Enigma 1720

Spoiler for New Scientist Enigma 1720: “Tables seating four”

(Follow the link to see the puzzle.)

Consider the second number in the left column (the one ending in 4). It must be ≥ 50, in order for the first number to have three digits, but if it were 64 or higher then the last number in the left column would be 2 and it should be 1. So it must be 54, and p must be 108 or 109.

The fifth number in the second column must be divisible by 4 (in fact by 16) so the last two digits must be divisible by 4. So they must be 40 or 48 (not 44 since the last digit is not shown to be 4). But if 48 then the fourth number must end in 4, and it doesn’t, so the fifth number ends in 40.

That means the first number, P, ends in 0 or 5. If it ended in 0, or if it ended in 5 and p were even, then ENIGMA would end with a 0, which is not allowed. So P ends in 5 and p is 109, and the second number in the second column (2P) must be even but not divisible by 4.

If the fifth number ends with 40 then the fourth number (divisible by 8) must end with 20; the third (divisible by 4) must end with 60; the second (divisible by 2 but not 4) must end with 30; the first (P) must end with 15 or 65.

Trying out possibilities for the last three digits of P, only 315, 815, 915, 065, 165, 565, and 665 give no 4 in the last three digits of all the numbers in the right column, except for the one in the fifth number.

Multiplying each of these by 109 we find we can rule out 315, 065, and 565 because they do not give three different nonzero digits in the last three places of the product.

In order for the fifth number to have six digits, P must be more than 6250. In order for 109P to have six digits, P must be under 9175. This reduces the number of possible values for P to twelve — four values for the last three digits, and three values in each of these cases for the first digit.

Of these, only three give six different nonzero digits when multiplied by 109:

109 * 6815 = 742835
109 * 7915 = 862735
109 * 6665 = 726485

Trying these out, we reject 6815 since 8 * 6815 = 54520 and 64 * 6665 = 426560, both of which contain the digit 4. We’re left with p = 109, P = 7915, and ENIGMA = 862735.

Which has no 4 in it either!

 

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