Problems & Puzzles: Puzzles

Puzzle 93.- Numbers such that the decimal and hexadecimal representation are the reverse of each other

The 12/01/2000 Jim Howell wrote:

"Some time ago, I noticed that there are a few numbers where the decimal and hexadecimal representations are the reverse of each other:

53 (decimal) = 35 (hex)
371 (decimal) = 173 (hex)
5141 (decimal) = 1415 (hex)
99481 (decimal) = 18499 (hex)

I believe these are the only such numbers (of more than one digit). [Now here is where the primes come in...]

53 = prime
371 = 7 * 53
5141 = 53 * 97
99481 = 53 * 1877

All four of these numbers are divisible by 53 (!)"

(1) are there any other numbers (of more than one digit) where the decimal and hexadecimal representations are reversals?
(2) is there any particular reason for all of the above numbers to be
divisible by 53?
(3) are there other pairs of bases for which something similar happens?

Solution

(1) Jud McCranie wrote "that is all of the solutions. If the number is >=1000000 then it has more decimal digits than hexadecimal digits, so they can't be reversals". But Enoch Haga sent the following additional solution: "8520280 (decimal) = 0820258 (hex)". Maybe Jud was not taking zero leading numbers as valid solutions...

(2) Enoch Haga wrote the following argument: "The reason that all of the decimal numbers above are divisible by 53 is that a sum of powers in one base must equal the sum of powers in another base. Example: 5*10^1=50 and 3*10^0=3, sum 53, and in hex: 3*16^1=48 and 5*16^0=5, sum 53. This is the necessary and sufficient condition for reversibility. The first such sum found will not always be prime, but if prime, will divide all subsequent numbers satisfying the same condition because all of the subsequent numbers are multiples of the first prime"...

(3) Other solutions of the same type are:

13 (10) = 31 (4) (J & E)

23 (10) =32 (7)   (J)
46=2*23
2116 = 2^2*23^2
15226 = 2*23*331

1527465 decimal = 5647251 octal. (J & E)

445 = 5*89 (10) = 544 (9) (J)
313725 = 3*5^2*47*89

315231 (10) = 132513 (12) (J)

43 (13) =34 (13) (J)
86 = 2*43
774 = 2*3^2*32

834 (10) = 438 (14) (J)

21 (10) = 12 (19) (J & E)
42
63
84
441
882
1540 (E)
3290 (E)
7721 = 7*1103 - not a multiple of 21
9471
etcetera

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