perplexus.info :: Just Math : Digits 0-9 In Reverse

Digits 0-9 In Reverse (Posted on 2008-11-07)

Think of this problem as a reverse of Digits 1-9.

N is a 10-digit positive integer which contains each of the ten base-10 digits 0 to 9 exactly once, such that:

(i) The number formed by the last two digits is divisible by 2.
(ii) The number formed by the last three digits is divisible by 3.
(iii) The number formed by the last four digits is divisible by 4.
and so on up to ten digits...

We will now consider the last three digits of each of the values of N that satisfy all the given conditions and regard them as 3- digit numbers. Let us denote the greatest common divisor of all these 3-digit numbers as G.

It will be observed that a certain 3-digit multiple of G (with no leading zero), each of whose digits are distinct, will never occur as the last three digits of any of the possible values of N.

What is that 3-digit multiple of G which fails to occur?

Note: Try to derive a non computer assisted method, although computer program/spreadsheet solutions are welcome.

See The Solution Submitted by K Sengupta

Rating: 5.0000 (1 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)

examples (spoiler)

Comment 2 of 2 |

9123567480

9123675840

9135648720

9138675240

9147528360

9345678120

end digits count
000         0
120        46
240        32
360         4
480        38
600         0
720        36
840        46
960         0

DEFDBL A-Z
DECLARE SUB permute (a$)
CLS
a$ = "123456789": h$ = a$
DO
  n = 10 * VAL(a$)
  tst = VAL(MID$(a$, 8))
  IF tst / 3 = INT(tst / 3) THEN
    n10$ = LTRIM$(STR$(n))
    tst = VAL(MID$(n10$, 8))
    IF tst / 40 = INT(tst / 40) THEN
      tst = VAL(MID$(n10$, 4))
      IF tst / 7 = INT(tst / 7) THEN
        tst = VAL(MID$(n10$, 2))
        IF tst / 9 = INT(tst / 9) THEN
          tst = VAL(MID$(n10$, 5))
          IF tst / 6 = INT(tst / 6) THEN
           l3 = n - 1000 * INT(n / 1000)
           IF ct(l3 / 120) = 0 THEN PRINT n: ct = ct + 1
           ct(l3 / 120) = ct(l3 / 120) + 1
          END IF
        END IF
      END IF
    END IF
  END IF
  permute a$
LOOP UNTIL a$ = h$

FOR i = 0 TO 8: PRINT ct(i): NEXT

Posted by Charlie on 2008-11-07 13:39:24

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