Applications of Bases: Part I, Problem 2

$\left ( 3322\right )_{3}=\left ( 2\times 3^{3} \right )+\left ( 2\times 3^{2} \right )+\left ( 1\times 3^{1} \right )+\left ( 1\times 3^{0} \right )=54+18+3+1=76$

I did it by counting the possible number of numbers formed with 0,2,3 upto 3322.

• Single digit numbers - 2,3 are 2 in all,

• Two digit numbers - 20,22,30,... are 6 in all,

• Three digit numbers - 200, 202, 203,.... are 18 in all,

• Four digit number starting with 2 - 2000, 2002, 2003, ..... are 27 in all,

• Four digit numbers starting with 30 - 3000, 3002,3003,... are 9 in all,

• Four digit numbers starting with 32 - 3200,3202,3203,... are 9 in all,

• Four digit numbers starting with 330 - 3300, 3302, 3303,.... are 3 in all,

• Only one number 3320 before 3322.

So the total numbers including 3322 are 2+6+18+27+9+9+3+1+1 = 76

It is just a trinary number 3322 -> 2211 ; 2 2 1 1 -> 27x2 + 9x2 + 3 +1 = 54 + 18 + 3 +1 = 76

This is the no. with base three where 3322 represents 2211 so answer is 1+1x3+2x9+2x27=76

We observe that we are actually counting in trinary! We are simply replacing the numbers 1 and 2 with 2 and 3, respectively. Therefore, 3322 is the 2211(base 3)-th number, or the 76th.