Michael's number

For positive integers b > 1 b>1 and n 1 , n\ge 1, let f n ( b ) f_n(b) be the largest number (in decimal notation) that has n n digits when expressed in base b b . Find the sum of all possible n n such that there exists a value of b b that satisfies f n ( b ) = 4095. f_n(b) = 4095.

This problem is posed by Michael T .

Details and assumptions

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The answer is 28.

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4 solutions

Daniel Chiu
Sep 1, 2013

The largest number in base b b with n n digits would have all digits as b 1 b-1 . Notice that this number would simply be b 1 b-1 times the number with n n 1's, which would have a value, in base 10, of the sum of the first n 1 n-1 powers of b b . This is, using the geometric series sum, ( b 1 ) ( b n 1 + b n 2 + + b + 1 ) = ( b 1 ) b n 1 b 1 = b n 1 (b-1)(b^{n-1}+b^{n-2}+\cdots+b+1)=(b-1)\dfrac{b^n-1}{b-1}=b^n-1 Therefore, we must find all b b such that b n = 4096 = 2 12 b^n=4096=2^{12} . Clearly, b b is a power of 2. The power of 2 must be a factor of 12. The possible b b are 2 1 , 2 2 , 2 3 , 2 4 , 2 6 , 2 12 2^1,2^2,2^3,2^4,2^6,2^{12} Then, n n would be the power on b b that equals 4096. The corresponding n n are 12 , 6 , 4 , 3 , 2 , 1 12,6,4,3,2,1 The sum is 28 \boxed{28} .

Nicely done!

Alexander Borisov - 7 years, 9 months ago

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.

John Lawrence - 4 years, 3 months ago
Daniel Liu
Sep 3, 2013

Note that f n ( b ) f_n(b) when written in base b b is ( b 1 ) ( b 1 ) ( b 1 ) \overline{(b-1)(b-1)(b-1)\cdots} where the digit which value is b 1 b-1 appears n n times. Note that when you add 1 1 to this number, you get 1000 1000\cdots where there are n n zeroes. The value of this number is therefore b n b^n . So now we know that f n ( b ) + 1 = b n f_n(b)+1=b^n . Seeing that f n ( b ) = 4095 f_n(b)=4095 , we see that b n = 4096 b^n=4096 . The possible values are 2 12 , 4 6 , 8 4 , 1 6 3 , 6 4 2 , 409 6 1 2^{12},4^{6},8^{4},16^3,64^2,4096^1 . Summing the exponents up, we get the answer is 28 \boxed{28} .

Moderator note:

Very nice!

Jonathon Capps
Sep 2, 2013

The largest number in a base b b with n n digits is simply b n 1 b^n-1

Therefore f n ( b ) = b n 1 = 4095 f_n(b)=b^n-1=4095

b = 4096 n = 2 12 n b=\sqrt[n]{4096}=\sqrt[n]{2^{12}}

b b will be a whole number if n n is any divisor of 12, or 1, 2, 3, 4, 6, 12 which when summed gives the answer 28.

Very nice!

Alexander Borisov - 7 years, 9 months ago
Tanishq Aggarwal
Sep 1, 2013

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 ) \displaystyle f_n(b)=(b-1)\sum_{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 f_n(b)=b^n-1 . We are thus looking for all integers b b and n n such that b n = 4096 b^n=4096 . We begin our search at the least b b and work our way upwards. 4096 = 2 12 = 4 6 = 8 4 = 1 6 3 = 6 4 2 = 409 6 1 \begin{aligned} 4096=2^{12} \\ =4^{6} \\ =8^4 \\ =16^3 \\ =64^2 \\ =4096^1 \end{aligned} The sum of all exponents is thus 28 \boxed{28} .

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