Math Competition Problems Problem of the Week 8 - Cryptogram Puzzle

 2BA + C6D  = 8AD. Find out A, B, C and D

Source: Adapted from CML math competition

Cryptarithm-style problems occasionally appear in competitions such as CML, MOEMS, and Noetic, and are excellent for developing logical reasoning skills.

What is a Cryptogram/ Cryptarithm?

A Cryptogram/ Cryptarithm is a puzzle where letters and symbols are used in place of numbers. Our goal is to figure out which number each letter and symbol corresponds to.

Rules:

1. Different letters represent different digits.

2. Same letters represent the same digits.

3. The leading digit represented by a letter can never be 0. For eg, if I have BAT, B can not be zero

This is a relatively accessible cryptarithm that many strong 3rd and 4th-grade students can solve independently.

Math Competition Problems - Problem of the Week 8

Solution:

Strategy Tip: Look for the column that places the strongest restrictions on the digits.

In this case, it is the unit's place

A + D = D, so A has to be 0 (zero).

This is the identity property of addition. Any number plus zero gives itself. No other number other than zero has this property.

Hence, A = 0 and D can be any number.

Common Mistake:

A common mistake is to look at the hundreds column first and conclude that 2+C=8 and so C=6. This ignores the possibility of a carryover from the tens column.

Even if we consider possible carryovers, we can only conclude that C is either 5 or 6. At this stage, there is not enough information to determine which one is correct.

So then A+D=D catches our eye, and that should make you decide that A has to be zero.

Next B+6=A and A=0.

Since A = 0, the tens column must produce a digit of 0. The only way for B + 6 to end in 0 is if B + 6 = 10. Therefore, B = 4, and there is a carry of 1 to the hundreds column.

Now 2+C+1 (carry over) = 8.

So, 3+C=8. Hence C=5

Final Answer:

A=0

B=4

C=5

D=Any digit other than 0, 4 and 5

Notice how finding A=0 immediately unlocked the rest of the puzzle. In many cryptarithms, a single strong clue can lead to several other deductions.

Now, since this is a question for practice, D could be any digit. If this were a question in MOEMS, it would be more worded like 2BA + C6D  = 8AD. Find A+B+C.

Notice that we never needed to determine D. Once we know that A = 0, B = 4, and C = 5, the value of A+B+C is fixed. So, A+B+C = 9. In a competition setting, recognizing which information matters can save both time and effort.

Try a Cryptogram Puzzle on your own now.

Post the answer in the comments section below

Challenge Cryptogram Problem 1:

CCB + C8D = 7D5. Find B, C and D

Challenge Cryptogram Problem 2:

Interesting Math Problems: ABC + BC + C = EF24. What number is ABC?

Want more Cryptarithm puzzles like these to solve?

Here's an entire blog on Cryptogram/ Cryptarithm puzzles

Cryptogram Puzzles Blog

My Facebook page also has a lot of puzzles and problems (including Cryptograms)

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Problem of the week-7   

 Click here for next week's problem

Problem of the week-9

About the Author

I'm Vasudha, and I'm the founder of V's Online Math Tutoring. I specialize in Beast Academy, AoPS, math competitions, and enrichment programs, and help students become confident and independent problem solvers.

To know more about me and how I work, click here.