Mystery Number 3
Logic puzzles require you to think. You will have to be logical in your reasoning.
There is a tendigit mystery number (no leading 0), represented by ABCDEFGHIJ, where each numeral, 0 through 9, is used once. Given the following clues, what is the number?
1) B + C + F + J = E + G + H + I = AD
2) B  H = J  G = 3
3) C  F = E  I = 5
4) B * I = AJ
Answer
2951804637
A = 2, B = 9, C = 5, D = 1, E = 8, F = 0, G = 4, H = 6, I = 3, J = 7
1) 9 + 5 + 0 + 7 = 8 + 4 + 6 + 3 = 21
2) 8  3 = 5  0 = 5
3) 7  4 = 9  6 = 3
4) 9 * 3 = 27
A) The sum of the ten digits is 45. Two digits (A and D) form a twodigit number. If A and D are 8 and 9, then the remaining eight digits sum to 28. If A and D are 0 and 1, then the remaining digits sum to 44. These sums must be split equally between two groups of four digits. Therefore, each group of four digits must sum to one of the following: 14, 15, 16, 17, 18, 19, 20, 21, 22.
B) If the sum of each group is 14, then A and D must be a combination of 1 and 4, which sum to 5. Subtract 5 from 45, leaving 40. Divide 40 by 2, leaving 20. 20 does not equal 14, so the sum of each group can not be 14. Repeating the preceding sequence for each of the possible sums leaves only two possible results: 18 (1 + 8 = 9, 45  9 = 36, 36 / 2 = 18) and 21 (2 + 1 = 3, 45  3 = 42, 42 / 2 = 21). A must be: 1, 2. D must be: 1, 8.
C) To satisfy equation #2, B and J must be 3 greater than H and G, respectively. B and J must each be one of the following: 3, 5, 6, 7, 8, 9. H and G must each be one of the following: 0, 2, 3, 4, 5, 6.
D) To satisfy equation #3, C and E must be 5 greater than F and I, respectively. C and E must each be one of the following: 5, 7, 8, 9. F and I must each be one of the following: 0, 2, 3, 4.
E) For equation #4, there are six possibilities for B and four possibilities for I, creating 24 possible equations. Eliminate any equation where A is not 1 or 2. Eliminate any equation where J is not 3, 5, 6, 7, 8, or 9. Eliminate any equation where any digit is duplicated. B must be: 6, 7, 8, 9. I must be: 2, 3, 4. J must be: 6, 7, 8.
F) Returning to equations #2 and #3, eliminate additional possibilities. E must be: 7, 8, 9. G must be: 3, 4, 5. H must be: 3, 4, 5, 6. This results in F = 0 and C = 5.
G) Since H can not be 5, B can not be 8. Since G can not be 5, J can not be 8. Going back to equation #4, eliminate those equations where either B or J is 8. B = 9, I = 3, A = 2, and J = 7.
H) The last four letters can now be assigned values. D = 1, E = 8, G = 4, and H = 6.
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