Puzzle for April 17, 2022 ( )
Scratchpad
Find the 6-digit number ABCDEF by solving the following equations:
A, B, C, D, E, and F each represent a one-digit non-negative integer.
* AB, CD, DE, and EF are 2-digit numbers (not A×B, C×D, D×E, or E×F).
Scratchpad
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Hint #1
Add A, B, C, and E to both sides of eq.2: F – A – B + A + B + C + E = A – C – E + A + B + C + E which becomes eq.2a) F + C + E = 2×A + B Add A to both sides of eq.3: C + D – A – E + A = A + B + A which becomes eq.3a) C + D – E = 2×A + B
Hint #2
In eq.3a, replace 2×A + B with F + C + E (from eq.2a): C + D – E = F + C + E In the above equation, subtract C from both sides, and add E to both sides: C + D – E – C + E = F + C + E – C + E which becomes D = F + 2×E which is the same as eq.3b) D = 2×E + F
Hint #3
In eq.6, replace AB with CD + EF (from eq.5): DE – F = CD + EF + F which may be written as 10×D + E – F = 10×C + D + 10×E + F + F which becomes 10×D + E – F = 10×C + D + 10×E + 2×F In the above equation, subtract D and E from both sides, and add F to both sides: 10×D + E – F – D – E + F = 10×C + D + 10×E + 2×F – D – E + F which becomes eq.6a) 9×D = 10×C + 9×E + 3×F
Hint #4
In eq.6a, substitute (2×E + F) for D (from eq.3b): 9×(2×E + F) = 10×C + 9×E + 3×F which becomes 18×E + 9×F = 10×C + 9×E + 3×F Subtract 9×E and 3×F from each side of the equation above: 18×E + 9×F – 9×E – 3×F = 10×C + 9×E + 3×F – 9×E – 3×F which becomes eq.6b) 9×E + 6×F = 10×C
Hint #5
Substitute 2×E + F for D (from eq.3b) into eq.4: 2×E + F + F – C – E = A – B + C which becomes eq.4a) E + 2×F – C = A – B + C
Hint #6
Substitute 2×E + F for D (from eq.3b) in eq.3a: C + 2×E + F – E = 2×A + B which becomes C + E + F = 2×A + B Subtract 2×A from each side of the above equation: C + E + F – 2×A = 2×A + B – 2×A which becomes eq.3c) C + E + F – 2×A = B
Hint #7
In eq.4a, substitute (C + E + F – 2×A) for B (from eq.3c): E + 2×F – C = A – (C + E + F – 2×A) + C which becomes E + 2×F – C = A – C – E – F + 2×A + C which becomes E + 2×F – C = 3×A – E – F In the above equation, add C and F to both sides, and subtract E from both sides: E + 2×F – C + C + F – E = 3×A – E – F + C + F – E which becomes eq.4b) 3×F = 3×A – 2×E + C
Hint #8
eq.6b may be written as: 9×E + 2×(3×F) = 10×C Substitute (3×A – 2×E + C) for 3×F (from eq.4b) into the above equation: 9×E + 2×(3×A – 2×E + C) = 10×C which becomes 9×E + 6×A – 4×E + 2×C = 10×C which becomes 5×E + 6×A + 2×C = 10×C Subtract 2×C from each side: 5×E + 6×A + 2×C – 2×C = 10×C – 2×C which becomes eq.6c) 5×E + 6×A = 8×C
Hint #9
eq.5 may be written as: 10×A + B = 10×C + D + 10×E + F which may be written as eq.5a) 8×A + 2×A + B = 10×C + D + 10×E + F
Hint #10
In eq.5a, replace 2×A + B with C + D – E (from eq.3a): 8×A + C + D – E = 10×C + D + 10×E + F Subtract 10×C, D, and 10×E from both sides of the above equation: 8×A + C + D – E – 10×C – D – 10×E = 10×C + D + 10×E + F – 10×C – D – 10×E which becomes eq.5b) 8×A – 9×C – 11×E = F
Hint #11
Substitute (8×A – 9×C – 11×E) for F (from eq.5b) in eq.6b: 9×E + 6×(8×A – 9×C – 11×E) = 10×C which becomes 9×E + 48×A – 54×C – 66×E = 10×C which becomes 48×A – 54×C – 57×E = 10×C Add 54×C to both sides of the above equation: 48×A – 54×C – 57×E + 54×C = 10×C + 54×C which becomes 48×A – 57×E = 64×C which may be written as eq.6d) 48×A – 57×E = 8×(8×C)
Hint #12
Substitute 5×E + 6×A for 8×C (from eq.6c) into eq.6d: 48×A – 57×E = 8×(5×E + 6×A) which becomes 48×A – 57×E = 40×E + 48×A In the above equation, subtract 48×A from both sides, and add 57×E to both sides: 48×A – 57×E – 48×A + 57×E = 40×E + 48×A – 48×A + 57×E which simplifies to 0 = 97×E which meaans 0 = E
Hint #13
Substitute 0 for E in eq.3b: D = 2×0 + F which becomes D = 0 + F which makes D = F
Hint #14
Substitute 0 for E in eq.6c: 5×0 + 6×A = 8×C which becomes 0 + 6×A = 8×C which makes 6×A = 8×C Divide both sides by 8: 6×A ÷ 8 = 8×C ÷ 8 which makes ¾×A = C
Hint #15
Substitute 0 for E, and ¾×A for C in eq.6b: 9×0 + 6×F = 10×(¾×A) which becomes 0 + 6×F = 7½×A which makes 6×F = 7½×A Divide both sides of the above equation by 6: 6×F ÷ 6 = 7½×A ÷ 6 which makes F = 1¼×A and also makes D = F = 1¼×A
Hint #16
Substitute ¾×A for C, 0 for E, and 1¼×A for F in eq.3c: ¾×A + 0 + 1¼×A – 2×A = B which makes 0 = B
Solution
Substitute 0 for B and E, ¾×A for C, and 1¼×A for D and F in eq.1: A + 0 + ¾×A + 1¼×A + 0 + 1¼×A = 17 which simplifies to 4¼×A = 17 Divide both sides of the above equation by 4¼: 4¼×A ÷ 4¼ = 17 ÷ 4¼ which means A = 4 making C = ¾×A = ¾ × 4 = 3 D = F = 1¼×A = 1¼ × 4 = 5 and ABCDEF = 403505