Number Pad Snake

Relevant wiki: Divisibility Rules (2,3,5,7,11,13,17,19,...)

Answer: No. ( Without Calculation we can derive the result using divisibility test )

The given number pad has digits from 1 to 9. Clearly, it has 5 odd digits and 4 even digits. The adjacent to even digits are odd & adjacent to odd digits are even. The valid moves are odd to even (or) even to odd. If it start with even digit then it is not possible to end with even digit(since number pad has only 4 even digits). So it must start with odd digit, and the generated number will end with odd digit.

The possible start digits are 1, 3, 5, 7, or 9. Suppose, it start with 1, 3 or 9 then the generated number is divisible by 1, 3 or 9 respectively. Since all intigers divisible by 1 and sum of digits 1+2+3+…+9 is 45, which is divisible by 3 and 9. The other two possible start digits are 5 or 7. If it start with 5 then the generated number not divisible by 5. Because it is not possible to get the end digit as 0 or 5. Hence the result always not an integer.

Relevant wiki: Divisibility Rules (2,3,5,7,11,13,17,19,...)

You can start with the number $5$ in the middle. ${\large 521478963 ÷ 5 = 104295792.6}$

The quotient is not an integer.

Many starting with 7 are not: Ex: $\frac {789654123}{7}\approx112807731.9$

For a number to be completely divisible by 5, it has to end with 5 or 0, which is ENTIRELY NOT POSSIBLE in the given scenario. So the answer is NO.

There is no need to do any test calculations For any chosen route beginning with a 5 the resultant number must end in a 5 or a 0 for there to be an integer result. We have already used the 5 and there is no 0. Therefore the method does not always end in an integer.

Just start moving from five , and move any direction the result wont lead to an integer .

E.g. the number 563214789 is obviously not divisible by 5. Hence the answer must be No.

As others have said, starting at 5 will never produce a multiple of 5, but also if you start at an even digit, then you won't even be able to get to all the digits, because the digits must alternate parity.

$\frac {789654123}{7} = 112807731.9$

Assume that the statement above is true for all cases. Disregarding the first case (the one completed above), the one counterexample that I found is this: 789654123

7 + 8 + 9 + 6 + 5 + 4 + 1 + 2 + 3 = 45 (All the numbers add to 45)

The quotient of $\frac{45}{7}$ is not an integer.

Contradiction.

Therefore the statement is false .

Really, the number is a counterexample if it starts with 7, regardless of the other numbers in it.

The sum of all the digits is forty five which is not divisible by 6 therefore if you were to start with six you will not end up with an integers |||

Always is too suspicious

start with 5 and end with a number that's not 5.

Start from 4, you will end up with an odd digit in last.

A number that is divisible by 5 has to end in a 5 or a 0, and since 0 is not on the number table and 5 can only be used once, if you start with 5 and then divide by 5 at the end, you will not get an integer

If we begin with 5 the number doesn't end in 5 and zero is not given. So the resulting number doesn't end in 5 or zero. So it's not divisible by 5.

Simply start on 5 and take any path through the remaining digits. You can't end on 5 or 0, so the result cannot be divisible by 5.

If you start with 5 the number can't end with 5 and since there's no 0 and any number not ending in a 5 or a 0 is not divisible by 5 no value you make will be divisible by 5.

Answer: No

When a number is made on the number pad, in order for that number to be divisible by the first integer, the last two digits have to divisible by the first integer.

For example, when 7 is used as the first number, the possible numbers are 789654123 and 741258369. The last two digits of both numbers aren't able to be divided equally by 7

(69/7 = 9.8571428571 and 23/7 = 3.2857142857).

Therefore no, the method proposed in the question does not always result in an integer.

Easy way to solve this is to note that a number being divisible by $5 \iff$ last digit is $0$ or $5$ . Therefore we start with $5$ in the middle and any number we make doesn't end with either $0$ or $5$ and so won't be divisible by $5$ .

First, the problem is wrong. You cannot go through all the digits if you start from one of the even numbers.

Second, you have no solution if you start from 5. For a number to be divisible by 5 it must end in 0 or 5. You don't have 0 on the number pad and you just started with 5 so you can't also end with it.

21 is not divisible by 2 if ignore the other digits or else go around via any of the other digits back to 1 since the last digit 1 divided by a 2 is 1/2 i.e. 0.5 which is not an integer. This is the simplest proof aside from start the with and divide by a 5 one which must end with a 5 or 0.

Begin by a even number and end by a odd number.

The is often divisible by 9.

If you start on 5, you cannot possibly achieve an integer division. All numbers divisible by 5 end in 0 or 5. This number pad has no 0, and starting on 5 removes it as a possible final digit.