How did they notice this?
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Exhaustion?
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How is this useful?
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Recreational mathematics.
It might become useful one day, though!
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Why didn't you expand the decimal representation of 2²³⁰⁷³⁹⁰¹⁴?

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70 000 digits...
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This Mathematica code (not optimized) tests 2^N: f[N_] := Module[(*Output: 0, zero in the expansion; 1 otherwise*) {n, m, s, i, t}, n = IntegerDigits[N, 10]; m = IntegerLength[N, 10]; t = True; Do[If[TrueQ[t], t = TrueQ[Not[n[[i]] == 0]], Break[]], {i, 1, m}]; s = If[t, 1, 0]; s]
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That looks incredibly inefficient. No way you can call IntegerDigits[2^k,10] for any k remotely near 230739014. Surely it must be better to work mod 10^n for increasing n and do an exhaustion instead
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This Mathematica code counts the no. of digit j, j=0,1,...,9, in the base-10 expansion of N: g[N_, j_] := Module[(*Output: No. of 0's in the b-10 expansion of N*){n, m, s, i}, n = IntegerDigits[N, 10]; m = IntegerLength[N, 10]; s = 0; Do[If[n[[i]] == j, s = s + 1], {i, 1, m}]; s]
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Cont.) The attached figure corresponds to N=2^n, for n=1, 2, ..., 5000, and j=0. Similar figures are obtained for j=1, 2, ..., 9. Thus no. of j scales like n, or ln(N). Note that number of digits of N=2^5000 is equal to 1506, which is around 10×150 (note the vertical scale).pic.twitter.com/EzQJWCtdPO
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