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© Copyright 2002, Jim Loy
This is an original puzzle which I stumbled upon. I have before me three numeric palindromes (numbers which read the same backwards and forewards, like 838). The first is two digits long; the second is three digits, and when we add those two numbers together we get the third number which is four digits long. What are the three numbers?
Solution:
We only need to deduce five digits (aa+bcb=deed), so trial and error should not be very time consuming. But it turns out that we can deduce all five of the digits. b must be 9 (aa+8c8 cannot be four digits long) and d must be 1 (aa+9c9 cannot be greater than 1098). So a must be 2 (aa+9c9=1ee1) and e must be 0, and the four digit number must be 1001. 1001-22=979, which is the only solution: 22+979=1001.