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 39 does it (Posted on 2012-02-19)
Prove that within any 39 consecutive natural numbers there must exist at least one number with the sum of its digits divisible by 11.
Bonus: Can the bound (39) be lowered?

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 computer exploration -- bonus only is shown | Comment 1 of 4

DEFDBL A-Z
FOR n = 1 TO 99999999
s\$ = LTRIM\$(STR\$(n))
sod = 0
FOR i = 1 TO LEN(s\$)
sod = sod + VAL(MID\$(s\$, i, 1))
NEXT
IF sod MOD 11 = 0 THEN
IF gap > max THEN
max = gap
PRINT prev, n, gap
END IF
prev = n
gap = 0
ELSE
gap = gap + 1
END IF
NEXT

finds that 999980 and 1000019 are two consecutive occurrences of numbers whose sod is a multiple of 11, and there is a gap of 38 where the sod is not a multiple of 11, so one of these two has to be included to get the congruence, bringing the set's cardinality to 39.

 Posted by Charlie on 2012-02-19 12:38:12

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