414 Digit problems
Project
Arrange the ten digits (0, 1,..., 9, norepetition) in a row
abcdef ghij
so that the following 3-digit numbers in the table below are divisible by
the prime below them.
bcd cde def ef g f gh ghi hij
2 3 5 7 11 13 17Solution: 1460357289 or 4160357289
Sincedefis divisible by 5,fmust be 0 or 5. Now, the numberfghmust
be divisible by 11. The only number that can be formed from the first
two digits of 13 mby appending a 0 ora5ontheleft to form a multiple
of 11 are 286, 390, 728, and 832. Clearly,f=5, andfghiis only one
of
5286 , 5390 , 5728 , 5832.
Now we want to find a multiple of 17 beginning with the last two digits
of these. This eliminates the last case.
52867 , 53901 , 57289.Sinceef g is a multiple of 7,mod(e,7) must satisfy2mod(e,7) +
mod(10f+g,7)≡0mod7. Necessarily,e=2, 1 , 3. Now we must
haveef ghij= 357289.
Now it remains to arrange 0, 1, 4, 6 asa,b,c,dsuch that the three
digit numbersbcd,cd 3 are 3 digit numbers divisible by 2, 3 respectively.
In particular,c+d+3is divisible by 3. The only choice isc=6and
d=0.Wehaveab 60357289 .aandbcan be 1, 4 or 4, 1.