13.6 Decoding Divisibility Tricks

Diya shows Aarav a trick:

  1. Choose a 2-digit number with different digits, say 38. Keep it hidden.
  2. Swap the digits: 83.
  3. Find the difference: 83 - 38 = 45.
  4. Divide by 9: 45 \div 9 = 5, no remainder!

It works every time. Why is the difference always divisible by 9?

Write the number using place value. If the tens digit is a and the units digit is b, the number is 10a + b; swapped it becomes 10b + a. Taking the larger minus the smaller (say b > a):

(10b + a) - (10a + b) = 10b - b - 10a + a = 9b - 9a = 9(b - a).

The difference is 9 times the gap between the digits, so it is forever a multiple of 9. (If a > b, the subtraction just turns around and gives 9(a - b), still a multiple of 9.)

The Secret Behind the Magic Nearly every digit trick hides behind place value. Writing a two-digit number as 10a + b (and a three-digit number as 100a + 10b + c) converts “magic” into ordinary algebra you can prove.

Enrichment: A Self-Working Card Number Trick Here is a brand-new trick to add to your collection, together with its proof.

The steps. Ask a friend to: 1. think of a number; 2. add 5; 3. double the result; 4. subtract 4; 5. halve it; 6. take away the number they first thought of.

The claim. The answer is always 3.

The proof. Let the chosen number be x:

x \xrightarrow{+5} x+5 \xrightarrow{\times 2} 2x+10 \xrightarrow{-4} 2x+6 \xrightarrow{\div 2} x+3 \xrightarrow{-x} 3.

The x vanishes at the last step, so 3 appears for every starting number. To make the answer land on any target k instead, replace “add 5” with “add k+2”: then step 4 leaves 2x + 2k, halving gives x + k, and subtracting x leaves k. You now own a whole family of tricks.

Figure it Out: Divisibility & Word Problems

Practice

  1. In the swap-and-subtract trick, what is the quotient after dividing by 9? How does it relate to the two digits?
  2. Instead of the difference, take the sum of the number and its reverse (e.g. 46 + 64 = 110, 27 + 72 = 99, 15 + 51 = 66). These are all divisible by 11. Always true? Justify with algebra.
  3. Take any 3-digit number abc = 100a + 10b + c. Cycle the digits to form bca and cab, then add all three. Show the sum is always divisible by 37. Is it always divisible by 3 too? (Hint: look at multiples of 37.)
  4. Take any 3-digit number abc and repeat it to make the 6-digit number abcabc. Divide by 7, then by 11, then by 13. What happens? Why? (Hint: multiply 7 \times 11 \times 13.)
  5. There are 3 shrines, each with a magical pond in front that doubles any flowers dipped in it. A pilgrim dips all his flowers in pond 1, leaves some at shrine 1; dips the rest in pond 2, leaves some at shrine 2; dips the rest in pond 3, leaves all at shrine 3. He left an equal number at each shrine and finished with none. How many flowers did he start with, and how many at each shrine?
  6. A pond has buffaloes and ducks. There are 70 heads in all and 196 legs. How many buffaloes and how many ducks? Can you solve it without letters? (Hint: if all 70 were ducks, how many legs?)
  7. A father is 7 times his son’s age. In 4 years he will be 3 times his son’s age. How old is the son now?
  8. Latika and Sunita are cowherds. Latika says, “You have twice as many cows as I do.” Sunita replies, “True, but if I gave you four of mine, we’d have the same number.” How many cows does each have?
  9. A chai stall costs ₹6000 per day to rent, and each cup of tea costs ₹5 to make.  (i) If you sell 200 cups a day, what selling price gives a profit of ₹3000?  (ii) If customers pay only ₹45 per cup, how many must you sell for a ₹3000 profit?
  10. Evaluate \dfrac{1}{3},\ \dfrac{1+3}{5+7},\ \dfrac{1+3+5}{7+9+11}. What do you notice, and why? (Hint: sum of the first n odd numbers.)
  11. Reena and the Genie. A genie tells Reena that each time she circles a banyan tree her coins double, but she must hand the genie 6 coins per round. Reena circles three times; after the third doubling she has exactly 6 coins, just enough to pay the genie, leaving her nothing.  (i) How many coins did Reena start with?  (ii) For what cost per round would Reena actually gain coins?  (iii) How should the genie set the cost so it ends up with all of Reena’s coins?
  12. A two-digit number is 4 times the sum of its digits. The units digit is 2 more than the tens digit. Find the number.
  1. The difference is 9(b-a), so the quotient is b - a, the (positive) gap between the two digits. For 83-38=45, the quotient 5 = 8 - 3.
  2. Sum = (10a+b) + (10b+a) = 11a + 11b = \mathbf{11(a+b)}, always a multiple of 11. ✓
  3. abc + bca + cab = (100a+10b+c) + (100b+10c+a) + (100c+10a+b) = 111(a+b+c). Since 111 = 3 \times 37, the sum is divisible by both 37 and 3.
  4. abcabc = abc \times 1001, and 1001 = 7 \times 11 \times 13. So dividing in turn by 7, 11, 13 lands you right back on the original 3-digit number abc.
  5. Let the start be s flowers and x the number left at each shrine. After pond 1: 2s, leave x → 2s - x. After pond 2: 2(2s-x) = 4s - 2x, leave x → 4s - 3x. After pond 3: 2(4s-3x) = 8s - 6x, leave x, end with 8s - 7x = 0. So 8s = 7x; the smallest whole-number solution is s = 7,\ x = 8. He started with 7 flowers and left 8 at each shrine (check: 7 \to 14, -8 = 6 \to 12, -8 = 4 \to 8, -8 = 0 ✓).
  6. Let b = buffaloes, d = ducks. Heads: b + d = 70. Legs: 4b + 2d = 196. From the first, d = 70 - b; substitute: 4b + 2(70-b) = 196 \Rightarrow 2b = 56 \Rightarrow b = 28, so d = 42. 28 buffaloes, 42 ducks. Without letters: if all 70 were ducks there’d be 140 legs; the extra 196 - 140 = 56 legs come from buffaloes having 2 legs more than a duck, so 56 \div 2 = 28 buffaloes.
  7. Let the son be s; father = 7s. In 4 years: 7s + 4 = 3(s + 4) = 3s + 12 \Rightarrow 4s = 8 \Rightarrow s = 2. The son is 2 (father 14; in 4 years 18 = 3 \times 6 ✓).
  8. Let Latika have g, so Sunita has 2g. After Sunita gives 4: g + 4 = 2g - 4 \Rightarrow g = 8. Latika 8 cows, Sunita 16 cows.
  9. Daily fixed cost ₹6000.  (i) Cost of 200 cups = 200 \times 5 = ₹1000; total cost ₹7000. For ₹3000 profit, revenue must be ₹10000, so price $= 10000 = $ ₹50 per cup.  (ii) Profit $= 45n - (6000 + 5n) = 40n - 6000 = 3000 40n = 9000 n = $ 225 cups.
  10. Each fraction equals \dfrac{1}{3}: \dfrac{4}{12} = \dfrac13, \dfrac{9}{27} = \dfrac13. The numerator is the sum of the first k odd numbers = k^2, and the denominator is the sum of the next k odd numbers = (2k)^2 - k^2 = 3k^2, so every fraction is \dfrac{k^2}{3k^2} = \dfrac13.
  11. Let Reena start with c coins. Round 1: 2c - 6. Round 2: 2(2c-6) - 6 = 4c - 18. Round 3: 2(4c-18) - 6 = 8c - 42, which equals 6: 8c - 42 = 6 \Rightarrow c = 6.  (i) She started with 6 coins.  (ii) Doubling c adds c coins, so she gains only if the cost per round is less than the current number of coins c (that is, less than what she held at the start of that round).  (iii) The genie should charge exactly c per round, the doubling adds c and the charge removes c, so Reena’s count never grows, and across the rounds the genie scoops up everything.
  12. Let the tens digit be a and the units digit be b, so the number is 10a + b with digit-sum a + b. The conditions give 10a + b = 4(a + b) and b = a + 2. The first becomes 10a + b = 4a + 4b \Rightarrow 6a = 3b \Rightarrow b = 2a. With b = a + 2 this means 2a = a + 2 \Rightarrow a = 2, so b = 4. The number is 24 (check: 24 = 4 \times 6, and 4 = 2 + 2 ✓).