Paidion wrote:Congratulations, Graeme! You have developed a method for finding my number 23.

Here is the method I use. I even wrote a computer program in BASIC to do the work for me.

For any remainders a,b,c, use the formula 70a + 21b + 15c. To make it easier subtract 105 whenever possible.

Suppose we use this procedure to calculate the number which yields the remainders I gave you: 2,3,2

We have 70×2 + 21×3 + 15×2

But 70×2=140. That's more than 105, so subtract 105 and get 35.

21×3=63. Add the 35 to it and get 98.

15×2 =30. Add the 98 to it and get 128. That's more than 105, so subtract 105 and get 23.

I still don't understand how to carry out your procedure. But is it harder or easier to work than this one?

I like the formula and imagined there would be a simple way to calculate the number. But I cannot figure out some of it. I understand the use of the figures 115,21, and 105. I haven't figured out why you use 70 and not 35 for the first part of the formula, except that it works. I recall the days when I wrote in BASIC as well and had fun doing it. The BASIC I wrote in was very, very,' basic' and I have not kept up with the modern versions.

Here is a more detailed explanation of how I solved the puzzle. I will use the remainders 2,3,2 to arrive at the number 23

1. There are 100 possible answers and the task is to eliminate all but one.

2. Solve the test of the multiplier (5). We know that any number that is a multiple of 5 ends with either 0 or 5 (5,10,15,etc). But because there is a remainder, then the 0 and 5 must be increased by that remainder. So for the remainder of 3, the answer must end in either 3 (0 + 3) or 8 (5 + 3). So now we know that the solution to the puzzle must have a last digit of either 3 or 8 and our options are limited to these numbers 3,13,23,33,43,53,63,73,83,93 and 8,18,28,38,48,58,68,78,88,98. There are no other numbers between 1 and 100 that pass the test of the multiplier (5).

3. Solve the test of the multiplier (3). We know that any number that is divisible by 3 also has a multiple of 3 for the sum of its digits. So, for example, 12, 72, 1212, are all divisible by 3 because in each case the sum of the digits that make up the number are also divisible by 3 (e.g. 1+2+1+2 = 6 and that is divisible by 3). [Incidentally this is also true for the number 9 as well. And 3^n.]

4. Because there is a remainder of 2 we have to factor that in as well. So instead of looking for digits that add up to 3,6,9,12,15 or 18, we must add to each of these the remainder and look for digits that add up to 5,8,11,14,17,20.

5. Of the list in point 2 only the following qualify 23 (2 + 3 = 5 etc), 53,83,38,68,98. So our list of candidates is now down to 6

7. Solve for the test of the multiplier (7). I do not know of any test for seven other than trial and error. But the easy way to do this is to deduct the remainder of 2 from each candidate (in point 5) and see if the result is divisible by 7. So this adjusted list ( above less 2 ) now becomes 21,51,81,36,66,96, for which we determine that only 21 is divisible by 7. Then we add back the 2 to get our final result 23.

8. Test the result 5 x 4 +3 = 23, 3 x 7 + 2 = 23, 7 x 3 + 2 = 23.

At first I processed the results in this order multiplier 5, then 7 then 3.

If I had done this on this example I would have determined that of the list in point 2, to find the possible results for the remainder of seven, I would have subtracted 2 from each to spot any numbers that divide by 7. This would have given me 21, 91 and 56, or finals of 23, 93 and 58. For the multiplier of three, (sum of the digits), only 23 works.

The results are the same. If I had the formula I would have used it. It looks easier, even if I don't know how it works.

Graeme

If you find yourself between a rock and a hard place, always head for the rock. Ps 62..