44944104494410 and friends

The number 4494410 has the property that when converted to base 16 it is 44944A_{16}, then if the A is expanded to 10 in the string we get back the original number.

3883544142410_{10}=3883544E24A_{16} is another.

These numbers are in OEIS A187829. They come in blocks of 6 or 10, depending on whether the one’s digit in hex is A-F or 0-9.

I suspect the list is complete but have not proven it. The largest is 806123145829415507126939101294137128298625241370656314360169_{10}=\\806C3E58294F507C6939AC94D7C829862524D706563E360169_{16}

If the number has m hex digits and n base 10 digits, we must have 16^{m-1} \gt 10^{n-1} and 16^{m-2} \lt 10^{n-2} which leads the hunt to m=6,n=7;\ m=11,n=13;\ m=16, n=19;\ m=50,n=60 guided by the convergents of \frac {\log 16}{\log 10}.

We can view finding these numbers as finding solutions to the subset-sum problem, where each hex digit contributes the difference between its value in base 16 and base 10 (depending on how many base 16 digits to the right are A-F and counting the two base 10 digits coming from one hex digit together). The sum then has to be zero.

My search program ran reasonably quickly even for the next convergent, m=535, n=644 and didn’t find any. I believe they just have too many ways to fail as the number gets long.

Can we prove that there are no more, or at least that there are no more with very high probability, in the sense of “proofs” of Goldbach that if the primes are “random” the chance of any large even number having no solution is very low?

Answer

I can prove a very narrow form.

Let’s consider numbers of the form:

(D_{n-1}D_{n-2}…..D_{2}D_{1}D_0)_{10} = (D_{n-1}D_{n-2}…..D_2[A..F])_{16}.

Here 2 least significant digits in decimal representation change
to [A..F]. For these numbers, conditions that need to satisfy are,

100*x + 10 = n ….(1)

16*y + 10 = n ….(2)

So,

y = 6.25*x ….(3)

Let’s assume,

x_{10} is of form ….N_{k-1}N_{k-2}…..N_{2}N_{1}N_{0}

or,

x_{10} = ….+ (N_{k-1}10^{k-1}) + (N_{k-2}10^{k-2}) + ….+ (N_{2}10^2) + (N_{1}10^1) + N_{0}

So,

y_{10} = ….+ (N_{k-1}16^{k-1}) + (N_{k-2}16^{k-2}) + ….+ (N_{2}16^2) + (N_{1}16^1) + N_{0}

Also from (3),

y_{10} = 6.25(….+ (N_{k-1}10^{k-1}) + (N_{k-2}10^{k-2}) + ….+ (N_{2}10^2) + (N_{1}10^1) + N_{0})

So,

(….+ (N_{k-1}16^{k-1}) + (N_{k-2}16^{k-2}) + ….+ (N_{2}16^2)
+ (N_{1}16) + N_{0}) =
6.25(….+ (N_{k-1}10^{k-1}) + (N_{k-2}10^{k-2}) + ….+ (N_{2}10^2)
+ (N_{1}10^1) + N_{0})

Using up to 6 digits for x,

(1048576N_5 + 65536N_4 + 4096N_3 + 256N_2 + 16N_1 + N_0) =
(625000N_5 + 62500N_4 + 6250N_3 + 625N_2 + 62.5N_1 + 6.25N_0)

or,

42357600N_5 + 303600N_4 – 215400N_3 – 36900N_2 – 4650N_1 – 525N_0 = 0

Value in Hex falls behind till N3 because initial modulus was 100 for decimal
and 16 for hex. But at and after N4, hex value overtakes decimal forever.

Only solution for N5 (and above) between 0 and 9 is 0. Also, there are no solutions possible less
than 5 digits for x in (1).

So essentially numbers of these forms are only 7-digits or 2 digits,
And only possible solutions are,

10-15
4494410-4494415
5660810-5660815
6784010-6784015
7950410-7950415

Attribution
Source : Link , Question Author : Ross Millikan , Answer Author : mj6174

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