# Ready to crack the code?

Thanks for visiting! Let’s get down to cracking the code. But first, the problem statement:

The first code is 63AC and it is transformed to FC19. The trick is that the numbers 6 and 3 get transformed to the 6th and 3rd letters of the alphabet series. And therefore, we have FC in place of 63.

Next, for AC, the trick is to note that the position of A in the alphabet series is 1 and that of C is 3. 19 is obtained from AC by squaring 1 and 3 respectively.

Now this can be verified using the second code transformation – 91EG is transformed to IA2549. Clearly, the 9th letter is I and the 1st letter is A. Also, E’s position in the series is 5 and G’s position is 7. Therefore, we have the squares of 5 and 7, i.e. 25 and 49.

Using this logic, the code 85DM is transformed into HE16169

Hope you enjoyed the puzzle!

At Cuemath, we encourage our students to think logically and understand the ‘Why’ behind ‘What’ of everything. Puzzle cards are an integral part of our foundation program (in addition to the worksheets that cover the core school curriculum; tab-based math activities; and creative reasoning activities). If you are interested, please book a FREE session with a Cuemath Math Expert near you.

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