Le Monde problem on PIN number

Now that the new school year has started, Christian Robert has picked up solving the Le Monde mathematical puzzles using R again, which leads me to solving them without R… Last week-end’s puzzle is:

Alice’s PIN number is made of four different non-zero digits. She remembers it by noting that when she sums all two-digit numbers that can be made out of those four digits, and multiplies this sum by 7, she gets her PIN number back.

If the PIN number is written \overline{abcd}, it is easy to check that the sum computed by Alice is equal to 33*(a+b+c+d). Thus \overline{abcd}=231*(a+b+c+d).

The sum (a+b+c+d) can take values between 10 and 30. Since 231 is divisible by 3, so is \overline{abcd}. Hence (by the rule used to check for divisibility by 3), (a+b+c+d) is divisible by 3. This makes \overline{abcd} divisible by 9, and so (by the rule used to check for divisibility by 9) (a+b+c+d) is also divisible by 9.

The only possible values for (a+b+c+d) are 18 and 27. It turns out that the only value which works is 18. Hence the PIN number is 18 \times 231 = 4158 (and the sum of those four digits is indeed equal to 18).



One Response to “Le Monde problem on PIN number”

  1. Mahendra Says:

    You can further restrict the possible values for (a+b+c+d) by noting that 11 divides abcd and so (this is the less well known rule to check for divisibility by 11 based on the fact that 10^k = (-1)^k mod 11) d-c+b-a=0. Hence a+b+c+d is even and the only even mutiple of 9 between 10 and 30 is 18.

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