Gelfand’s question
Jaap Top
Bernoulli Institute & DIAMANT
June 25th, 2018
(PAM symposium, Groningen)
1
Israel M. Gel’fand (1913–2009)
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3
Note that mathworld already provides the answer to the first
question: (does 9 occur as the most significant decimal digit of
some 2n?). Here is a small table.
n 1 2 3 4 5 6 7 8 9 102n 2 4 8 16 32 64 128 256 512 1024
11 12 13 14 15 16 172048 4096 8192 16384 32768 65536 131072
18 19 20 21 22262144 524288 1048576 2097152 4194304
So, not with 1 ≤ n ≤ 22.
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Staring at the table, one might be tempted to think that the
array of blue digits is periodic with period 10.
In other words: do 2n and 2n+10 have the same most significant
decimal digit?
Example: 21 and 211 and 221 and 231 and 241 and 251 and 261
all have leftmost decimal digit 2.
The same for 271, 281, . . . ,2161,2171
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But then
2181 =
3064991081731777716716694054300618367237478244367204352
and 2301 has most significant decimal digit 4,
and 2391 even has leftmost decimal digit 5.
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Concerning the occurrence of a leftmost decimal digit 7 or 9,
246 = 70368744177664
253 = 9007199254740992
7
8
9
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So how about the remaining questions 2) and 3) of Gel’fand:
2) Does n > 1 exist such that the array of leftmost decimal digits
of (2n,3n,4n, . . . ,8n,9n) equals (2,3,4,5,6,7,8,9)?
3) Do n > 0 and ` ∈ {1,2,3, . . . ,8,9} exist such that the ar-
ray of leftmost decimal digits of (2n,3n,4n, . . . ,8n,9n) equals
(`, `, `, `, `, `, `, `)?
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Blogger / mathematician John D. Cook used the summer of2013 to verify:
For no n with 1 < n < 1010 is (2,3,4,5,6,7,8,9), or is any(`, `, `, `, `, `, `, `) the array of leftmost digits of (2n,3n,4n, . . . ,9n).
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Answering Gel’fand’s question turns out to be surprisingly simple!
It was done by Jaap Eising as a small part of his bachelor’s
project (2013, Groningen):
Questions 2) and 3) have the same answer: NO.
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Jaap Eising
(and his fellow board members of FMF, 2014)
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To answer question 2): write
2n = αn × 10m
for a nonnegative integer m, and 1 < αn < 10.
With this notation, the leading decimal digit of 2n equals bαnc.
Since 2n = 10τn with τ = log(2)/ log(10), we have m = bτnc and
αn = 10τn−bτnc.
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We treat 5n in a similar way: so 5n = γn × 10k where k is a
nonnegative integer and 1 < γn < 10.
As 2n × 5n = 10n, we have αn × γn = 10.
Now suppose that for n > 1 it holds that bαnc = 2 and also
bγnc = 5.
That means 2 ≤ αn < 3 and 5 ≤ γn < 6.
Of course αn 6= 2, because 2n 6= 2×10m (remember that n > 1).
So 10 = αn × γn > 10, a contradiction!.
Conclusion: there is no n > 1 for which the leading decimal digits
of 2n and 5n are 2 and 5, respectively.
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A similar argument works for Question 3):
if the leading decimal digit of each of 2n to 9n were the same,
then using αn × γn = 10 and bαnc = bγnc,
it follows that 3 < αn, γn < 4.
However, in this case 4n = 2n×2n = α2n×102m will have leading
decimal digit either 9 or 1, because 9 < α2n < 16. As we saw that
2n must have leading decimal digit equal to 3,
we conclude: all of 2n, . . . ,9n having the same leading decimal
digit, does not occur!
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We return to question 1).
Here 2n = αn × 10m, and with τ = log(2)/ log(10) we have
m = bτnc and αn = 10τn−bτnc.
So, if we want 2n to have a certain leading decimal digit (say `),
then we want ` ≤ αn < `+ 1,
which means log(`)/ log(10) ≤ τn− bτnc < log(`+ 1)/ log(10).
Leopold Kronecker showed in 1884 that for any σ ∈ R\Q it holds
that the numbers (σn− bσnc)n≥1 are dense in the interval [0,1].
Piers Bohl, Wac law Sierpinski, and Hermann Weyl independently
proved in 1909-1910 a stronger result:
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Thm. For every σ ∈ R \ Q and every integrable f : [0,1] → C itholds that ∫ 1
0f(t) dt = lim
N→∞1
N
N∑n=1
f(σn− bσnc).
Applying this to
f(t) :=
{1 if log(`)/ log(10) ≤ t < log(`+ 1)/ log(10);0 otherwise,
shows that
limN→∞
#{n ≤ N : 2n has leading digit `}N
=
log(`+ 1)/ log(10)− log(`)/ log(10) = log(1 + 1`)/ log(10).
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So this tells which ‘proportion’ of all n yield leading digit of 2n
equal to `.
Note that the answer is the same if we consider 3n, or 4n, or 5n,
et cetera.
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Details and more can be found in a paper published in 2015 in
the American Mathematical Monthly:
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