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 N & n (Posted on 2014-01-27)
N is a m-th power of an integer.
It consists of n distinct digits, averaging n.

Find n,m and N, using your head + pen and paper only.

Check, whether there exist additional solutions (computer allowed).

Clearly, n>1; m>1 - to exclude trivial cases.

 No Solution Yet Submitted by Ady TZIDON No Rating

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 computer solution for all the solutions | Comment 3 of 7 |

DEFDBL A-Z
CLS

DIM SHARED used(9), h(6), n, sod, tot, value, scount

FOR n = 2 TO 6
sod = n * n
FOR first = 1 TO 9
used(first) = 1
h(1) = first
value = first
tot = first

used(first) = 0
NEXT
NEXT
PRINT scount

FOR new = 0 TO 9
IF tot + new <= sod THEN
IF used(new) = 0 THEN
used(new) = 1
h(upto) = new
svalue = value
value = 10 * value + new
tot = tot + new

IF upto = n THEN
IF tot = sod THEN

FOR rt = 2 TO 20
r = INT(value ^ (1 / rt) + .5)
IF value = INT(r ^ rt + .5) THEN
PRINT value, r; "^"; rt
scount = scount + 1
END IF
NEXT rt

END IF
ELSE
END IF

used(new) = 0
value = svalue
tot = tot - new
END IF
END IF
NEXT
END SUB

finds 12 solutions:

216      =    6 ^ 3
243      =    3 ^ 5
324      =    18 ^ 2
3481     =    59 ^ 2
9025     =    95 ^ 2
12769    =    113 ^ 2
30976    =    176 ^ 2
37249    =    193 ^ 2
85264    =    292 ^ 2
96721    =    311 ^ 2
287496   =    66 ^ 3
751689   =    867 ^ 2

More than 6 distinct digits would be impossible as 7 digits would require a total of 7*7=49 and the highest that 7 distinct digits could be would be 9+8+7+6+5+4+3=42.

 Posted by Charlie on 2014-01-27 13:33:02

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