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Where 2^n leads, 3^n cannot be far behind... (Posted on 2012-08-24) Difficulty: 3 of 5

Carrying on in the spirit of '86 at most' here is a pair of 'statistical' conjectures about 3^n: (see A060956 in Sloane, particularly the table, for the values up to 3^1000)

To start with a definition; if the first digit of 3^n is a 9 ( e.g. 3^23 = 94,143,178,827) then we say that 3^n is 'good'.

Conjecture 1: If 3^n is good, then either 3^(n+21), or 3^(n+23) is also good.

Conjecture 2: If 3^n is good, then 3^(m+n) is also good, for some constant, m, and n greater than 2.

True or false?

See The Solution Submitted by broll    
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Solution solution | Comment 1 of 2

Part 1 is easy:

3^21 =  10460353203
3^23 =  94143178827

The former, when multiplied by a good number whose significand is lower than 9.5599... will produce another good number. The latter, when multiplied by a good number whose significand is greater than that same number will be good. Only if the significand of the number were exactly this (10^11 / 3^21) would the result fail to be good. But such an original number would not be a power of 3, as it would be a rational number with some power of 10 in the numerator.

BTW, the significand is the a in the scientific notation a x 10^b, and was previously called the mantissa, but the name was changed to avoid conflict of the use of "mantissa" as the fractional part of a common logarithm.

So, yes, conjecture 1 holds.

For Part 2:

If the common log of 3 is taken to be a normal number, with any sequence of digits equally likely for any count of digits anywhere in the expansion, then, again using common logs:

log(3^n)=n*log(3)

The fractional part of the resulting logarithm, called the mantissa, will then be expected to be some "random", that is, normal, number as well, equally likely to be anywhere on the unit number line.

The antilog of this mantissa is called the significand of the scientific notation for the given power of 3.

If we continue to higher and higher powers of 3 we would expect to get numbers ever closer, on both sides, to having a significand of 9 (only 3^2 has one exactly 9), as demonstrated by:

  10   point 255
  20   L3=log(3)/log(10):L9=2*L3:Lp=2*L3:Lpmin=99
  30   for Pwr=3 to 10000000
  40      Lp=Lp+L3
  50      Lp=Lp-int(Lp)
  60      if Lp-L9>=0 then if Lp-L9<Lpmin then Lpmin=Lp-L9:gosub *Report
  70      if L9-Lp>=0 then if Lp>Lpmax then Lpmax=Lp:gosub *Report
  80   next
  90   end
 100   *Report
 110     print Pwr;tab(20);using(3,15),10^Lp
 120   return

n                    significand of 3^n
3                    2.700000000000000
4                    8.100000000000000
23                   9.414317882700000
25                   8.472886094430000
46                   8.862938119652501
67                   9.270946314789784
111                  9.129758166511361
155                  8.990720186353585
264                  9.120344560464466
417                  9.110940660696444
570                  9.101546457199243
723                  9.092161939975133
876                  9.082787099036689
1029                 9.073421924406787
1182                 9.064066406118588
1335                 9.054720534215532
1488                 9.045384298751323
1641                 9.036057689789921
1794                 9.026740697405532
1947                 9.017433311682595
2100                 9.008135522715775
2253                 8.998847320609947
4351                 9.006981801364682
6602                 9.005828227777013
8853                 9.004674801933845
11104                9.003521523816254
13355                9.002368393405320
15606                9.001215410682127
17857                9.000062575627758
20108                8.998909888223301
37963                8.998972456271678
55818                8.999035024755081
73673                8.999097593673513
91528                8.999160163026977
109383               8.999222732815476
127238               8.999285303039014
145093               8.999347873697593
162948               8.999410444791216
180803               8.999473016319886
198658               8.999535588283607
216513               8.999598160682381
216513               8.999598160682381
234368               8.999660733516211
252223               8.999723306785101
270078               8.999785880489053
287933               8.999848454628070
305788               8.999911029202156
323643               8.999973604211313
341498               9.000036179655545
665139               9.000009783760749
988780               8.999983387943368
1653917              8.999993171686058
2319054              9.000002955439384
3972969              8.999996127123199
6292021              8.999999082561311
8611073              9.000002038000393

It seems that whenever we put x digits into an exponent for 3, we can get at least x-1 signficant digits of closeness to 9 in the significand, on both the high side and the low side.

Any given single value for m will have its power of 3 have either a significand some finite amount above 9 or below 9.  Either way, it will destroy the goodness of some number that's sufficiently close to 9 in the same way, and this sufficient degree of closeness is always achievable.

So conjecture 2 would be disconfirmed.


  Posted by Charlie on 2012-08-24 15:33:51
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