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What number equals the average of its digits?
How many can you find if you allow for trailing zeros?
Previous iteration: #932106
0, 1, 2, 3, 4, 5, 6, 7, 8, and 9. A number n is equal to the average of its digits if:
n= sum of digits of n/ number of digits of n
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Those are indeed some of the solutions, albeit trivial ones.
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For me only 0 can solve this question. 0 0,0 0,00 0,000 but probably I'm wrong
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Including a fractional part is indeed a good idea...
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4.5
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Allowing for trailing zeroes, there is an infinite number of solutions:
sum 1, count 10, avg 0.100000000
sum 1, count 100, avg 0.010000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000
This construction obviously works for larger power-of-tens as count.
Other interesting ones are:
sum 1, count 10, avg 0.100000000
sum 2, count 10, avg 0.200000000
sum 3, count 10, avg 0.300000000
sum 3, count 25, avg 0.120000000000000000000000
sum 4, count 10, avg 0.400000000
sum 5, count 10, avg 0.500000000
sum 6, count 4, avg 1.500
sum 6, count 10, avg 0.600000000
sum 6, count 25, avg 0.240000000000000000000000
sum 7, count 10, avg 0.700000000
sum 7, count 28, avg 0.250000000000000000000000000
sum 8, count 10, avg 0.800000000
sum 9, count 2, avg 4.5
sum 9, count 4, avg 2.250
sum 9, count 5, avg 1.8000
sum 9, count 8, avg 1.1250000
sum 9, count 10, avg 0.900000000
sum 9, count 20, avg 0.4500000000000000000
sum 9, count 25, avg 0.360000000000000000000000
sum 12, count 16, avg 0.750000000000000
sum 12, count 25, avg 0.480000000000000000000000
sum 15, count 4, avg 3.750
sum 18, count 32, avg 0.5625000000000000000000000000000
sum 27, count 16, avg 1.687500000000000
sum 27, count 32, avg 0.8437500000000000000000000000000
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