What is the answer to this hard problem for 4th graders?

Question

Here is the problem. Peter's password for his mail is a 6 digit number. The first two digits are his house number. Next to them is the sum of the digits of his phone number. Next to them is the sum of the last two digits of the preceding digits and this is the complete password. What is the password if 3 divides the house number and the password does not have 1 among its digits?

Answer 1

Is it possible to modify the problem so that it has only one solution? Without writing $6$ as $06$ of course. For example: "Next to them is the sum of the last two digits of the preceding digits and so on until we get a $6$ digit number."

Yes, with this version of the problem reading as follows:

Peter's password for his mail is a six-digit number. The first two digits are his house number, next is the sum of those two digits, next is the sum of the last two of all the preceding digits, next is the sum of the last two of all the preceding digits, and so on, until we get a six-digit number, where all numbers are written with no leading $0$s. What is the password if $3$ divides the house number and the password does not have $1$ among its digits?

This has a unique solution, which can be found either by "brute force" inspection of all the relevant cases, or by algebra.

Brute force method: Here are the relevant cases, each continued only until a $1$ (or a solution) appears:

24 6 10
27 9 16
30 3 3 6 9  <-- 303369
33 6 9 15
36 9 15
39 12
42 6 8 14
45 9 14
48 12
54 9 13
57 12
60 6 6 12
63 9 12
66 12
69 15
72 9 11
75 12
78 15
84 12
87 15
90 9 9 18
93 12
96 15
99 18

Algebraic method: First note that if the sum of two digits is a two-digit number, then the first digit of that sum is a $1$, because the sum can only range from $0+0 = 0$ up to $9+9= 18$. Therefore, denoting the six-digit password by $abcdef$, it can be formed only by the following steps, each of which adds the last two preceding digits:

$$ab \to c \to d \to e \to f$$

where

$$c = a + b,\quad d = b + c, \quad e = c + d, \quad f = d + e,$$

such that $3$ divides $ab$, none of the digits is a $1$, and $a > 0$ (so $a \ge 2$).

Thus, substituting to find $f$ in terms of $a$ and $b$ alone: $$d = b + (a + b) = a + 2b$$ $$e = (a+b) + (a+2b) = 2a + 3b$$ $$f = (a + 2b) + (2a + 3b) = 3a + 5b.$$

Now $f \le 9$ and $a \ge 2$, so $3 \cdot 2 + 5b \le 3a + 5b \le 9$; hence $b \le \frac{9-3\cdot 2}{5} = \frac{3}{5}$, which implies $b = 0$. But then $3a + 5b \le 9$ implies $a \le 3$, so $a$ is either $2$ or $3$. Since $3$ must divide $ab = a0$, it follows that $a = 3$, and the entire password is now determined to be $303369$:

$$30 \to 3 \to 3 \to 6 \to 9.$$

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NB: I strongly suspect that the original question was "mis-edited". The present unique-solution version would have resulted if the only changes to the original had been "home" $\to$ "house", and "this is the complete password" $\to$ "this is repeated until the password is complete".

Answer 2

This is impossible. The sum of two one-digit numbers is between $0$ and $18$. Since the digit $1$ doesn't appear, the two sums in the password must each be single-digit numbers. But then the password has only $4$ digits, not $6$.

Answer 3

If the question is interpreted as Eric proposes, then there are sixteen solutions. Specifically, take any two-digit number divisible by 3 whose digits do not contain 1 and whose sum of digits do not contain 1. These are: 00, 03, 06, 09, 24, 27, 30, 33, 36, 42, 45, 54, 60, 63, 72, 90. Then that determines the rest of the number:

000000
030303
060606
090909
240606
270909
300303
330606
360909
420606
450909
540909
600606
630909
720909
900909