The decimal expansion of 1/271 repeats with a period of length 5:
.003690036900369 ...
However, it is not the smallest number q for which the decimal expansion of 1/q has a repetition length of 5.
Find the smallest q so that the decimal expansion of 1/q has repetition length n for each of {1, 2, ..., 10}
Is there a simple way of finding such a number?
When a fraction repeats with a length of n, it can be represented as a fraction with the denominator consisting of n 9's. That fraction is of course not necessarily in its reduced form. For example, 1/271 = .0036900369... = 369/99999, and the 1/271 is the fraction in its simplest (reduced) form.
Therefore q must be a factor of the repeated-9 number. The question is how to determine which factor will work.
The case of length 1 is ambiguous, as any terminating decimal can be said to have length 1, with its repeating 0's, as in q=1, 1/1 = 1.0000.... Otherwise the smallest q would be 3, giving .333....
Here's a table for the first 12 lengths:
length prime factors of rep-9 smallest q prime factors of smallest q
2 3 3 11 11 11
3 3 3 3 37 27 3 3 3
4 3 3 11 101 101 101
5 3 3 41 271 41 41
6 3 3 3 7 11 13 37 7 7
7 3 3 239 4649 239 239
8 3 3 11 73 101 137 73 73
9 3 3 3 3 37 333667 81 3 3 3 3
10 3 3 11 41 271 9091 451 11 41
11 3 3 21649 513239 21649 21649
12 3 3 3 7 11 13 37 101 9901 707 7 101
I can't see a pattern of the choice of prime factor(s) that make up the smallest q.
DEFDBL A-Z
DECLARE SUB factor (num, s$)
DIM SHARED nFct
DIM SHARED fct(30)
CLS
FOR i = 2 TO 12
nbr = VAL(STRING$(i, "9"))
factor nbr, f$
f$ = LTRIM$(f$) + " "
PRINT i, f$; TAB(41);
nFct = 0
DO
ix = INSTR(f$, " ")
IF ix > 0 THEN
n = VAL(LEFT$(f$, ix - 1))
f$ = MID$(f$, ix + 1)
nFct = nFct + 1
fct(nFct) = n
END IF
LOOP UNTIL f$ = ""
lowestP = 999999999999999#
FOR combo = 1 TO INT(2 ^ nFct - .5)
prod = 1: pwr2 = 1
FOR j = 1 TO nFct
IF pwr2 AND combo THEN prod = prod * fct(j)
pwr2 = pwr2 * 2
NEXT
q$ = LTRIM$(STR$(1 / prod))
IF LEFT$(q$, 1) = "0" THEN q$ = MID$(q$, 2)
q2$ = ""
FOR j = 1 TO LEN(q$)
ts$ = MID$(q$, j, 1)
tst = INSTR("0123456789", ts$)
IF tst > 0 THEN q2$ = q2$ + ts$
IF ts$ = "E" OR ts$ = "D" THEN EXIT FOR
NEXT
FOR l = 1 TO i
good = 1
FOR j = l + 1 TO LEN(q2$) - 1
IF MID$(q2$, j - l, 1) <> MID$(q2$, j, 1) THEN good = 0: EXIT FOR
NEXT
IF good THEN replen = l: EXIT FOR
NEXT
IF replen = i THEN
IF prod < lowestP THEN lowestP = prod
END IF
NEXT
PRINT lowestP;
factor lowestP, fc$
PRINT TAB(50); fc$
NEXT
SUB factor (num, s$)
s$ = "": n = ABS(num): IF n > 0 THEN limit = SQR(n): ELSE limit = 0
IF limit <> INT(limit) THEN limit = INT(limit + 1)
dv = 2: GOSUB DivideIt
dv = 3: GOSUB DivideIt
dv = 5: GOSUB DivideIt
dv = 7
DO UNTIL dv > limit
GOSUB DivideIt: dv = dv + 4 '11
GOSUB DivideIt: dv = dv + 2 '13
GOSUB DivideIt: dv = dv + 4 '17
GOSUB DivideIt: dv = dv + 2 '19
GOSUB DivideIt: dv = dv + 4 '23
GOSUB DivideIt: dv = dv + 6 '29
GOSUB DivideIt: dv = dv + 2 '31
GOSUB DivideIt: dv = dv + 6 '37
IF INKEY$ = CHR$(27) THEN s$ = CHR$(27): EXIT SUB
LOOP
IF n > 1 THEN s$ = s$ + STR$(n)
EXIT SUB
DivideIt:
DO
q = INT(n / dv)
IF q * dv = n AND n > 0 THEN
n = q
s$ = s$ + STR$(dv)
IF n > 0 THEN limit = SQR(n): ELSE limit = 0
IF limit <> INT(limit) THEN limit = INT(limit + 1)
ELSE
EXIT DO
END IF
LOOP
RETURN
END SUB
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Posted by Charlie
on 2006-09-25 11:03:49 |
computer exploration
