[Guest]

N is a ten digit base ten positive integer having the form PQRSTUVWXY that uses each of the digits 0 to 9 exactly once, and satisfies all of these conditions:

(i) XY is divisible by 10.

(ii) WX is divisible by 9.

(iii) VW is divisible by 8.

(iv) UV is divisible by 7.

(v) TU is divisible by 6.

(vi) ST is divisible by 5.

(vii) RS is divisible by 4.

(viii) QR is divisible by 3.

(ix) PQ is divisible by 2.

Determine all possible value(s) that N can assume.

[Glycereine]

2 Possible solutions:

7812549630

1872549630

[Guest]

2 Possible solutions:

7812549630

1872549630

Congratulations!!!

Indeed, 7812549630, or 1872549630 are the only two values that N can assume.

The next step is a proof that these are the only two values that N can assume.

Edited June 6, 2009 by K Sengupta

[Glycereine]

The next step is a proof that these are the only two values that N can assume.

Congratulations!!!

Indeed, 7812549630, or 1872549630 are the only two values that N can assume.

[spoiler='Proof

']First, Q,S,U,W and Y must be even numbers (they are the second digit of numbers that must be divisible by a number with a factor of 2.

Therefore the others must be odd.

- Y must = 0 as any number divisible by 10.

- T must = 5 as any number divisible by 5 must end in 0 or 5 (0 is taken by Y)

- U must = 4, for working down, the only number divisible by 6 that starts with 5 is 54

- V must = 9 for the only two numbers divisible by 7 that begin with 4 are 42 and 49, but V must be odd (see top)

- W must = 6 as the only number beginning with 9 that is divisible by 8 is 96

- X must = 3 because the only number divisible by 9 beginning with 6 is 63

Simple part is done.

Now starting back at S, the lowest digit we do not yet know. RS must be divisible by 4 and the only 2 even numbers remaining are 8 and 2. S cannot be 8 because the two odd numbers left, 1 and 7 combined with 8 are not divisible by 4. 18/4 = 4.5 and 78/4 = 19.5

- S must = 2

- Q must = 8 as it is the only even number left

- R can be either 1 or 7 because both 12 and 72 are divisible by 4. Also both 81 and 87 are divisible by 3.

- P can be anything because no matter what the number starts with PQ will be divisible by 2 if Q is even. Therefore P = 7 or 1 (as well as R = 1 or 7).

Swapped with each other they make the two possibilities

[Guest]

[spoiler='Proof

']First, Q,S,U,W and Y must be even numbers (they are the second digit of numbers that must be divisible by a number with a factor of 2.

Therefore the others must be odd.

- Y must = 0 as any number divisible by 10.

- T must = 5 as any number divisible by 5 must end in 0 or 5 (0 is taken by Y)

- U must = 4, for working down, the only number divisible by 6 that starts with 5 is 54

- V must = 9 for the only two numbers divisible by 7 that begin with 4 are 42 and 49, but V must be odd (see top)

- W must = 6 as the only number beginning with 9 that is divisible by 8 is 96

- X must = 3 because the only number divisible by 9 beginning with 6 is 63

Simple part is done.

Now starting back at S, the lowest digit we do not yet know. RS must be divisible by 4 and the only 2 even numbers remaining are 8 and 2. S cannot be 8 because the two odd numbers left, 1 and 7 combined with 8 are not divisible by 4. 18/4 = 4.5 and 78/4 = 19.5

- S must = 2

- Q must = 8 as it is the only even number left

- R can be either 1 or 7 because both 12 and 72 are divisible by 4. Also both 81 and 87 are divisible by 3.

- P can be anything because no matter what the number starts with PQ will be divisible by 2 if Q is even. Therefore P = 7 or 1 (as well as R = 1 or 7).

Swapped with each other they make the two possibilities

Well done and congratulations once again !!!!!

This indeed completes the solution to the given puzzle.

[Glycereine]

Thanks!! I encourage others to do it though without looking, it doesn't take too long but you can make it harder than it should be (I almost did)

[Guest]

First, Q,S,U,W and Y must be even numbers (they are the second digit of numbers that must be divisible by a number with a factor of 2.

Therefore the others must be odd.


- Y must = 0 as any number divisible by 10.
- T must = 5 as any number divisible by 5 must end in 0 or 5 (0 is taken by Y)
- U must = 4, for working down, the only number divisible by 6 that starts with 5 is 54
- V must = 9 for the only two numbers divisible by 7 that begin with 4 are 42 and 49, but V must be odd (see top)
- W must = 6 as the only number beginning with 9 that is divisible by 8 is 96
- X must = 3 because the only number divisible by 9 beginning with 6 is 63


Simple part is done.

Now starting back at S, the lowest digit we do not yet know. RS must be divisible by 4 and the only 2 even numbers remaining are 8 and 2. S cannot be 8 because the two odd numbers left, 1 and 7 combined with 8 are not divisible by 4. 18/4 = 4.5 and 78/4 = 19.5

- S must = 2
- Q must = 8 as it is the only even number left
- R can be either 1 or 7 because both 12 and 72 are divisible by 4. Also both 81 and 87 are divisible by 3.
- P can be anything because no matter what the number starts with PQ will be divisible by 2 if Q is even. Therefore P = 7 or 1 (as well as R = 1 or 7).

Swapped with each other they make the two possibilities Oh so that's why. At first I only got 7812549630, but that was because I worked backwards starting from Y moving to P. If you switch the first 3 numbers, you will still get a viable answer of 1872549630.