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Ternary Sum Travail (Posted on 2014-10-14) Difficulty: 3 of 5
G(n) represents the sum of the digits of the ternary representation of n, where n is a positive integer randomly chosen from 1 (base ten) to 2014 (base ten) inclusively.

Determine the probability that G(n) ≥ 7.

No Solution Yet Submitted by K Sengupta    
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Luckily enough, it is 2014 and not 20140, so straightforward bean-counting is feasible, albeit quite tedious.
 I have decided to post it after I saw that my result does not contradict computer findings.


There are 3^7 (=2187 )ternary 7-digit numbers between 0000000 and 2222222 inclusive.

The average G (=sum of digits) is 7 and clearly there are as many numbers   with G>7  as there are numbers   with G<7  ( consider 2-s complement:  - to any number with  G=7-K  corresponds exactly one number with  G=7+K ) .

So once we know the quantity of numbers with G=7, say Q,   then the quantity of numbers   with G>7

Will be  y=(2187-Q)/2.

Let see how many numbers (still up to 2187) have G=7:

All ones  ……………………………..… 1 way to do it
one 2, one zero, 5 ones…………..7*6=42 ways
two 2’s two zeroes,3 ones ……..21*10= 210 ways
three 2’s three zeroes,1 one ……..35*4= 140 ways

 Ergo Q=393 and our number    y=(2187-393)/2=897


From this number we have to deduce  the quantity falling  in the range between  2202122(2015 dec) to  2222222 (2186)  i.e. to examine 172 numbers for their s.o.d.

Instead of counting the numbers with G>7 let us show that only 6 numbers over 2014   yield  G=6 or less.
Indeed those numbers are  2210000, 2210001, 2210010, 2210100, 2211000 and 2220000.

So the quantity of numbers with G=7 in the range 0 to 2014 is 393-(172-6)=393-166=233

The total of 233 and  897 is 1124  and the desired probability    p=1124/2014=.558

Answer:   p=0.558

  Posted by Ady TZIDON on 2014-10-14 17:42:09
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