# Numbers 1-9 All used up?

There are nine numbers from one to nine.

Each are to be used only once. Take two numbers from them to make a 2-digit number. It is multiplied by a different single digit no. to get a 2-digit number containing different numbers from the previously used numbers. Then it is added to a 2-digit number to get a different 2-digit number, in which all numbers should be distinct.

That is to say something like this mathematically :

$$ab \ \times \ c = de$$ $$de +fg = hi$$

You have to tell all the numbers used in the process. I was asked this puzzle and I couldn't find the answer to it. I hope people here can solve it :-)

I found that:

There is one answer (at the bottom)

My reasoning (some steps may be redundant in an optimal solution):

c is not 1 (ab * 1 = ab [used a and b twice])

de is at most 86 (88 + 12 > 99, 87 + 12 = 99 [used 9 twice])

ab is at most 43 (c is not 1, and 44 * 2 > 86)

c is at most 7 (12 * 8 > 86)
c is not 5 (if b is even, then e=0. if b is odd, then e=5 and c=5)

if c is 6, then ab is 13, because:
12 * 6 = 72 [used 2 twice]
13 * 6 = 78
14 * 6 = 84 [used 4 twice]
15 * 6 = 90 [need a 0]
16 * 6 = 96 [used 6 three times]

if c is 4, then ab is 13, 17, 18 or 19, because:
12 * 4 = 48 [used 4 twice]
13 * 4 = 52
14 * 4 = 56 [used 4 twice]
15 * 4 = 60 [need a 0]
16 * 4 = 64 [used 4 twice]
17 * 4 = 68
18 * 4 = 72
19 * 4 = 76
20 * 4 = 80 [need a 0]
21 * 4 = 84 [used 4 twice]

if c is 3, then ab is 16, 18, 19 or 26, because:
12 * 3 = 36 [used 3 twice]
13 * 3 = 39 [used 3 twice]
14 * 3 = 42 [used 4 twice]
15 * 3 = 45 [used 5 twice]
16 * 3 = 48
17 * 3 = 51 [used 1 twice]
18 * 3 = 54
19 * 3 = 57
20 * 3 = 60 [need a 0]
21 * 3 = 63 [used 3 twice]
22 * 3 = 66 [used 2 twice]
23 * 3 = 69 [used 3 twice]
24 * 3 = 72 [used 2 twice]
25 * 3 = 75 [used 5 twice]
26 * 3 = 78
27 * 3 = 81 [f must be 1]
28 * 3 = 84 [used 8 twice]

if c is 2, then ab is 17, 18, 19, 34, 38 or 39, because:
12 * 2 = 24 [used 2 twice]
13 * 2 = 26 [used 2 twice]
14 * 2 = 28 [used 2 twice]
15 * 2 = 30 [need a 0]
16 * 2 = 32 [used 2 twice]
17 * 2 = 34
18 * 2 = 36
19 * 2 = 38
20 * 2 = 40 [used 2 twice]
21 * 2 = 42 [used 2 twice]
22 * 2 = 44 [used 2 twice]
23 * 2 = 46 [used 2 twice]
24 * 2 = 48 [used 2 twice]
25 * 2 = 50 [used 2 twice]
26 * 2 = 52 [used 2 twice]
27 * 2 = 54 [used 2 twice]
28 * 2 = 56 [used 2 twice]
29 * 2 = 58 [used 2 twice]
30 * 2 = 60 [need a 0]
31 * 2 = 62 [used 2 twice]
32 * 2 = 64 [used 2 twice]
33 * 2 = 66 [used 3 twice]
34 * 2 = 68
35 * 2 = 70 [need a 0]
36 * 2 = 72 [used 2 twice]
37 * 2 = 74 [used 7 twice]
38 * 2 = 76
39 * 2 = 78
40 * 2 = 80 [need a 0]
41 * 2 = 82 [used 2 twice]
42 * 2 = 84 [used 2 twice]
43 * 2 = 86 [ef must be 12 or 13, needing 2 or 3 twice]

The possible combinations of ab and c are:
ab * c = de [ impossible because ]
13 * 6 = 78, with 2,4,5,9 remaining [ 78 + 24 > 95 ]
13 * 4 = 52, with 6,7,8,9 remaining [ 52 + 67 > 98 ]
17 * 4 = 68, with 2,3,5,9 remaining (#1)
18 * 4 = 72, with 3,5,6,9 remaining [ 72 + 35 > 96 ]
19 * 4 = 76, with 2,3,5,8 remaining [ 78 + 24 > 85 ]
16 * 3 = 48, with 2,5,7,9 remaining (#2)
18 * 3 = 54, with 2,6,7,9 remaining (#3)
19 * 3 = 57, with 2,4,6,8 remaining (#4)
26 * 3 = 78, with 1,4,5,9 remaining (#5)
17 * 2 = 34, with 5,6,8,9 remaining (#6)
18 * 2 = 36, with 4,5,7,9 remaining (#7)
19 * 2 = 38, with 4,5,6,7 remaining [ 38 + 45 > 76 ]
34 * 2 = 68, with 1,5,7,9 remaining (#8)
38 * 2 = 76, with 1,4,5,9 remaining (#9)
39 * 2 = 78, with 1,4,5,6 remaining [ 78 > 65 ]

Now we look at the possible last digit of de + {one of remaining digits}
#1: 68 + 5 ends with 3
#2: 48 + 7 ends with 5, 48 + 9 ends with 7
#3: 54 + 2 ends with 6
#4: impossible
#5: 78 + 1 ends with 9
#6: 34 + 5 ends with 9
#7: 36 + 9 ends with 5
#8: 68 + 1 ends with 9, 68 + 7 ends with 5, 68 + 9 ends with 7
#9: 76 + 5 ends with 1

Making:
#N: ab * c = de, (de) + fg = hi
#1: 17 * 4 = 68, (68) + 25 = 93 [OK! (not with 95 and 23)]
#2: 16 * 3 = 48, (48) + f7 = h5 [impossible with 2 and 9]
#2: 16 * 3 = 48, (48) + f9 = h7 [impossible with 2 and 5]
#3: 18 * 3 = 54, (54) + f2 = h6 [impossible with 7 and 9]
#5: 26 * 3 = 78, (78) + f1 = h9 [impossible with 4 and 5]
#6: 17 * 2 = 34, (34) + f5 = h9 [impossible with 6 and 8]
#7: 18 * 2 = 36, (36) + f9 = h5 [impossible with 4 and 7]
#8: 34 * 2 = 68, (68) + f1 = h9 [impossible with 5 and 7]
#8: 34 * 2 = 68, (68) + f7 = h5 [impossible with 1 and 9]
#8: 34 * 2 = 68, (68) + f9 = h7 [impossible with 1 and 5]
#9: 38 * 2 = 76, (76) + f5 = h1 [impossible with 4 and 9]

So the only answer is:
17 * 4 = 68
68 + 25 = 93

You mean something like this?

$$17 \times 4 = 68$$ $$68 + 25 = 93$$

(Method: tried a couple of multiplications, trying to save some small numbers for the addition in order to not go over 100. Found this on the fourth attempt, so I've no idea if the answer is unique.)

I can confirm that P1storius' answer is the only solution. Here is the (matlab) code I used to search:

permutations = perms(1:9); % 362,880 permuatations

solutions = NaN(0, 9);
i = 1;
tic;
solutionCount = 0;
while(i <= size(permutations, 1))
row = permutations(i, :);
ab = 10 * row(1) + row(2);
c = row(3);
de = 10 * row(4) + row(5);
if(ab * c == de)
fg = 10 * row(6) + row(7);
hi = 10 * row(8) + row(9);
if(de + fg == hi)
solutions(end + 1, :) = row;
solutionCount = solutionCount + 1;
fprintf('%d * %d = %d; %d + %d = %d\n', ab, c, de, de, fg, hi);
end
end
i = i + 1;
end

fprintf('\nfound %d total solutions\n', size(solutions, 1));
toc;


Which yields the output:

[Redacted]

found 1 total solutions
Elapsed time is 0.177841 seconds.


This is the kind of problem that just screams "use a computer" to me. You can run the code in an online engine to play with it yourself, although that one took 15 seconds to run for me.