Puzzle for January 24, 2020 ( )
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, BC, and EF are 2-digit numbers (not A×B, B×C, or E×F).
** "B ^ F" means "B raised to the power of F".
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Hint #1
eq.5 may be written as: A + 10×B + C = 10×A + B + 10×E + F Subtract 10×A, B, and C from each side of the above equation: A + 10×B + C – 10×A – B – C = 10×A + B + 10×E + F – 10×A – B – C which becomes –9×A + 9×B = 10×E + F – C which may be written as 9×(B – A) = 10×E + F – C In the above equation, replace (B – A) with D (from eq.2): eq.5a) 9×D = 10×E + F – C
Hint #2
In eq.5a, substitute (E + F) for C (from eq.3): 9×D = 10×E + F – (E + F) which is equivalent to 9×D = 10×E + F – E – F which becomes 9×D = 9×E Divide both sides by 9: 9×D ÷ 9 = 9×E ÷ 9 which makes D = E
Hint #3
In eq.3, substitute D for E: D + F = C Substitute D + F for C in eq.4: D + F + F = D Subtract D from each side: D + F + F – D = D – D which makes 2×F = 0 which means F = 0
Hint #4
In eq.4, substitute 0 for F: C + 0 = D which makes C = D and which also makes E = D = C
Hint #5
Substitute 0 for F, and D for E in eq.6: B ^ 0 = A – D which means 1 = A – D (assumes B ≠ 0) Add D to both sides of the above equation: 1 + D = A – D + D which makes eq.6a) D + 1 = A
Hint #6
Substitute B – A for D (from eq.2) in eq.6a: B – A + 1 = A Add A to both sides: B – A + 1 + A = A + A which becomes B + 1 = 2×A Substitute (D + 1) for A (from eq.6a): B + 1 = 2×(D + 1) which becomes B + 1 = 2×D + 2 Subtract 1 from each side: B + 1 – 1 = 2×D + 2 – 1 which becomes eq.6b) B = 2×D + 1
Solution
Substitute D + 1 for A (from eq.6a), 2×D + 1 for B (from eq.6b), D for C and E, and 0 for F in eq.1: D + 1 + 2×D + 1 + D + D + D + 0 = 20 which becomes 6×D + 2 = 20 Subtract 2 from each side: 6×D + 2 – 2 = 20 – 2 which becomes 6×D = 18 Divide both sides by 6: 6×D ÷ 6 = 18 ÷ 6 D = 3 making A = D + 1 = 3 + 1 = 4 (from eq.6a) B = 2×D + 1 = 2×3 + 1 = 7 (from eq.6b) C = E = D = 3 and ABCDEF = 473330