The numbers 160, 161 and 162 form a set of three consecutive integers (in order) that are divisible respectively by 5, 7 and 9.
Find the first set of four consecutive integers (in order) that are respectively divisible by 5, 7, 9, and 11.
How about first set of five consecutive integers (in order) that are respectively divisible by 5, 7, 9, 11 and 13.
CLS
FOR n = 5 TO 999999
m11 = n MOD 11
IF m11 = 0 THEN
m9 = n MOD 9
IF m9 = 1 THEN
m7 = n MOD 7
IF m7 = 2 THEN
m5 = n MOD 5
IF m5 = 3 THEN PRINT "Part a:"; n  3; "thru"; n: GOTO p2
END IF
END IF
END IF
NEXT
p2:
FOR n = 5 TO 999999
m13 = n MOD 13
IF m13 = 0 THEN
m11 = n MOD 11
IF m11 = 1 THEN
m9 = n MOD 9
IF m9 = 2 THEN
m7 = n MOD 7
IF m7 = 3 THEN
m5 = n MOD 5
IF m5 = 4 THEN PRINT "Part b:"; n  4; "thru"; n: END
END IF
END IF
END IF
END IF
NEXT
finds
Part a: 1735 thru 1738
Part b: 22525 thru 22529
Of course the above has somewhat silly programming (though speed wasn't a factor here, it is still sloppy coding). The outer loop for part a should have started
FOR n = 11 to 999999 STEP 11
so that the test for divisibility by 11 could have been eliminated.
A similar consideration would be for part b:
FOR n = 13 to 999999 STEP 13
and the test for divisibilty by 13 eliminated.
But the way it was done made copying one set of code to the other easier.

Posted by Charlie
on 20130204 16:34:33 