# 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]

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