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Lesson 3 of 4

Number Play · Lesson 3 of 4

Digits in Disguise

“Solve addition and multiplication cryptarithms using place value, carries and systematic reasoning.”

Learning Objectives

• Translate a cryptarithm into place-value equations. • Use column sums and carries to restrict possible digits. • Use the size of a product and its units digit to solve multiplication puzzles. • Check consistent assignments, distinct letter digits and leading-zero restrictions. • Find all solutions when a puzzle admits more than one assignment.

Imagine finding an addition on a page where some digits have been replaced by letters. The arithmetic is still correct, but the letters hide its meaning. You could try random digits, yet each column already gives you clues: a sum ending in zero may need a carry, and multiplying a two-digit number by eight cannot produce just any two-digit answer. Your task is to read these clues and use them to uncover the digits.

These puzzles combine ideas you have already used: place value, parity, divisibility and clear reasoning. Work from the strongest restriction, keep a record of each choice, and check the completed calculation. A correct numerical equality is necessary, but it must also obey every letter rule.

Definition
Cryptarithm

An arithmetic puzzle in which letters stand for digits. Every occurrence of the same letter has the same digit, different letters have different digits, and the leading digit of a multi-digit number cannot be zero.

Read the symbols carefully

AB denotes the two-digit number 10A + B, not A × B. A written digit such as the 2 in L2N stays fixed. The distinct-digit rule applies to different letters; a letter is not automatically forbidden from taking a digit printed elsewhere in the puzzle unless an extra rule says so. Carry symbols are working quantities, not additional puzzle letters.

Begin with the rightmost column when addition makes it useful. If the total is below 10, write that digit and carry nothing. If it is 10 or more, write its units digit and carry the remaining tens to the next column. Adding two digits creates a carry of at most 1; adding three digits can create a carry of 0, 1 or 2.

It helps to write an equation for each column. A units total of 1 + B ending in 0 means 1 + B = 10c, where c is the carry. The digit restrictions make c and B much easier to identify than if you treated the whole calculation as a guess.

A units column that forces the answer

Problem
Solve A1 + 1B = B0.

  1. 1.Units: 1 + B must end in 0. Since B is a digit, the only possible total is 10, giving B = 9 and carry 1.
  2. 2.Tens: A + 1 + 1 = B. Substitute B = 9 to obtain A + 2 = 9, so A = 7.
  3. 3.Check the whole addition: 71 + 19 = 90. A and B are different and neither leading digit is zero.
  4. 4.The units column forced B and the tens column then forced A, so the solution is unique.
A1 + 1B = B0carry 171+1990Units: 1 + 9 = 10 → write 0, carry 1Tens: 7 + 1 + 1 = 9
Column addition and the carry in A1 plus 1B
Use the carry to connect the columns

Problem
Solve AB + 37 = 6A.

  1. 1.Let c be the carry from the units. Then B + 7 = A + 10c, with c equal to 0 or 1.
  2. 2.The tens column gives A + 3 + c = 6, so A = 3 − c.
  3. 3.If c = 0, then A = 3 and B + 7 = 3, giving B = −4. That is not a digit, so reject this case.
  4. 4.If c = 1, then A = 2 and B + 7 = 12, giving B = 5.
  5. 5.Check 25 + 37 = 62. Both carry cases have been considered, so there are no other solutions.

A repeated-addition puzzle can be rewritten as multiplication. ON + ON + ON = PO means 3(10O + N) = 10P + O. This connection reduces the amount you need to write, but the letter positions still matter.

Do not assume a puzzle has only one solution because you found a working assignment. To find every solution, identify a small set of possible leading digits or carries and examine each one. A useful restriction for a two-digit product of three equal two-digit numbers is that the original number cannot exceed 33.

A puzzle with three solutions

Problem
Find all solutions of ON + ON + ON = PO.

  1. 1.The product is two-digit, so ON ≤ 33 and the leading digit O must be 1, 2 or 3. Let c be the carry from 3N.
  2. 2.The column equations are 3N = O + 10c and P = 3O + c, where c is 0, 1 or 2.
  3. 3.If O = 1, only c = 2 makes O + 10c divisible by 3: N = 21/3 = 7, and P = 5. This gives 17 + 17 + 17 = 51.
  4. 4.If O = 2, only c = 1 works: N = 12/3 = 4, and P = 7. This gives 24 + 24 + 24 = 72.
  5. 5.If O = 3, c = 0 works: N = 1 and P = 9. This gives 31 + 31 + 31 = 93.
  6. 6.All three assignments use distinct letter digits and valid leading digits. No other O is possible, and each carry was checked, so these are all the solutions.
Repeated units in a three-digit answer

Problem
Solve QR + QR + QR = PRR.

  1. 1.Write 3(10Q + R) = 100P + 11R. Rearranging gives 30Q − 8R = 100P, or 15Q − 4R = 50P.
  2. 2.Alternatively, inspect the units column: 3R = R + 10c, so 2R = 10c. Thus R = 0 with c = 0, or R = 5 with c = 1.
  3. 3.If R = 0, then 3Q = 10P. No Q from 1 to 9 makes this possible: it would need Q to be a multiple of 10.
  4. 4.If R = 5, the tens column requires 3Q + 1 = 10P + 5, or 3Q = 10P + 4.
  5. 5.The sum of three two-digit numbers is at most 297, so P is 1 or 2. P = 1 gives Q = 14/3, not a digit. P = 2 gives Q = 8.
  6. 6.Thus Q = 8, R = 5, P = 2. Check 85 + 85 + 85 = 255.

For multiplication puzzles, check the size of the answer before doing any column work. If PQ × 8 has only two digits, then PQ × 8 ≤ 99, so PQ ≤ 12. That leaves only 10, 11 and 12 as possible two-digit inputs.

The units column then provides another restriction. In GH × H = 9K, the units digit of the multiplicand must be the same digit as the multiplier. A perfectly correct calculation such as 16 × 6 = 96 still fails this puzzle: G = 1, H = 6, and K = 6, so H and K collide. Matching arithmetic alone is not enough.

A narrow size range

Problem
Solve PQ × 8 = RS.

  1. 1.Because the result is two-digit, PQ can only be 10, 11 or 12.
  2. 2.10 × 8 = 80 would give Q = 0 and S = 0, violating the different-letter rule.
  3. 3.11 × 8 = 88 already has P = Q, which is not allowed.
  4. 4.12 × 8 = 96 gives P = 1, Q = 2, R = 9 and S = 6. All four digits are different.
  5. 5.13 × 8 = 104 is already three-digit, and every larger input gives an even larger product. Therefore 12 × 8 = 96 is the unique solution.
Check a repeated letter in multiplication

Problem
Choose the solution of GH × H = 9K from 11 × 9 = 99, 12 × 8 = 96, 46 × 2 = 92, 24 × 4 = 96, 47 × 2 = 94, 31 × 3 = 93 and 16 × 6 = 96.

  1. 1.The multiplier must equal the units digit of GH. This immediately rejects 11 × 9, 12 × 8, 46 × 2, 47 × 2 and 31 × 3.
  2. 2.24 × 4 = 96 matches H = 4 in both places and uses G = 2, K = 6, all distinct.
  3. 3.16 × 6 = 96 matches the repeated H, but it gives H = K = 6. Different letters cannot share a digit.
  4. 4.Thus the valid choice is 24 × 4 = 96. To check beyond the list: H = 1 would require 91 × 1 = 91 and repeat H as K; H = 2 would require an input from 45 to 49, none ending in 2; H = 3 gives 33 × 3 = 99 but repeats G as H. H = 4 yields 24 × 4. H = 5 would need an input from 18 to 19, neither ending in 5. H = 6 yields only 16 × 6 and fails distinctness. H ≥ 7 makes the smallest possible GH × H at least 17 × 7 = 119.
A useful solving routine

Read the letter rules. Bound the leading digits using the number of digits in the result. Write units and carry equations. List the remaining candidates systematically. Reject a candidate as soon as it breaks a rule. Finally, substitute every letter into the complete calculation, and decide whether you have found one solution or every solution.

A three-digit multiplication with several clues

Problem
Solve BYE × 6 = RAY.

  1. 1.The input is three-digit and its product with 6 is below 1000. Therefore 100 ≤ BYE ≤ 166, so B = 1 and Y is at most 6.
  2. 2.The units digit Y of 6E is even, so Y can be 0, 2, 4 or 6. We now examine these four cases.
  3. 3.Y = 0: E is 0 or 5. E = 0 would repeat Y, so E = 5. This gives 105 × 6 = 630, with R = 6 and A = 3. The five letters have digits 1, 0, 5, 6, 3, all distinct.
  4. 4.Y = 2: E is 2 or 7. Reject E = 2 because E = Y. With E = 7, 127 × 6 = 762, which gives R = E = 7 and is invalid.
  5. 5.Y = 4: E is 4 or 9. Reject E = 4. With E = 9, 149 × 6 = 894, which gives A = E = 9 and is invalid.
  6. 6.Y = 6: E is 1 or 6. E = 1 repeats B, and E = 6 repeats Y. Both fail.
  7. 7.Therefore the only solution is 105 × 6 = 630. Zero is allowed for Y because it is not the leading digit of either number.
105 × 6 = 630Units: 5 × 6 = 30 → write 0, carry 3Tens: 0 × 6 + 3 = 3 → write 3Hundreds: 1 × 6 = 6 → write 6
A carry in multiplication links the next column

Sometimes the same last digits appear on both sides of a multiplication. Instead of testing all digits, write the numbers using place value and subtract their shared part. For UT × 3 = PUT, the number PUT is 100P + UT. This makes the repeated two-digit ending especially useful.

The same two-digit ending

Problem
Solve UT × 3 = PUT.

  1. 1.Let x be the two-digit number UT. Then 3x = 100P + x.
  2. 2.Subtract x: 2x = 100P, so x = 50P.
  3. 3.Since 10 ≤ x ≤ 99 and P is a positive digit, only P = 1 is possible. Then x = 50.
  4. 4.Thus U = 5, T = 0, P = 1, and 50 × 3 = 150. All letter digits are different.
A leading digit that reappears

Problem
Solve AB × 5 = BC.

  1. 1.The answer is two-digit, so 10 ≤ AB ≤ 19. Hence A = 1.
  2. 2.Expand: 5(10 + B) = 10B + C, giving C = 50 − 5B.
  3. 3.For C to lie from 0 to 9, B must be 9: B = 8 would give C = 10, and smaller B gives still larger values.
  4. 4.Therefore C = 5, and 19 × 5 = 95. The digits for A, B and C are distinct.
A fixed printed digit and two solutions

Problem
Find all solutions of L2N × 2 = 2NP.

  1. 1.The result begins with 2 and is three-digit, so the input lies from 100 to 149. Thus L = 1.
  2. 2.Expand both numbers: 2(120 + N) = 200 + 10N + P. Rearranging gives P = 40 − 8N.
  3. 3.Since 0 ≤ P ≤ 9, we need 31 ≤ 8N ≤ 40. The only integer digits possible for N are 4 and 5.
  4. 4.N = 4 gives P = 8 and 124 × 2 = 248. N = 5 gives P = 0 and 125 × 2 = 250.
  5. 5.Both use distinct digits for L, N and P and valid leading digits. Hence both are solutions.
A units digit moves to the end of the product

Problem
Solve XY × 4 = ZX.

  1. 1.The result is two-digit, so XY ≤ 24. Therefore X is 1 or 2.
  2. 2.If X = 1, then 4(10 + Y) must end in 1. This is impossible because a multiple of 4 is even.
  3. 3.So X = 2 and Y lies from 0 to 4. The units product 4Y must end in 2, which forces Y = 3 in this range.
  4. 4.Then 23 × 4 = 92, so Z = 9. All three letter digits are different.
Repeated-digit factors

Problem
Find all solutions of PP × QQ = PRP.

  1. 1.PP = 11P and QQ = 11Q. Also PRP = 101P + 10R. Thus 121PQ = 101P + 10R.
  2. 2.The units digit of the product is P. Therefore P(Q − 1) is divisible by 10. Since 1 ≤ P ≤ 9 and 0 ≤ R ≤ 9, the equation also gives 121Q = 101 + 10R/P ≤ 191. Hence Q = 1.
  3. 3.Substitute Q = 1: 121P = 101P + 10R, giving 20P = 10R and R = 2P.
  4. 4.The digit restriction R ≤ 9 gives P = 1, 2, 3 or 4. P = 1 repeats Q, so reject it.
  5. 5.The remaining answers are 22 × 11 = 242, 33 × 11 = 363 and 44 × 11 = 484. Each uses distinct P, Q and R.
  6. 6.The order of the factors matters to the letter pattern: swapping PP and QQ also changes which letter must occupy the first and last positions of the answer.
A product with three identical digits

Problem
Solve JK × 6 = KKK.

  1. 1.Expand: 6(10J + K) = 111K.
  2. 2.This gives 60J + 6K = 111K, so 60J = 105K. Divide by 15 to obtain 4J = 7K.
  3. 3.Thus K must be a multiple of 4. Since K is a nonzero digit, try 4 and 8.
  4. 4.K = 4 gives J = 7. K = 8 gives J = 14, which is not a digit.
  5. 5.The unique solution is 74 × 6 = 444.

The next two challenges use the same tools, with a little more organisation. When a repeated-digit answer is divisible by 3, 9 or 11, that divisibility can restrict a letter. In other cases, the units column is so restrictive that it determines the leading digit immediately.

Use a repeated-digit product to limit the multiplier

Problem
Solve EF × E = GGG.

  1. 1.Since GGG = 111G, write E(10E + F) = 111G. The product is three-digit, and E is a nonzero digit.
  2. 2.If E = 1 or 2, the largest products are 19 and 58, so neither works. For E from 3 to 9, we can test the narrow product interval E(10E) through E(10E + 9).
  3. 3.E = 3 gives products from 90 to 117. The only repeated-digit candidate is 111, reached by 37 × 3.
  4. 4.E = 4 gives 160–196 and E = 5 gives 250–295; neither interval contains a repeated-digit number. E = 6 gives 360–414, also none.
  5. 5.E = 7 gives 490–553, also none. E = 8 gives 640–712, containing 666, but 666 is not divisible by 8. E = 9 gives 810–891, containing 888, but 888 is not divisible by 9.
  6. 6.Thus E = 3, F = 7, G = 1 is the only assignment: 37 × 3 = 111.
A four-digit product with a repeated outer letter

Problem
Solve WOW × 5 = MEOW.

  1. 1.The units column requires 5W to end in W. Thus 4W is divisible by 10, giving W = 0 or 5. W leads WOW, so W = 5.
  2. 2.Units: 5 × 5 = 25, so write 5 and carry 2. The tens column requires 5O + 2 to end in O. Hence 4O + 2 is divisible by 10.
  3. 3.Testing the possible digits gives O = 2 or 7. If O = 2, the number is 525 and its product is 2625; this makes M = O = 2, so reject it.
  4. 4.If O = 7, the number is 575 and 575 × 5 = 2875. This gives M = 2 and E = 8.
  5. 5.The digits W = 5, O = 7, M = 2 and E = 8 are distinct, so 575 × 5 = 2875 is the solution.
Check your reasoning, not just the answer

After solving a puzzle, ask: Did I preserve every repeated letter? Did any different letters receive the same digit? Is every leading digit nonzero? Did I include each carry? Did I examine all candidates before saying the solution is unique? A rejected trial is useful when you can explain exactly why it fails.

Quiz

Quick check

In a cryptarithm, what does AB mean?

Quick check

Which use of zero is forbidden by the usual cryptarithm rules?

Quick check

Why does 16 × 6 = 96 fail the puzzle GH × H = 9K?

Quick check

Why can PQ in PQ × 8 = RS not exceed 12?

Quick check

What should you conclude after finding one valid assignment?

Answer: 10A + B. The letters occupy the tens and units positions, so place value gives 10A + B.

Practice Problems

Practice Problems
  1. Solve A2 + 2B = B1.
  2. Solve AB × 4 = BC.
  3. Find all solutions of A3 + 3B = BA.
  4. Find all solutions of AA × B = CCC.
  5. Find all solutions of AB + AB + AB = CA and explain why your list is complete.

Units: 2 + B must end in 1, so it must be 11. Thus B = 9 and the carry is 1. Tens: A + 2 + 1 = 9, so A = 6. Check 62 + 29 = 91. A and B differ and no leading digit is zero.

Key Takeaways

Key Takeaways

• A cryptarithm must satisfy both arithmetic and the rules for repeated, distinct and leading digits. • Translate a digit string using place value: AB = 10A + B. • Column equations reveal carries and sharply reduce the possibilities. • The number of digits in a product often bounds its leading digits before any trial is needed. • Use parity and divisibility to reject impossible cases with reasons. • One valid solution does not prove uniqueness; list and check every remaining case.

Previous · Lesson 2

Checking Divisibility Quickly

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Chapter Summary and Practice