar for any month to a spectator, and asks him to secretly choose one date from each of the five rows. There are certain months for which the calendar has six rows. In those cases, the spectator is simply told to ignore the sixth row.

The magician asks the spectator 7 questions. How many Sundays are among the dates chosen? (Answer: 1). How many Mondays? How many Tuesdays? And so on, up to Saturday. The answers are 1 Sunday, 0 Mondays, 3 Tuesdays, 0 Wednesdays, 0 Thursdays, 1 Friday, and 0 Saturdays. From these answers, the magician derives the sum and announces it: 76.

How does he do it? The process requires a set of key numbers, and a series of calculations. First, the keys. The weekday keys are fixed: Monday = 1, Tuesday = 2, Wednesday = 3, Thursday = 4, Friday = 5, Saturday = 6. For Sunday, the key is different for different calendars. The magician determines it as follows. The column with the date 1 is assigned a value of 75 (in this case, it’s Tuesday). The magician goes left, one column at a time, reducing this value successively by 5 until he reaches Sunday. Thus, after taking Tuesday as 75, Monday is 70 and Sunday is 65.

When the spectator announces the number of selected days falling on each day of the week, the magician ignores the Sunday count. Sunday is already 65 in his mental checklist. Every other count is multiplied by the key for that date. Thus, three Tuesdays give him 3 x 2 = 6, and one Friday gives him 1 x 5 = 5. The total comes to 65 + (3 x 2) + (1 x 5) = 76. It matches the sum of the dates.

One example should be enough, but let me illustrate just one more, in the way a student revises what he or she has learnt. Say you are the spectator, and I am the magician who gives you the January 2027 calendar. You choose 2 from the first row, 5 from the second row, 13 from the third row, 18 from the fourth row, and 30 from the fifth row. The total is 2 + 5 + 13 + 18 + 30 = 68.

The weekday counts are 0 Sundays, 1 Monday, 1 Tuesday, 1 Wednesday, 0 Thursdays, 0 Fridays and 2 Saturdays. As a magician, I need only the 1 Monday, 1 Tuesday, 1 Wednesday, and 2 Saturdays. After multiplying them by their respective keys, I get (1 x 1) + (1 x 2) + (1 x 3) + (2 x 6) = 18. Next, I determine the Sunday key. The first date in January 2027 is Friday, so this column is assigned a value of 75. Moving left, the Thursday column is 70, Wednesday is 65, Tuesday is 60, Monday is 55, and Sunday is 50. So, as the magician, I add 50 + (1 x 1) + (1 x 2) + (1 x 3) + (2 x 6) = 68. This matches.

Why does this calculation with keys work each time? See if you can work it out mathematically.

#Puzzle 220.1

A chemist wants a diluted solution with 12% acid and 88% water. Her lab does not have concentrated acid but has two diluted solutions: one with 3% acid (for whatever purpose that may serve) and the other with 30% acid. She wants to create the 12% solution from these two alone, without using external amounts of acid or water.

In what proportions must the two solutions be mixed?

MAILBOX: LAST WEEK’S SOLVERS

#Puzzle 219.1

Hi,

There were originally 8 pairs of rabbits per colony pen. After six months, the number increased to 12 pairs per pen. The courtyard has 25 pens and the garden has 35 pens. This is determined as follows:

Let x be the original number of pairs in each colony pen, and n be the number of pens in the courtyard. Since there are 480 pairs altogether and 35 pens in the garden, the total number of pens is n + 35. After six months, with each pen having gained 4 pairs, the courtyard has 300 pairs in total. So:

x(n + 35) = 480 (equation 1)

n(x + 4) =300 (equation 2)

From (1), we get n = 480/x – 35. Substituting this for n in (2), we get:

(480/x − 35)(x + 4) = 300

Dividing by 5, multiplying throughout by x:

(96 – 7x)(x + 4) = 60x²

=> 7x² − 8x – 384 = 0

=> (7x + 48)(x − 8)=0

Thus, x = 8 (the negative solution is not meaningful), and

n = 300/(8 + 4) = 25

Total pens = 25 + 35 = 60, initial number of pairs = 60 x 8 = 480, and number of pairs in each pen after six months = 8 + 4 = 12.

— Ajay Ashok, New Delhi

#Puzzle 219.2

{3(280 + x}² = 72y904

=> 9(280 + x)² = 72y904

Since x is an integer, the number on the right must be divisible by 9. The sum of the given digits on the right = 7 + 2 + 9 + 0 + 4 = 22. To be divisible by 9, the sum of all six digits has to be the nearest multiple of 9, i.e. 27, so y = 27 - 22 = 5. Thus,

9(280 + x}² = 725904

=> (280 + x}² = 80656 = 284²

=> 280 + x = 284

=> x = 4

Therefore, x = 4 and y = 5.

— Kanwarjit Singh, Chief Commissioner of Income-tax, retired

Solved both puzzles: Ajay Ashok (Delhi), Kanwarjit Singh (Chief Commissioner of Income-tax, retired), Dr Sunita Gupta (Delhi), Yadvendra Somra (Sonipat), Professor Anshul Kumar (Delhi), Shishir Gupta (Indore), Sabornee Jana (Mumbai)

Solved Puzzle 219.2 : Anil Khanna (Fremont), Rajendra Gade (ex-ISRO, Ahmedabad), YK Munjal (Delhi), Kishore Rao (Bengaluru)

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