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Cryptarithms, Lecture notes of Algebra

Cryptarithms are puzzles where letters stand for different digits in an arithmetic problem. The problem is usually to figure out which letters stand for ...

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2021/2022

Uploaded on 08/05/2022

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Download Cryptarithms and more Lecture notes Algebra in PDF only on Docsity! Cayuga Heights 5th Grade Math Club Solutions for February 26, 2016 Cryptarithms Cryptarithms are puzzles where letters stand for different digits in an arithmetic problem. The problem is usually to figure out which letters stand for which digits. Often it helps to write equations using the letters. Algebra can then be used to solve for the letters. Sometimes you can narrow the possibilities for a given letter down to a small number of choices, and you just have to try them and see if there is a way to get a solution. 1. In this addition problem, different letters stand for different digits. What digit does A stand for? 4 A + A 4 B C B 2. HA and AH represent two-digits numbers. If HA − AH = 18, what is the value of the expression H −A? Answer: 9 3. In this multiplication problem, different letters stand for different digits. What digit does H stand for? A H A × A T A D A 4. When the six-digit number 3456X7 is divided by 8, the remainder is 5. Give both possible values of the digit X . Answer: 3 or 7. Notice 345600 is divisible by 8. 5. In this multiplication problem, A and B represent different digits. What is the 4-digit product? A B × B A  8       Answer: 42× 24 = 1008 6. If 23AB3 is divisible by 99, what is the two-digit number AB? Answer: 46 7. The digits 1,2,3,4 and 5 are each used once to write a five-digit number ABCDE. The 3-digit number ABC is divisible by 4, BCD is divisible by 5, and CDE is divisible by 3. Find the five-digit number ABCDE. Answer: 12,453 . Note D must be 5, so C +E must have a remainder of 1 when divided by 3. So one of (C,E) must be 3 and the other is either 1 or 4. Since ABC is divisible by 4, C must be even—it must be 4. Then E is 3. Then B must be 2 since 14 is not divisible by 4, and A is 1 by elimination.
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