Consider the three digit numbers 301, 431, 602, 715, 842,
856, 973, 986
They ALL satisfy the following properties:
1) All the digits are different
2) When the digits are
placed along a cirlce, each digit is three times the previous digit, or one
unit more, or two units more (while doing this, keep in mind that if this
number comes out to be more than 9, only the units digit is taken). Hence 7 is followed by 1, 2 or 3. 3 is followed by 9, 0 or 1. If the number ends with the digit 8, then the
first digit can be 4, 5 or 6.
Not only the above listed numbers, their cyclic permutations
also satisfy the above properties.
Hence the above list gives 22 three digit numbers. We know, no number starts with a zero.
Now, try for four digited numbers that satisfy the above
properties. How many such numbers are
there, if we consider their cyclic permutations also as different number?
2 comments:
There are 243 such possibilities. Am I right?
There are 44 such numbers. Create a tree of digits to help you make out the sequence of digits.
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