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# Sunday Times Teaser 2672 – 11, 12, 13 …

### by Victor Bryant

To celebrate the upcoming date of 11/12/13, I have selected three integers whose sum is 2013.

The smallest of these integers is divisible by 11 and the other two are divisible by 12 and 13 respectively.

Between them they use a total of nine digits, none more than once.

What are my numbers in increasing order?

4 Comments Leave one →
1. brian gladman permalink

• geoffrounce permalink

Here is a slower permutation solution:

2. ahmet cetinbudaklar permalink

Taking the three numbers to be a 3-digit number, the smallest one cannot be larger than 594 since the sum of the three numbers must be equal to 2013 and the first digit of the other two numbers can be 6,7,8 or even 9.
Thus we can approach the etaser by trial and error as seen from the table below:

/11 /12 /13
594 612 807x
583 672 x
572 648 x
561 708 x
539 840 x
528 924 x
517 804 692x
506 732 x
495 816 702 which gives us the solution.

3. brian gladman permalink

Hi Geoff,

You might be interested to know that a permutation based solution can be made very fast by handling the units, ten and hundreds columns of the sum separately as follows:

This is significantly faster than my first solution above.