   SEARCH HOME Math Central Quandaries & Queries  Question from Greg, a student: show that there is no five-digit number which uses each of the digits 1,2,3,4,5 such that the numbers formed by the first digit is divisible by 1, by the first two digits is divisible by 2 by the first three digits is divisible by 3, by the first four digits is divisible by 4, by the first five digits is divisible by 5 We have three responses for you

Hi Greg,

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.

Victoria

Hi Greg.

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
or 4 and 2.

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.

Cheers,
Stephen La Rocque.

Neat! I can prove that there are only two six-digit number which uses each of the
digits 1,2,3,4,5,6 such that the numbers formed

a) by the first digit (from the left) is divisible by 1,
b) by the first two digits (from the left) is divisible by 2
c) by the first three digits (from the left) is divisible by 3,
d) by the first four digits (from the left) is divisible by 4,
e) by the first five digits (from the left) is divisible by 5,
f) by the first six digits (from the left) is divisible by 6.

a) is no constraint at all.
b), d) and f) say that the second, fourth and sixth digits are even
e) says that the fifth digit is 5.
So, the first and third digits must be 1 and 3 in some order. Now, c) says that the sum of the first three digits must be a multiple of 3, so the second digit must be a 2. So, now d) says that the fourth digit must be a 6, so the last digit will be a 4.

The two possible numbers are 123654 and 321654.

Claude     Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences.