BTT Mathematics / Primary Mathematics Learning Hub / Cryptarithms and Digit Puzzles
A cryptarithm hides one or more digits inside an arithmetic relationship. The puzzle is solved not by guessing a pretty digit but by keeping place value, operation rules and every stated restriction active at the same time. When the same letter appears twice, it represents the same digit each time unless the puzzle explicitly says otherwise.
For example, if 3A + A = 44, the first term means thirty plus A, not three times A. The equation is 30+A+A=44. Therefore 2A=14 and A=7. A quick return check gives 37+7=44.
In an alphametic, letters usually stand for digits and different letters are normally required to represent different digits. A leading letter in a multi-digit number cannot be zero unless the rules explicitly allow a leading zero. These conventions must be stated rather than silently assumed. This guide uses them only when the question says so.
The deeper purpose is place-value reasoning. A two-digit number AB means 10A+B. Its reversal BA means 10B+A. A repeated digit AA means 11A. Those translations reveal relationships that are hard to see when a puzzle is treated as a visual trick.
This guide is Primary Mathematics enrichment. It builds on Place Value and Regrouping, Patterns, Early Algebra and Equations and Simultaneous Conditions. It is not a statement that formal alphametics are required in every Primary syllabus. Use the MOE Primary curriculum page and the learner’s school programme to separate required work from extension.
Translate the digits · Use columns and carries · Use constraints · Search systematically · 24 questions · Worked answers · Teaching and transfer
1. Translate a digit pattern before calculating
If A is a digit, then A can take an integer value from zero through nine. If AB is stated to be a two-digit number, A must be from one through nine and B may be zero through nine. The place-value translation is AB=10A+B.
Worked example A: A hidden units digit
Suppose 4A−17=26. Translate 4A as 40+A. Then 40+A−17=26, so 23+A=26 and A=3. Check 43−17=26.
Worked example B: A repeated digit
If AA+27=82, then AA=11A. Thus 11A=55 and A=5. The repeated digit forms fifty-five, not ten plus five or five times five.
Worked example C: Reversal
AB−BA=27. Translating gives (10A+B)−(10B+A)=9A−9B=27. Therefore A−B=3. Several pairs are possible: 41−14, 52−25, 63−36 and so on. The equation gives a relationship, not automatically one unique answer.
That distinction matters. A puzzle can be correctly solved by reporting a complete set of possibilities when the conditions do not determine one value.
Worked example D: A condition makes the answer unique
Suppose A+B=10 and A−B=2. Add the two relationships to get 2A=12, so A=6. Then B=4. The pair is unique because two independent conditions are active.
2. Column arithmetic reveals carries and borrows
In a longer cryptarithm, solve from the units column while recording any carry into the next place. The carry is part of the state. Ignoring it can make a digit seem possible in one column while breaking the whole sum.
Worked example E: A carry appears
Consider 58+6A=1B3, where A and B are digits. In the units column, 8+A must end in 3. The only digit possibility is A=5 because 8+5=13, giving a carry of one. In the tens column, 5+6+1=12, so B=2 and the hundreds digit is one. The completed sum is 58+65=123.
Worked example F: A carry is impossible
Suppose a puzzle says 3A×2=A6. Direct translation gives 2(30+A)=10A+6. This simplifies to 54=8A, so A would have to be 6.75. Since A must be a digit, there is no solution.
A no-solution result is legitimate. Do not round a digit or alter the printed arithmetic to manufacture an answer.
Borrowing in subtraction
For AB−BA=27 with A>B, the units column B−A needs borrowing. After borrowing, B+10−A is the units digit seven. The tens column becomes A−1−B=2. Both conditions reduce to A−B=3, matching the place-value algebra. Two representations verify the same structure.
3. Each restriction removes candidates
A well-formed digit puzzle may state that different letters represent different digits, that a letter is non-zero, that a result is even or that a sum has a fixed number of digits. Treat each statement as a constraint.
Worked example G: Reversal plus a total
Suppose AB+BA=121 and A>B, with both letters non-zero. Translation gives 11(A+B)=121, so A+B=11. The ordered possibilities with A>B are (9,2), (8,3), (7,4) and (6,5). Four answers remain.
If another clue says A is even, only (8,3) and (6,5) remain. If A is also a multiple of three, only (6,5) remains. A clue is useful because of how it changes the candidate set.
Worked example H: Distinct letters
AA+BB=99 with A and B different non-zero digits gives 11(A+B)=99, so A+B=9. If order does not matter, the digit pairs are (1,8), (2,7), (3,6) and (4,5). Pair (0,9) is excluded because AA would not be a two-digit number under the stated non-zero convention.
Leading zero needs an explicit rule
If AB is described as a two-digit number, A=0 is not permitted because 0B is just the one-digit number B in ordinary notation. However, a two-character code such as 04 may legitimately begin with zero. The object being represented determines whether leading zero is allowed.
Literal numerals are not automatically different from letter digits
In 7A+A7=154, the letter A may equal seven unless a rule says letters must differ from printed numeral digits. Substitution A=7 gives 77+77=154. Do not invent a distinctness condition that the puzzle never stated.
4. A systematic search is evidence, not random guessing
Some puzzles reduce immediately to one equation. Others leave a short candidate set. When a search is needed, organise it by a useful digit and stop only after every permitted value has been tested.
Worked example I: Consecutive reversal pairs
AB+9=BA. Translation gives 10A+B+9=10B+A, so B=A+1. With both AB and BA two-digit numbers, A can be one through eight. The complete solutions are 12, 23, 34, 45, 56, 67, 78 and 89.
Finding 12 is not enough because the question’s conditions allow seven more. The equation itself supplies the stopping rule.
Worked example J: A digit search collapses to one value
4B+B4=99. Translation gives 44+11B=99, so B=5. The completed arithmetic is 45+54=99. A search from zero through nine would also find five, but the place-value equation explains why no other digit can work.
Worked example K: Check the answer in the original layout
Suppose A3+2B=57. Translation gives 10A+B=34, so A=3 and B=4. Return to the original: 33+24=57. This catches a common mistake in which A and B are solved correctly but attached to the wrong positions.
Uniqueness needs an argument
If a puzzle asks “find the digit,” a complete solution should explain why only one digit survives. Solving an equation, exhausting a finite candidate set or proving that all alternatives violate a carry condition can establish uniqueness. “This digit works” establishes only existence.
Cross-check with bounds
A two-digit number multiplied by three lies between thirty and 297. If the printed result is two digits, that immediately restricts the first digit. Simple magnitude checks can remove impossible cases before exact calculation.
5. Practice: 24 original questions
Unless a question says otherwise, letters represent digits from 0 to 9, and a leading letter in a two-digit number is non-zero. The same letter represents the same digit each time. Different letters are not required to be different unless stated.
Questions 1–8: Translate the notation
1. A+7=15. Find A.
2. 3A+A=44. Here 3A is the two-digit number with tens digit 3. Find A.
3. AB+BA=99, with A>B and both digits non-zero. List all ordered pairs (A,B).
4. 2A×3=81, where 2A is a two-digit number. Find A.
5. A5+27=82. Find A.
6. 4A−17=26. Find A.
7. AA+27=82. Find A.
8. A+B=10 and A−B=2. Find A and B.
Questions 9–16: Reversal, repetition and feasibility
9. AB+9=BA, with both numbers two-digit. List every possible AB.
10. AB+BA=121, with A>B and both digits non-zero. List every ordered pair (A,B).
11. AB−BA=27, with A>B and both numbers two-digit. List every possible AB.
12. 1A+A1=66. Find A.
13. 2A+A2=99. Find A.
14. 3A×2=A6. Does a digit A exist? Justify.
15. A+A+A=27. Find A.
16. AB+AB=144. Find the two-digit number AB.
Questions 17–24: Multiple letters and complete solution sets
17. A3+2B=57. Find A and B.
18. 4A+B6=100. Find A and B.
19. A7+2A=80. Does a digit A exist?
20. 4B+B4=99. Find B.
21. AB+BA=110, with A and B distinct non-zero digits. List all ordered pairs (A,B).
22. AA+BB=99, with A and B distinct non-zero digits. List the unordered pairs with A<B.
23. A3+2B=61. Find A and B.
24. 7A+A7=154. Find A. No rule says A must differ from the printed digit 7.
6. Worked answers
Answers 1–8
1. A=8. Undo the addition: 15−7=8. Check 8+7=15.
2. A=7. 30+A+A=44, so 2A=14. Check 37+7=44.
3. (5,4), (6,3), (7,2), (8,1). AB+BA=11(A+B)=99, so A+B=9. With A>B and both positive, these four pairs exhaust the possibilities.
4. A=7. The number 2A is 27 because 81÷3=27.
5. A=5. A5=82−27=55.
6. A=3. 4A=26+17=43.
7. A=5. AA=55, so the repeated digit is five.
8. A=6, B=4. Adding the equations gives 2A=12. Then B=10−6=4.
Answers 9–16
9. 12, 23, 34, 45, 56, 67, 78, 89. The equation reduces to B=A+1. A can be one through eight.
10. (6,5), (7,4), (8,3), (9,2). A+B=11 and A>B. These are the four positive digit pairs.
11. 41, 52, 63, 74, 85, 96. AB−BA=9(A−B)=27, so A−B=3. B must be non-zero because BA is two-digit.
12. A=5. 10+A+10A+1=66 gives 11A=55.
13. A=7. 20+A+10A+2=99 gives 11A=77.
14. No solution. 2(30+A)=10A+6 gives 54=8A, so A=6.75, not a digit.
15. A=9. Three equal A digits total twenty-seven, so A=9.
16. AB=72. Two copies of AB make 144, so AB=72. Thus A=7 and B=2.
Answers 17–24
17. A=3, B=4. 10A+3+20+B=57 gives 10A+B=34.
18. A=4, B=5. 40+A+10B+6=100 gives 10B+A=54.
19. No solution. 10A+7+20+A=80 gives 11A=53, which has no digit solution.
20. B=5. 40+B+10B+4=99 gives 11B=55. Check 45+54=99.
21. (1,9), (2,8), (3,7), (4,6), (6,4), (7,3), (8,2), (9,1). A+B=10. Pair (5,5) is excluded because the digits must be distinct.
22. (1,8), (2,7), (3,6), (4,5). A+B=9 and A<B. Pair (0,9) is excluded by the non-zero condition.
23. A=3, B=8. 10A+3+20+B=61 gives 10A+B=38.
24. A=7. 70+A+10A+7=154 gives 11A=77. The completed sum is 77+77=154.
7. Teaching and transfer
If a learner treats AB as A×B, return to place-value cards. Build a two-digit number physically as A tens and B ones. Then translate it to 10A+B. The puzzle notation becomes compressed place value rather than a mysterious code.
When the learner guesses repeatedly
Ask what one column or one equation already restricts. A systematic table of A from zero through nine is acceptable when the set is small, but each candidate should be tested against the same complete conditions. Randomly trying digits without a stopping rule makes it difficult to know whether alternatives have been missed.
When a carry is forgotten
Write the carry above the next column as soon as it appears. In a two-column cryptarithm, treat the carry as another small known quantity. The carry can turn an apparently possible tens digit into an impossible one.
When the first working answer is declared unique
Ask whether another digit or pair could satisfy the same relationship. Questions 3, 9, 10, 11, 21 and 22 deliberately have several answers. A complete solution set is a stronger mathematical habit than assuming every school puzzle must end with one digit.
When no solution appears
Use Questions 14 and 19. A contradiction or non-digit requirement is evidence that the stated puzzle has no solution under its rules. Do not round, change a sign or introduce an unstated leading zero.
Create a new puzzle forwards
Choose a digit first, construct a correct arithmetic statement and then replace selected occurrences with letters. For example, start with 52+25=77 and hide the tens digit of the first number and units digit of the second. Check whether the hidden version remains uniquely solvable before giving it to another learner.
This construction habit reveals how constraints create or fail to create uniqueness. It also connects to Magic Squares, Number Grids and Balanced Sums, where every row and column condition must agree.
Continue through the puzzle-enrichment collection
For balancing several lines at once, continue to Magic Squares, Number Grids and Balanced Sums. For geometric construction constraints, use Tessellations, Tiling and Spatial Construction Puzzles. For finding the shortest or most efficient valid construction, use Optimisation, Minimum Moves and Efficient Constructions.
Return to the BTT Primary Mathematics Learning Hub.
Original enrichment guide with 24 original practice questions and separate worked answers. Digit conventions are stated locally and should not be imported from one puzzle into another without checking. No official examination coverage or performance guarantee is implied.