All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > Just Math
Curious Cyclic Circumstance (Posted on 2009-02-07) Difficulty: 2 of 5
Three distinct 3-digit positive decimal integers P, Q and R, each having no leading zeroes and with P > Q > R, are such that:

(i) The product P*Q*R contains precisely one 8, one 6, one 5, one 4, two 3ís and three 2ís (albeit not necessarily in this order), and:

(ii) P*Q*R consists of precisely 9 digits with the last digit being 6, and:

(iii) P, Q and R are obtained from one another by cyclic permutation of digits.

Determine all possible triplet(s) (P, Q, R) that satisfy the given conditions.

Note: This problem can be solved without using a computer program, however computer program/spreadsheet solutions are welcome.

See The Solution Submitted by K Sengupta    
No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution computer solution Comment 1 of 1

DEFDBL A-Z
FOR n1 = 100 TO 999
  ns$ = LTRIM$(STR$(n1))
  n2 = VAL(MID$(ns$, 2) + LEFT$(ns$, 1))
  n3 = VAL(MID$(ns$, 3) + LEFT$(ns$, 2))
  prod$ = LTRIM$(STR$(n1 * n2 * n3))
  IF LEN(prod$) = 9 THEN
    t$ = "865433222"
    FOR i = 1 TO 9
     ix = INSTR(t$, MID$(prod$, i, 1))
     IF ix = 0 THEN EXIT FOR
     t$ = LEFT$(t$, ix - 1) + MID$(t$, ix + 1)
    NEXT i
    IF t$ = "" THEN PRINT n1; n2; n3, prod$
  END IF
NEXT n1

produces

398  983  839              328245326
839  398  983              328245326
983  839  398              328245326

Since P > Q > R, the triplet is (983, 839, 398) and the product is 328,245,326.


  Posted by Charlie on 2009-02-07 17:00:51
Please log in:
Login:
Password:
Remember me:
Sign up! | Forgot password


Search:
Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (9)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2017 by Animus Pactum Consulting. All rights reserved. Privacy Information