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Round Results (Posted on 2013-08-05) Difficulty: 2 of 5
In how many ways can you select a set of six distinct integers, each having absolute value below 51 to yield a product which is an integer power of 10?

No Solution Yet Submitted by Ady TZIDON    
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Solution re: a few more Comment 4 of 4 |
(In reply to a few more by Steve Herman)

Indeed I had neglected that allowing negative numbers also allows duplicates of a given absolute value. There are 1852 ways altogether.

By power of 10 they are

power of 10   number
    3 20
    4 152
    5 440
    6 632
    7 458
    8 140
    9 10

DEFDBL A-Z
DIM lct(25)
CLS
FOR a = -50 TO 45
    IF a MOD 3 <> 0 AND a MOD 7 <> 0 AND a MOD 11 <> 0 AND a MOD 13 <> 0 THEN
        FOR b = a + 1 TO 46
            IF b MOD 3 <> 0 AND b MOD 7 <> 0 AND b MOD 11 <> 0 AND b MOD 13 <> 0 THEN
                FOR c = b + 1 TO 47
                    IF c MOD 3 <> 0 AND c MOD 7 <> 0 AND c MOD 11 <> 0 AND c MOD 13 <> 0 THEN
                        abc = a * b * c
                        FOR d = c + 1 TO 48
                            IF d MOD 3 <> 0 AND d MOD 7 <> 0 AND d MOD 11 <> 0 AND d MOD 13 <> 0 THEN
                                FOR e = d + 1 TO 49
                                    IF e MOD 3 <> 0 AND e MOD 7 <> 0 AND e MOD 11 <> 0 AND e MOD 13 <> 0 THEN
                                        FOR f = e + 1 TO 50
                                            IF f MOD 3 <> 0 AND f MOD 7 <> 0 AND f MOD 11 <> 0 AND f MOD 13 <> 0 THEN
                                                prod = abc * d * e * f
                                                s$ = LTRIM$(STR$(prod))
                                                IF LEFT$(s$, 1) = "1" THEN
                                                    good = 1
                                                    FOR i = 2 TO LEN(s$)
                                                        IF MID$(s$, i, 1) <> "0" THEN good = 0: EXIT FOR
                                                    NEXT i
                                                    IF good THEN pow10 = LEN(s$) - 1: lct(pow10) = lct(pow10) + 1: ct = ct + 1: PRINT a; b; c; d; e; f, prod
                                                END IF
                                            END IF
                                        NEXT f
                                    END IF
                                NEXT e
                            END IF
                        NEXT d
                    END IF
                NEXT c
            END IF
        NEXT b
    END IF
NEXT a

FOR i = 0 TO 10: PRINT i, lct(i): NEXT
PRINT ct

 

 


  Posted by Charlie on 2013-08-05 22:17:25
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