To decide whether a number of n digits is divisible by 7, we can define a process by which its magnitude is reduced as follows: (i_{1}, i_{2}, i_{3},..... are the digits of the number, starting from the most significant digit). i_{1} i_{2} ... i_{n} => i_{1}.3^{n-1} + i_{2}.3^{n-2} + ... + i_{n}.3^0. e.g. 259 => 2.3^2 + 5.3^1 + 9.3^0 = 18 + 15 + 9 = 42 Ultimately the resulting number will be seven after repeating the above process a certain number of times. After how many such stages, does the number 203 reduce to 7?

Question 49

To decide whether a number of n digits is divisible by 7, we can define a process by which its magnitude is reduced as follows: $$(i_{1}, i_{2}, i_{3}$$,..... are the digits of the number, starting from the most significant digit). $$i_{1} i_{2} ... i_{n} => i_{1}.3^{n-1} + i_{2}.3^{n-2} + ... + i_{n}.3^0$$.
e.g. $$259 => 2.3^2 + 5.3^1 + 9.3^0 = 18 + 15 + 9 = 42$$
Ultimately the resulting number will be seven after repeating the above process a certain number of times. After how many such stages, does the number 203 reduce to 7?

Solution

For 203 :
first step = $$2\times 3^2 + 0 \times 3^1 + 3 \times 3^0$$ = 21
second step = $$2 \times 3^1 + 1 \times 3^0$$ = 7
So two steps needed to reduce it to 7

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