Strategy with regard to how to approach this algorithm?

Question

I was asked this question in a test and I need help with regards to how I should approach the solution, not the actual answer. The question is

You have been given a 7 digit number(with each digit being distinct and 0-9). The number has this property

product of first 3 digits = product of last 3 digits = product of central 3 digits

Identify the middle digit.

Now, I can do this on paper by brute force(trial and error), the product is 72 and digits being

8,1,9,2,4,3,6

Now how do I approach the problem in a no brute force way?

Answer 1

You have been given a 7 digit number(with each digit being distinct and 0-9). The number has this property

product of first 3 digits = product of last 3 digits = product of central 3 digits Identify the middle digit.

Now, I can do this on paper by brute force(trial and error), the product is 72 and digits being

8,1,9,2,4,3,6 Now how do I approach the problem in a no brute force way?

use linq and substring functions

example var item = array.Skip(3).Take(3) in such a way that you have a loop

for(f =0;f<charlen.length;f++){
var xItemSum = charlen[f].Skip(f).Take(f).Sum(f => f.Value);
}

// untested code

Answer 2

Here is how you can consider the problem:

Let's note the final solution N1 N2 N3 N4 N5 N6 N7 for the 3 numbers N1N2N3, N3N4N5 and N5N6N7

1. 0, 5 and 7 are to exclude because they are prime and no other ciphers is a multiple of them. So if they had divided one of the 3 numbers, no other number could have divided the others.

2. So we get the 7 remaining ciphers : 1234689 where the product of the ciphers is 2^7*3^4

3. (N1*N2*N3) and (N5*N6*N7) are equals so their product is a square number. We can then remove, one of the number (N4) from the product of the previous point to find a square number (i.e. even exponents on both numbers)

   • N4 can't be 1, 3, 4, 6, 9.
   • We conclude N4 is 2 or 8

If N4 is 8 and it divides (N3*N4*N5), we can't use the remaining even numbers (2, 4, 6) to divides both (N1*N2*N3) and (N6*N7*N8) by 8. So N4 is 2 and 8 does not belong to the second group (let's put it in N1).

Now, we have: 1st grp: 8XX, 2nd group: X2X 3rd group: XXX

Note: at this point we know that the product is 72 because it is 2^3*3^2 (the square root of 2^6*3^4) but the result is not really important. We have made the difficult part knowing the 7 numbers and the middle position.

4. Then, we know that we have to distribute 2^3 on (N1*N2*N3), (N3*N4*N5), (N5*N6*N7) because 2^3*2*2^3=2^7

We already gave 8 to N1, 2 to N4 and we place 6 to N6, and 4 to N5 position, resulting in each of the 3 numbers being a multiple of 8.

Now, we have: 1st grp: 8XX, 2nd group: X24 3rd group: 46X

5. We have the same way of thinking considering the odd number, we distribute 3^2, on each part knowing that we already have a 6 in the last group. Last group will then get the 3. And first and second ones the 9.

Now, we have: 1st grp: 8X9, 2nd group: 924 3rd group: 463

And, then 1 at N2, which is the remaining position.

Answer 3

You have your 7 numbers - instead of looking at it in groups of 3 divide up the number as such:

AB | C | D | E | FG

Get the value of AB and use it to get the value of C like so: C = ABC/AB

Next you want to do the same thing with the trailing 2 digits to find E using FG. E = EFG/FG

Now that you have C & E you can solve for D

Since CDE = ABC then D = ABC/CE

Remember your formulas - instead of looking at numbers create a formula aka an algorithm that you know will work every time.

ABC = CDE = EFG However, you have to remember that your = signs have to balance. You can see that D = ABC/CE = EFG/CE Once you know that, you can figure out what you need in order to solve the problem.

Made a quick example in a fiddle of the code: http://jsfiddle.net/4ykxx9ve/1/

var findMidNum = function() {
    var num = [8, 1, 9, 2, 4, 3, 6];

    var ab = num[0] * num[1];
    var fg = num[5] * num[6];
    var abc = num[0] * num[1] * num[2];
    var cde = num[2] * num[3] * num[4];
    var efg = num[4] * num[5] * num[6];
    var c = abc/ab;
    var e = efg/fg;
    var ce = c * e
    var d = abc/ce;

    console.log(d); //2
}();

Answer 4

7 digits: 8,1,9,2,4,3,6

say XxYxZ = 72

1) pick any two from above 7 digits. say X,Y

2) divide 72 by X and then Y.. you will get the 3rd number i.e Z.

we found XYZ set of 3-digits which gives result 72.

now repeat 1) and 2) with remaining 4 digits.

this time we found ABC which multiplies to 72.

lets say, 7th digit left out is I.

3) divide 72 by I. result R

4) divide R by one of XYZ. check if result is in ABC.

if No, repeat the step 3)

if yes, found the third pair.(assume you divided R by Y and the result is B)

YIB is the third pair.

so... solution will be.

XZYIBAC

Answer 5

Let the number is: a b c d e f g

So as per the rule(1):

axbxc = cxdxe = exfxg

more over we have(2):

axb = dxe and cxd = fxg

This question can be solved with factorization and little bit of hit/trial. Out of the digits from 1 to 9, 5 and 7 can rejected straight-away since these are prime numbers and would not fit in the above two equations.

The digits 1 to 9 can be factored as:

1 = 1, 2 = 2, 3 = 3, 4 = 2X2, 6 = 2X3, 8 = 2X2X2, 9 = 3X3

After factorization we are now left with total 7 - 2's, 4 - 3's and the number 1. As for rule 2 we are left with only 4 possibilities, these 4 equations can be computed by factorization logic since we know we have overall 7 2's and 4 3's with us.

1: 1X8(2x2x2) = 2X4(2x2)

2: 1X6(3x2) = 3X2

3: 4(2x2)X3 = 6(3x2)X2

4: 9(3x3)X2 = 6(3x2)X3

Skipping 5 and 7 we are left with 7 digits. With above equations we have 4 digits with us and are left with remaining 3 digits which can be tested through hit and trial. For example, if we consider the first case we have:

1X8 = 2X4 and are left with 3,6,9. we have axbxc = cxdxe we can opt c with these 3 options in that case the products would be 24, 48 and 72.

24 cant be correct since for last three digits we are left with are 6,9,4(=216)

48 cant be correct since for last three digits we are left with 3,9,4(=108)

72 could be a valid option since the last three digits in that case would be 3,6,4 (=72)

Answer 6

My answer actually extends @Ansh's answer.

Let abcdefg be the digits of the number. Then

ab=de

cd=fg

From these relations we can exclude 0, 5 and 7 because there are no other multipliers of these numbers between 0 and 9. So we are left with seven numbers and each number is included once in each answer. We are going to examine how we can pair the numbers (ab, de, cd, fg).

What happens with 9? It can't be combined with 3 or 6 since then their product will have three times the factor 3 and we have at total 4 factors of 3. Similarly, 3 and 6 must be combined at least one time together in response to the two factors of 9. This gives a product of 18 and so 9 must be combined at least once with 2.

Now if 9x2 is in a corner then 3x6 must be in the middle. Meaning in the other corner there must be another multiplier of 3. So 9 and 2 are in the middle.

Let's suppose ab=3x6 (The other case is symmetric). Then d must be 9 or 2. But if d is 9 then f or g must be multiplier of 3. So d is 2 and e is 9. We can stop here and answer the middle digit is

2

Now we have 2c = fg and the remaining choices are 1, 4, 8. We see that the only solutions are c = 4, f = 1, g = 8 and c = 4, f = 8, g = 1.

So if is 3x6 is in the left corner we have the following solutions:

3642918, 3642981, 6342918, 6342981

If 3x6 is in the right corner we have the following solutions which are the reverse of the above:

8192463, 1892463, 8192436, 1892436

Answer 7

Prime factor of distinct digit (if possible)

0 = 0
1 = 1
2 = 2
3 = 3
4 = 2 x 2
5 = 5
6 = 2 x 3
7 = 7
8 = 2 x 2 x 2
9 = 3 x 3

In total: 7 2's + 4 3's + 1 5's + 1 7's

With the fact that When A=B=C, composition of prime factor of A must be same as composition of prime factor of B and that of C, 0 , 5 and 7 are excluded since they have unique prime factor that can never match with the fact.

Hence, 7 2's + 4 3's are left and we have 7 digit (1,2,3,4,6,8,9). As there are 7 digits only, the number is formed by these digits only.

Recall the fact, A, B and C must have same composition of prime factors. This implies that A, B and C have same number of 2's and 3's in their composition. So, we should try to achieve (in total for A and B and C):

1. 9 OR 12 2's AND
2. 6 3's

(Must be product of 3, lower bound is total number of prime factor of all digits, upper bound is lower bound * 2)

Consider point 2 (as it has one possibility), A has 2 3's and same for B and C. To have more number of prime factor in total, we need to put digit in connection digit between two product (third or fifth digit). Extract digits with prime factor 3 into two groups {3,6} and {9} and put digit into connection digit. The only possible way is to put 9 in connection digit and 3,6 on unconnected product. That mean xx9xx36 or 36xx9xx (order of 3,6 is not important)

With this result, we get 9 x middle x connection digit = connection digit x 3 x 6. Thus, middle = (3 x 6) / 9 = 2

Answer 8

This problem is pretty easy if you look at the number 72 more carefully.

We have our number with this form abcdefg

and abc = cde = efg, with those digits 8,1,9,2,4,3,6

• So, first, we can conclude that 8,1,9 must be one of the triple, because, there is no way 1 can go with other two numbers to form 72.

• We can also conclude that 1 must be in the start/end of the whole number or middle of the triple.

So now we have 819defg or 918defg ...

Using some calculations with the rest of those digits, we can see that only 819defg is possible, because, we need 72/9 = 8,so only 2,4 is valid, while we cannot create 72/8 = 9 from those 2,4,3,6 digits, so -> 81924fg or 81942fg and 819 must be the triple that start or end our number.

So the rest of the job is easy, we need either 72/4 = 18 or 72/2 = 36, now, we can have our answers: 8192436 or 8192463.

Answer 9

This question is good to solve with Relational Programming. I think it very clearly lets the programmer see what's going on and how the problem is solved. While it may not be the most efficient way to solve problems, it can still bring desired clarity and handle problems up to a certain size. Consider this small example from Oz:

fun {FindDigits}
   D1 = {Digit}
   D2 = {Digit}
   D3 = {Digit}
   D4 = {Digit} 
   D5 = {Digit}
   D6 = {Digit}
   D7 = {Digit}
   L = [D1 D2 D3] M = [D3 D4 D5] E= [D5 D6 D7] TotL in
   TotL = [D1 D2 D3 D4 D5 D6 D7]
   {Unique TotL} = true
   {ProductList L} = {ProductList M} = {ProductList E}
   TotL
end

(Now this would be possible to parameterize furthermore, but non-optimized to illustrate the point).

Here you first pick 7 digits with a function Digit/0. Then you create three lists, L, M and E consisting of the segments, as well as a total list to return (you could also return the concatenation, but I found this better for illustration).

Then comes the point, you specify relations that have to be intact. First, that the TotL is unique (distinct in your tasks wording). Then the next one, that the segment products have to be equal.

What now happens is that a search is conducted for your answers. This is a depth-first search strategy, but could also be breadth-first, and a solver is called to bring out all solutions. The search strategy is found inside the SolveAll/1 function.

{Browse {SolveAll FindDigits}}

Which in turns returns this list of answers:

[[1 8 9 2 4 3 6] [1 8 9 2 4 6 3] [3 6 4 2 9 1 8] 
 [3 6 4 2 9 8 1] [6 3 4 2 9 1 8] [6 3 4 2 9 8 1] 
 [8 1 9 2 4 3 6] [8 1 9 2 4 6 3]]

At least this way forward is not using brute force. Essentially you are searching for answers here. There might be heuristics that let you find the correct answer sooner (some mathematical magic, perhaps), or you can use genetic algorithms to search the space or other well-known strategies.