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 thought on “Le Monde problem on PIN number

  1. 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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