Pandigital Sum (Posted on 2006-09-19)

	Submitted by Charlie
Rating: 5.0000 (1 votes)
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The 2-, 3- and 4-digit numbers that are divisible by each of their digits are: 12 124 1236 3915 15 126 1248 3924 24 128 1296 4128 36 132 1326 4172 48 135 1362 4236 162 1368 4368 168 1395 4392 175 1632 4632 184 1692 4872 216 1764 4896 248 1824 4932 264 1926 4968 312 1935 6132 315 1962 6192 324 2136 6312 384 2184 6324 396 2196 6384 412 2316 6432 432 2364 6912 612 2436 6984 624 2916 8136 648 3126 8496 672 3162 8736 728 3168 9126 735 3195 9135 784 3216 9162 816 3264 9216 824 3276 9315 864 3492 9324 936 3612 9432 3624 9612 3648 9648 3816 9864 3864 The solution is: 48 + 672 + 3195 = 3915 DIM num(2 TO 4, 80) col = 1: row = 1: pLen = 2 CLS FOR i = 12 TO 9876 n$ = LTRIM$(STR$(i)) good = 1 FOR j = 1 TO LEN(n$) d$ = MID$(n$, j, 1) IF INSTR(MID$(n$, j + 1), d$) THEN good = 0: EXIT FOR IF d$ = "0" THEN good = 0: EXIT FOR IF i MOD VAL(d$) <> 0 THEN good = 0: EXIT FOR NEXT IF good THEN pCt = pCt + 1 IF LEN(n$) > pLen THEN col = col + 1: row = 1: pCt = 1 num(LEN(n$), pCt) = i howMany(LEN(n$)) = pCt LOCATE row, col * 10 - 9 PRINT i; : ct = ct + 1 row = row + 1 IF row > 34 THEN row = 1: col = col + 1 pLen = LEN(n$) END IF NEXT PRINT ct LOCATE 38, 1 FOR a = 1 TO howMany(2) FOR b = 1 TO howMany(3) FOR c = 1 TO howMany(4) b$ = LTRIM$(STR$(num(2, a))) + LTRIM$(STR$(num(3, b))) + LTRIM$(STR$(num(4, c))) good = 1 FOR i = 1 TO LEN(b$) - 1 IF INSTR(MID$(b$, i + 1), MID$(b$, i, 1)) THEN good = 0: EXIT FOR NEXT IF good THEN dVal = num(2, a) + num(3, b) + num(4, c) good = 0 FOR i = 1 TO howMany(4) IF dVal = num(4, i) THEN good = 1: EXIT FOR NEXT IF good THEN PRINT num(2, a); num(3, b); num(4, c); dVal END IF NEXT NEXT NEXT From New Scientist, 19 August 2006, Enigma number 1405.

	Subject	Author	Date
	Numbers	Math Man	2017-03-10 14:25:06
	answer	K Sengupta	2008-11-24 00:45:17
	re(2): Solution	Penny	2006-09-19 10:15:26
	re: Solution	bumble	2006-09-19 08:30:29
	Solution	Penny	2006-09-19 07:54:02