Clock arithmetic
Tonight's idea
Start at 9 o'clock and wait 5 hours. It is 2 o'clock. The clock throws away every full turn of 12 and keeps only what's left. Mathematicians write this as \(9 + 5 \equiv 2 \pmod{12}\), read "9 plus 5 is 2, mod 12". The word mod just means "keep the remainder after dividing by". We will mostly write the short form: \(17 \bmod 5 = 2\), because 17 is three 5s with 2 left over.
Days of the week are a mod 7 clock; minutes are mod 60.
The useful part: you can do the mod before multiplying or adding, not just after, and the answer comes out the same. That lets you reason about enormous numbers by working only with their remainders.
The last digit of \(7^{100}\). The last digit is the number mod 10, so multiply by 7 and keep only the last digit each time:
| \(7^1\) | \(7^2\) | \(7^3\) | \(7^4\) | \(7^5\) |
|---|---|---|---|---|
| 7 | 9 | 3 | 1 | 7 |
The last digits cycle 7, 9, 3, 1 with period 4. Since 100 is a multiple of 4, \(7^{100}\) lands at the end of a cycle: last digit 1.
Watch one: what day is it 1000 days after a Monday?
Every 7 days the weekday comes back around, so throw away the 7s: \(1000 = 142 \times 7 + 6\), so \(1000 \bmod 7 = 6\).
Six days after Monday is Sunday.
Here is the same trick explaining something you have probably noticed: \(365 \bmod 7 = 1\). That is why your birthday moves forward one weekday each year (two when a 29 February falls in between, because \(366 \bmod 7 = 2\)).
Your turn
- Work out \(38 \bmod 7\), \(100 \bmod 12\), and \(365 \bmod 7\).
- It is 10 o'clock. What time will it be in 100 hours? (Don't worry about a.m. or p.m.)
- Your birthday fell on a Tuesday this year. There is no 29 February before your next birthday. What weekday is your next birthday? And the one after that, if a 29 February comes in between?
- Find the last digit of \(7^4\) by actually multiplying it out, then find the last digit of \(2^{100}\) using the cycle trick. (Write out the cycle of last digits of \(2, 4, 8, 16, 32, \ldots\) first.) Then do \(3^{100}\).
- Casting out nines. Add the digits of a number, and keep adding until one digit is left. Call that its digit root. Fact: a number and its digit root leave the same remainder mod 9. Someone claims \(4321 \times 1234 = 5{,}332{,}124\). Check it: (a) find the digit root of 4321; (b) the digit root of 1234; (c) multiply (a) by (b) and take the digit root of that; (d) find the digit root of 5,332,124; (e) do (c) and (d) agree? If not, the multiplication is wrong.
- Why does casting out nines work? Start from this: \(10 = 9 + 1\), \(100 = 99 + 1\), \(1000 = 999 + 1\). So what is \(347 \bmod 9\) compared to \(3 + 4 + 7\)?
Stretch: a trick that reads a crossed-out digit
Ask a friend to write any number with several digits, say 7,415. Have them scramble its digits into a new number, say 1,547, and subtract the smaller from the larger. Tell them to cross out any nonzero digit of the answer and read you the digits that are left, in any order. You name the crossed-out digit instantly.
Why is the difference always a multiple of 9? (Casting out nines is the whole secret.) How do you recover the missing digit from the ones you hear? Why did we ban crossing out a zero?
Answers
- \(38 \bmod 7 = 3\), \(100 \bmod 12 = 4\), \(365 \bmod 7 = 1\).
- \(10 + 100 = 110\), and \(110 \bmod 12 = 2\). So 2 o'clock.
- Wednesday (one day on). Then Friday, because a leap year moves it two days.
- \(7^4 = 2401\), last digit 1. Powers of 2 end in 2, 4, 8, 6, 2, 4, 8, 6, … with period 4. 100 is a multiple of 4, so \(2^{100}\) is at the end of a cycle: last digit 6. Powers of 3 end in 3, 9, 7, 1, so \(3^{100}\) ends in 1.
- (a) 1 (4+3+2+1 = 10, then 1+0 = 1). (b) 1. (c) \(1 \times 1 = 1\). (d) 2 (5+3+3+2+1+2+4 = 20, then 2). (e) They disagree, so the claim is wrong. (The real product is 5,332,114, whose digit root is 1.)
- \(347 = 3 \times 100 + 4 \times 10 + 7 = 3 \times 99 + 4 \times 9 + (3 + 4 + 7)\). The 99 and 9 parts are multiples of 9, so \(347\) and \(3+4+7 = 14\) leave the same remainder mod 9 (both leave 5). The same works for any number because every power of 10 is one more than a multiple of 9.
Stretch. A number and its scrambled version have the same digits, so the same digit sum, so the same remainder mod 9. Their difference therefore has remainder 0: it is a multiple of 9, and so its digit root is 9 (or the number is 0). Add up the digits you are told; the missing digit is whatever brings the total up to the next multiple of 9. If the digits you hear already sum to a multiple of 9, the missing digit would be 0 or 9 and you could not tell which, so we ban zero and answer 9. Example: \(7415 - 1547 = 5868\); cross out the 6, hear 5, 8, 8 which sum to 21, next multiple of 9 is 27, missing digit 6.