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The Minimum Digit Arithmetic Progression Puzzle

The Minimum Digit Arithmetic Progression Puzzle

Find six numbers with difference between each pair of consecutive numbers constant. The condition in this puzzle is to use only 3 digits in the six numbers.

The puzzle

The four numbers 5, 55, 105 and 155 are special in two ways: first, in the arithmetic progression, difference between each pair of consecutive numbers is constant 50, and second, only three digits 0, 1 and 5 are used in the four numbers.

The problem: Can you think of six such numbers in an arithmetic progression using just three digits?

Time to solve the riddle: 30 minutes.

It is true, no arithmetic progression can have only two digits used in the numbers. Indeed the given series is special as not only is it an arithmetic progression with constant increment of 50, it also have only the 3 digits used. But such a six number arithmetic progression? It needs some thinking.

A hint: more than mathematical knowledge, discovery of necessary pattern in the digits and mathematical reasoning will play primary role in solving the riddle.

Solution to the minimum digit arithmetic progression puzzle

The first realization is:

To maximize efficiency of digit use in the series, the first in the series must be a single digit number. This will have the big advantage of reuse as the most significant or the least significant digit in the following five other members.

Ignoring the four given numbers, our focus is on knowing the most important question that needs to be answered for smooth solution.

It doesn't take long to hit upon the critical question:

How to minimize the number of digits used in the most significant digits of the six numbers in the series?

Reasons for analyzing the most significant digit first are:

  1. If the numbers are spread all over the range between 1 to 99, no way a minimum digit arithmetic progression can be formed. The range of numbers must further be constrained. For this, analyzing the most significant digits will play the critical role.
  2. Alternately, if the series crosses 100, it will be absolutely impossible to control the number of digits used in a six number series.

With these insights, the most promising answer to the critical question is:

Restrict the six numbers in the series below 30 so that only two digits 1 and 2 (tens and twenties) are used as most significant digits and the third digit, the starting single digit number, used in forming the second digit (from left) of the (five) two digit members.

This realization implies with certainty:

The numbers in the series must have constant increment of 5 (six numbers, each pair of consecutive numbers separated by 5).

This only will ensure the largest in the series of 6 numbers to remain less than 30 (within twenties, satisfying most significant digit analysis).

Next logical conclusion follows:

1 and 2 being used as the most significant digits in the last five members of the series, the starting single digit number should be either of the two to maximize efficiency of use of digits.

So, follows the first solution starting from 1,

1, 6, 11, 16, 21, 26 : three digits 1, 2 and 6 are used with series increment of 5.

Any more solution?

The second solution starting the series from 2 is also immediate,

2, 7, 12, 17, 22, 27 : three digits 1, 2 and 7 are used with series increment of 5.

These are the only two solutions of the puzzle for the simple reason of any other starting digit such as 0, 4 or 5 introducing a fourth digit in the series and starting from 6 onwards exceeding 29.

Check yourself.

Food for more thought

How many four member minimum digit arithmetic progression possible with constant increment 50?

Or, any minimum digit arithmetic progression possible with increment other than 5 or 50?


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