Puzzle for November 13, 2021 ( )
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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, and EF are 2-digit numbers (not A×B, C×D, or E×F).
Our thanks go out to Abby S (age 10) for this interesting puzzle. Thank you, Abby!
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Hint #1
Add F to both sides of eq.4: C – F + F = B + E + F which becomes eq.4a) C = B + E + F In eq.3, replace C with B + E + F (from eq.4a): E + F = B + B + E + F which becomes E + F = 2×B + E + F Subtract E and F from both sides of the above equation: E + F – E – F = 2×B + E + F – E – F which makes 0 = 2×B which means 0 = B
Hint #2
In eq.3, replace B with 0: E + F = 0 + C which becomes eq.3a) E + F = C
Hint #3
In eq.5, replace B with 0: 0 + D – A = C + E – 0 which becomes D – A = C + E Add A to both sides of the above equation: D – A + A = C + E + A which becomes eq.5a) D = C + E + A
Hint #4
In eq.2, substitute 0 for B, and C + E + A for D (from eq.5a): 0 + C + E + A = A + F which becomes C + E + A = A + F Subtract A from each side of the equation above: C + E + A – A = A + F – A which becomes eq.5b) C + E = F
Hint #5
Substitute E + F for C (from eq.3a) into eq.5b: E + F + E = F which becomes 2×E + F = F Subtract F from each side of the above equation: 2×E + F – F = F – F which makes 2×E = 0 which means E = 0
Hint #6
Substitute 0 for E in eq.5b: C + 0 = F which makes C = F
Hint #7
eq.6 may be written as: 10×A + B = 10×C + D + 10×E + F Substitute 0 for B and E, and C for F in the above equation: 10×A + 0 = 10×C + D + 10×0 + C which becomes eq.6a) 10×A = 11×C + D
Hint #8
Substitute 0 for E in eq.5a: D = C + 0 + A which makes eq.5c) D = C + A
Hint #9
Substitute C + A for D (from eq.5c) in eq.6a: 10×A = 11×C + C + A which becomes 10×A = 12×C + A Subtract A from each side of the equation above: 10×A – A = 12×C + A – A which becomes 9×A = 12×C Divide both sides by 12: 9×A ÷ 12 = 12×C ÷ 12 which makes ¾×A = C and also makes F = C = ¾×A
Hint #10
Substitute ¾×A for C in eq.5c: D = ¾×A + A which makes D = 1¾×A
Solution
Substitute 0 for B and E, ¾×A for C and F, and 1¾×A for D in eq.1: A + 0 + ¾×A + 1¾×A + 0 + ¾×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 = F = ¾×A = ¾ × 4 = 3 D = 1¾×A = 1¾ × 4 = 7 and ABCDEF = 403703