We have three responses for you
Maybe it will help to know the rules for testing divisibility.
Every integer is divisible by 1.
An integer is divisible by 2 if and only if its last digit is 0, 2, 4, 6, or 8.
An integer is divisible by 3 if and only if the sum of its digits is divisible by 3.
An integer is divisible by 4 if and only if the number obtained from its last two digits is a multiple of 4. For example 7724 is divisible by because 24 is divisible by 4.
An integer is divisible by 5 if and only if it its last digit is 0 or 5.
If you use these rules starting with the first digit, the given conditions tell you a lot about the digits of the number. For example, the last digit has to be 5 and the 2nd digit has to be 2 or 4. You might have to try each possibility and work from there.
The 5 digit number is divisible by 5, and only a number ending in zero or five is divisible by 5, so the last digit is a 5. We can now think of a four digit number with 1, 2, 3, 4 each used once.
The last digit is a 4, so it must be an even number. The only choices are 2 and 4. As well, the first two digits end in an even number, because that's divisible by 2. So the 2nd and 4th digits are 2 and 4
That leaves 1 and 3 or 3 and 1 for the first and third digits.
Can you write out the four possibilities and test each one? Then you can prove your hypothesis.
Neat! I can prove that there are only two six-digit number which uses each of the
a) by the first digit (from the left) is divisible by 1,
a) is no constraint at all.
The two possible numbers are 123654 and 321654.
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