For positive integers b > 1 and n ≥ 1 , let f n ( b ) be the largest number (in decimal notation) that has n digits when expressed in base b . Find the sum of all possible n such that there exists a value of b that satisfies f n ( b ) = 4 0 9 5 .
This problem is posed by Michael T .
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Nicely done!
Why can't b be 65? 63•65^1 + 0•65^0 = 4095
In fact, b can be any base between 2 and 65. For example, b =11 4095 = 3•11^3 + 0•11^2+ 9•11^1 + 3•11^0 Therefore when b=11, n=4. f12(2), f8(3), f6(4),.... f2(64), f2(65) I think the answer is 191.
Note that f n ( b ) when written in base b is ( b − 1 ) ( b − 1 ) ( b − 1 ) ⋯ where the digit which value is b − 1 appears n times. Note that when you add 1 to this number, you get 1 0 0 0 ⋯ where there are n zeroes. The value of this number is therefore b n . So now we know that f n ( b ) + 1 = b n . Seeing that f n ( b ) = 4 0 9 5 , we see that b n = 4 0 9 6 . The possible values are 2 1 2 , 4 6 , 8 4 , 1 6 3 , 6 4 2 , 4 0 9 6 1 . Summing the exponents up, we get the answer is 2 8 .
Very nice!
The largest number in a base b with n digits is simply b n − 1
Therefore f n ( b ) = b n − 1 = 4 0 9 5
b = n 4 0 9 6 = n 2 1 2
b will be a whole number if n is any divisor of 12, or 1, 2, 3, 4, 6, 12 which when summed gives the answer 28.
Very nice!
Since f n ( b ) = ( b − 1 ) i = 1 ∑ n b i − 1 = ( b − 1 ) ( b n − 1 + b n − 2 + b n − 3 . . . + b 2 + b + 1 ) , f n ( b ) = b n − 1 . We are thus looking for all integers b and n such that b n = 4 0 9 6 . We begin our search at the least b and work our way upwards. 4 0 9 6 = 2 1 2 = 4 6 = 8 4 = 1 6 3 = 6 4 2 = 4 0 9 6 1 The sum of all exponents is thus 2 8 .
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The largest number in base b with n digits would have all digits as b − 1 . Notice that this number would simply be b − 1 times the number with n 1's, which would have a value, in base 10, of the sum of the first n − 1 powers of b . This is, using the geometric series sum, ( b − 1 ) ( b n − 1 + b n − 2 + ⋯ + b + 1 ) = ( b − 1 ) b − 1 b n − 1 = b n − 1 Therefore, we must find all b such that b n = 4 0 9 6 = 2 1 2 . Clearly, b is a power of 2. The power of 2 must be a factor of 12. The possible b are 2 1 , 2 2 , 2 3 , 2 4 , 2 6 , 2 1 2 Then, n would be the power on b that equals 4096. The corresponding n are 1 2 , 6 , 4 , 3 , 2 , 1 The sum is 2 8 .