Puzzle for January 8, 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.
* CD is a 2-digit number (not C×D).
** "C ^ (A – F)" means "C raised to the power of (A – F)".
Scratchpad
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Hint #1
Add B and F to both sides of eq.3: E – B + B + F = C + D – F + B + F which becomes E + F = C + D + B In eq.4, replace E + F with C + D + B: A – B + C + D + B = B – D which becomes A + C + D = B – D Add D to both sides of the equation above: A + C + D + D = B – D + D which becomes eq.4a) A + C + 2×D = B
Hint #2
In eq.4a, replace A + C with B + D (from eq.2): B + D + 2×D = B which becomes B + 3×D = B Subtract B from each side of the equation above: B + 3×D – B = B – B which becomes 3×D = 0 which means D = 0
Hint #3
In eq.1, substitute 0 for D: F = A + 0 which becomes F = A
Hint #4
In eq.2, substitute 0 for D: B + 0 = A + C which becomes eq.2a) B = A + C
Hint #5
Substitute A for F in eq.6: C ^ (A – A) = B – E which becomes eq.6a) C ^ 0 = B – E which becomes 1 = B – E (implies C ≠ 0) In the above equation, subtract 1 from both sides, and add E to both sides: 1 – 1 + E = B – E – 1 + E which makes eq.6b) E = B – 1
Hint #6
Substitute B – 1 for E (from eq.6b), A for F, and 0 for D in eq.4: A – B + B – 1 + A = B – 0 which becomes eq.4b) 2×A – 1 = B
Hint #7
Substitute 2×A – 1 for B (from eq.4b) into eq.2a: 2×A – 1 = A + C In the equation above, subtract A from both sides, add 1 to both sides: 2×A – 1 – A + 1 = A + C – A + 1 which makes A = C + 1 which also makes eq.2b) F = A = C + 1
Hint #8
Substitute C + 1 for A (from eq.2b) in eq.2a: B = C + 1 + C which becomes eq.2c) B = 2×C + 1
Hint #9
Substitute 2×C + 1 for B (from eq.2c) in eq.6b: E = 2×C + 1 – 1 which makes E = 2×C
Hint #10
eq.5 may be written as: 10×C + D = E × F Substitute 0 for D, 2×C for E, and (C + 1) for F in the equation above: 10×C + 0 = 2×C × (C + 1) which becomes eq.5a) 10×C = 2×C × (C + 1)
Solution
Since C ≠ 0 (from eq.6a), divide both sides of eq.5a by 2×C: 10×C ÷ 2×C = (2×C × (C + 1)) ÷ 2×C which becomes 5 = C + 1 Subtract 1 from each side of the above equation: 5 – 1 = C + 1 – 1 which means 4 = C making A = F = C + 1 = 4 + 1 = 5 (from eq.2b) B = 2×C + 1 = 2×4 + 1 = 8 + 1 = 9 (from eq.2c) E = 2×C = 2×4 = 8 and ABCDEF = 594085