60’s Puzzler

Can you solve this puzzle?

Text from the image:

And Let’s Consider Another Quotient. It takes shape as men’s minds go deep into space. One center of these calculations is Martin-Denver, where successful professional enjoy full scope from basic research to vehicle design. You will find range, reward, and stature here…

CLUE: The sum of the digits of the divisor leaves a remainder of seven when divided by nine, and the sum of the digits of the quotient leaves a remainder of three when divided by nine.

Some clues to help you out, based on *my* solution:

  1. The color of the chess pieces is not important.
  2. The type of chess piece is not important.
  3. None of the numbers represented have leading zeros.
  4. The “clues” are required to solve the puzzle.

Ready for my solution?  Ok, here goes.

First, let’s rewrite the base equation as A / B = C.  A is seven digits, so it is in the range 1000000 <= A <= 9999999.  Likewise, 100<= B <= 999 and 1400 <= C <= 9499.

Upon inspection of the first line of the long division (pawn pawn pawn pawn minus knight pawn knight) it can be certain that the first two digits of A are 1,0.  And the first digit knight-pawn-knight is a 9 (no other combination would result in a difference with a zero in the first two places).

Thus 1000000 <= A <= 1099999.  At this point we can refine the ranges for B and C : The largest value of A divided by the smallest B gives the upper range of C, etc…

So C is at most: 1099999 / 100 = 11000.  Previously we’d shown C <= 9499, so this doesn’t help.

C is at least 1000000 / 999 = 1001.  Again no help.

B is at most 1099999 / 1400 = 785.  Progress.

B is at least 1000000 / 9499 = 106.

However, we know that 4 * B is a three digit number (blue bishop – red king – blue bishop).  So if 4 * B < 1000, B must be less than 250.  Revising again we restrict the ranges to

1000000 <= A <= 1099999, 106 <= B <= 249, 4416 <= C <= 9499

At this point we can apply the rule about the quotient (C).  In the range 4416 to 9499 with the second digit a 4, only these values meet the rule: 8409, 8400, 7401, 6402, 5403.  From here is a simple matter to further restrict the ranges for B, e.g.

for C = 8409, 119 <= B <= 130.  However, applying the clue for B shows that only 124 is a valid value for B in that range.

It is then a simple matter to calculate A for B=124, and C=8409 : 1042716.

Then do the long division and see if the number of digits line up.  For each line in the figure they are:

  1. 992
  2. 507
  3. 496
  4. 1116
  5. 1116

Success!  Crank through the other values for B (8400, 7401, 6402, 5403) and you’ll see this is the only answer with the correct number of digits on each line.

3 thoughts on “60’s Puzzler

  1. Brilliant! I can’t tell you how many countless hours and wasted paper it took me to come to essentially the same conclusion. I might quible with the whole “9408 issue” but I got to send a shout out and props to you!

  2. Pingback: A tricky puzzle from the past | The Universal Machine

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