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Divisible by 198 (Posted on 2016-05-12) Difficulty: 3 of 5
N is a positive integer entirely formed by the digits 0,1,2,3 and 5 such that N is evenly divisible by 198.(Each of these five digits must occur at least once in N. None of the remaining five digits can occur in N.)

Determine the minimum value of N.

No Solution Yet Submitted by K Sengupta    
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analytical solution | Comment 1 of 2
198=2*9*11

in order for N to be divisible by 2 the last digit must be even
which leaves 0 or 2 as the last digit

in order for N to be divisible by 9 all the digits must total a multiple of 9

in order for N to be divisible by 11, the alternating addition/subtraction of the digits
must sum to a multiple of 11.

so starting from the left number the digits 0,1,2,...
let the sum of the even indexed numbers be x
and the sum of the odd indexed numbers be y

then we need
x+y=9a
x-y=11b
which gives

x=(9a+11b)/2
y=(9a-11b)/2

in order for x,y to be integers we need
a,b to either both be even or both be odd.

so let us try with a=1
that means the total of the digits must be 18
0+1+2+3+5=11, so we need to add more digits to get the total up to 18
the fewer the digits the better and this is done optimally with 2,5
thus we need to take the digits 0,1,2,2,3,5,5 and partition them into sets of 3 and 4
such that the difference in their totals is a multiple of 11
this is the only possible ways are
(5,3,1,0)(5,2,2)
and
(5,2,2,0)(5,3,1)
now we need an even digit as the last digit and we want the smallest possible leading digit
this is achieved with
(5,3,1,0)(5,2,2)
so the smallest number formed from these is done by taking the smallest digit from each group
into alternating positions (with exception of placing zero last)
this gives us the number
1232550=198*6225

  Posted by Daniel on 2016-05-12 09:54:54
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