Number Play · Lesson 9 of 10
Digits in Disguise
“Solve letter-and-digit puzzles with place value and carrying.”
• State the digit rules used in a cryptarithm. • Use units and tens columns to infer unknown digits. • Check a solution by substituting every letter consistently. • Explain why a guess is impossible using place value or a carry.
When a letter stands for a digit
A cryptarithm replaces digits in an arithmetic calculation with letters. Each letter represents one fixed digit from 0 to 9 throughout that puzzle, and different letters represent different digits. A multi-digit number cannot begin with 0. We read a letter pair such as UT as the two-digit number with tens digit U and units digit T; it is not U multiplied by T.
A number puzzle in which letters stand for digits in a correctly arranged arithmetic calculation.
Problem
Solve T + T + T = UT.
- 1.The result ends in T, so three T units and one T unit differ by a multiple of 10: 2T is a multiple of 10.
- 2.T cannot be 0 because UT is a two-digit result and T must be a nonzero digit. With 1 ≤ T ≤ 9, the only possibility is 2T = 10, giving T = 5.
- 3.Three copies of 5 total 15, so U = 1 and UT = 15. The check is 5 + 5 + 5 = 15.
Read the columns and carries
In K2 + K2 = HMM, start at the units column. Two 2s make 4, so M = 4 and there is no carry from units. The tens column must therefore produce a tens digit 4 and a nonzero hundreds digit H. Trying the feasible two-digit sum 2K with ones digit 4 in the tens calculation gives 2K = 14, so K = 7 and H = 1. Indeed, 72 + 72 = 144.
Problem
Find K, H, and M in K2 + K2 = HMM.
- 1.Units: 2 + 2 = 4, so M = 4 and no unit carry occurs.
- 2.Tens: K + K must give a 4 in the tens place and a carry to hundreds. Thus 2K = 14, so K = 7 and the carry H = 1.
- 3.Check: 72 + 72 = 144. The distinct letters represent distinct digits.
Use a whole-number check
Column clues are powerful, but a complete answer must satisfy every column and use each letter consistently. If the final sum has more digits than the addends, its first digit comes from a carry. A useful strategy is to record deductions as they arise, check that no two different letters received the same digit, and substitute into the original calculation.
Problem
Solve UT + TA = TAT.
- 1.The sum of two two-digit numbers cannot exceed 198. Since TAT has three digits, its leading digit T must be 1.
- 2.Units: T + A ends in T, so A = 0, with no carry from units.
- 3.Tens: U + T must give tens digit A = 0 and carry 1 to hundreds. With T = 1, U + 1 = 10, so U = 9.
- 4.Check: UT = 91, TA = 10, and TAT = 101; indeed 91 + 10 = 101.
A letter cannot be 4 in one column and 7 in another. Also, a two-digit number such as TA cannot have T = 0, and distinct letters must not share a digit in this lesson’s puzzles.
Quiz
In UT, which letter is the tens digit?
What is T in T + T + T = UT?
What is M in K2 + K2 = HMM?
Why is T = 1 in UT + TA = TAT?
Which complete substitution solves UT + TA = TAT?
Practice Problems
- Solve T + T + T = UT again by testing the possible values of T and explaining why only one works.
- Verify each column of 72 + 72 = 144 and identify K, H, and M.
- Solve UT + TA = TAT. Check your digits in the original addition.
- Try the chapter puzzle YY + Z = ZOO. Begin with the carry into the hundreds column and record deductions.
- Try B5 + 3D = ED5. Use the units digit first and check your completed addition.
- Try KP + KP = PRR. Explain the hundreds carry before choosing the remaining digits.
- Try C1 + C = 1FF. Which digit must C be? Explain using the tens and hundreds columns.
- Explain why a solution with a zero as the first digit of a two-digit addend must be rejected.
Key Takeaways
• The same letter keeps the same digit throughout a cryptarithm; different letters use different digits. • A multi-digit number cannot have a leading zero. • Work from units to tens and hundreds, keeping track of carries. • Substitute the final digits and check the original calculation.