Find all numbers

[Rushil]

Find all 6-digit numbers formed by using the digits once each such that the number is divisible by for .

[Singular]

a_1 is always divisible by 1.

a_1a_2 is divisible by 2 if a_2 is even, so it is 2,4,6. (2)

a_1a_2a_3 is divisible by 3 if a_1 + a_2 + a_3 is 0 mod 3. (3)

a_1a_2a_3a_4 is divisible by 4 if a_3a_4 is, in particular a_4 must be even. (4)

a_1a_2a_3a_4a_5 is div. by 5 if a_5 = 0 or 5, so a_5 = 5. (5)

a_1a_2a_3a_4a_5a_6 is div by 6 if a_6 is even and is 0 mod 3, so a_4+a_5+a_6 must be. (6)


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Try a_6 = 2. From (6) we must have 5+2+a_4 0 mod 3, which cant happen.

Try a_6 = 4. From (6) a_4 is 0 mod 3, so it must be 3 or 6.
From (4) a_4 is even, so it must be 6, and then a_3 must be 1.
In (2) the only even number left is 2 = a_2. Then a_1 = 3 which checks out.

Try a_6 = 6; from (6) we get a_4 = 1 or 4 ; must be 4 by (4). Then from (2) we have a_2 = 2, and we can mix the choices.

In total, 321654, 123456, 321456

[varun]

sorry for bringing this old post now.
I was searching for RMO problems.

SINGULAR, you should have checked your solution. Only 1 of your answers is correct - 321654
Also this one is a answer - 123654

[Attila the Hun]

a_5 = 5(fixed).

a_2, a_6,a_4 have to be 2, 4, 6 in some order.

that leaves us with 1 and 3 for a_1 and a_3.

if a_3 = 1, a_4 = 2 and a_2 = 4 and a_6 = 6

if a_3 = 3, a_4 = 6 and a_2 = 2 or 4 and a_6 = 4 or 2.

thus, the only numbers are

123654, 143652, 341256.

[murgi]

thus, the only numbers are

123654, 143652, 341256. [/quote]

ur 2nd answer is incorrect......143 is not divisible by 3
ooooo also the third one.....

[Attila the Hun]

oh, yes I was wrong

if is , can be or .

[Maharjun]

Highly intuitive->

Click to see solution

1. is divisible by 5
   
   
2. will always be divisible by 3 as which is divisible by three.
   if is even then it will be divisible by also and hence by .
   any even number
   
   
3. also for to be divisible by 2, any even number
   
   
4. Now we can see that and belong to either of the odd numbers 1 or 3 since all the three even numbers are used up in , (which is also ovbiously even), . and
   
   
5. now is divisible by 4 is even. also its value will depend on wether the preceding digit is even or odd. in this case, the preceding digit is odd by statement 4. hence
   
   
6. this means that can come only in or using statement 4
   
   
7. now is divisible by 3 multiple of three. now
   and using 3.
   
   thus only if we have multiple of 3
   
   
   
8. using statements 5,7 we have
   
   
9. using statements 6,7 we have
   
   
10. hence we are left only with {a_1} and {a_3} which can have values 1 or 3. hence we arrage 1,3 among and to get two answers
    

[saubhik mukherjee]

Statement 7: a little mistake; it should be