Puzzle for May 14, 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.
* "F ^ F" means "F raised to the power of F". "D ^ D" means "D raised to the power of D". "D ^ F" means "D raised to the power of F".
One more time, we extend our thanks to Judah S (age 15) for contributing another puzzle, his 6th of the week. Thank you, Judah!
Scratchpad
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Hint #1
In eq.1, subtract D from both sides, and add F to both sides: D + F – D + F = C – F – D + F which becomes eq.1a) 2×F = C – D In eq.4, replace C – D with 2×F (from eq.1a): eq.4a) 2×F = F ^ F
Hint #2
To make eq.4a true, check several possible values for F: If F = 0, then 2×0 = 0 ^ 0 which would make 0 = undetermined If F = 1, then 2×1 = 1 ^ 1 which would make 2 = 1 If F = 2, then 2×2 = 2 ^ 2 which would make 4 = 4 If F = 3, then 2×3 = 3 ^ 3 which would make 6 = 27 If F > 3, then 2×F < F ^ F From the above equations, the only value for F that makes eq.4a true is: F = 2
Hint #3
In eq.1a, replace F with 2: 2×2 = C – D which becomes 4 = C – D Add D to both sides of the equation above: 4 + D = C – D + D which becomes eq.1b) 4 + D = C
Hint #4
In eq.3, substitute 4 + D for C (from eq.1b), and 2 for F: D × D = 4 + D + 2 which becomes D × D = 6 + D Subtract D from each side of the equation above: D × D – D = 6 + D – D which becomes D × D – D = 6 which may be written as eq.3a) D × (D – 1) = 6
Hint #5
To make eq.3a true, check several possible values for D: If D = 0, then 6 = 0 × (0 – 1) = 0 × (–1) = 0 If D = 1, then 6 = 1 × (1 – 1) = 1 × 0 = 0 If D = 2, then 6 = 2 × (2 – 1) = 2 × 1 = 2 If D = 3, then 6 = 3 × (3 – 1) = 3 × 2 = 6 If D = 4, then 6 = 4 × (4 – 1) = 4 × 3 = 12 If D > 4, then 6 < D × (D – 1) From the above equations, the only value for D that makes eq.3a true is: D = 3
Hint #6
Substitute 3 for D in eq.1b: 4 + 3 = C which makes 7 = C
Hint #7
Substitute 2 for F, and 3 for D in eq.2: 2 – 3 = B – 2 which becomes –1 = B – 2 Add 2 to both sides of the equation above: –1 + 2 = B – 2 + 2 which makes 1 = B
Hint #8
Substitute 7 for C, 3 for D, 2 for F, and 1 for B in eq.6: 7 + (3 × E) = (3 ^ 2) – (2 ÷ 1) which becomes 7 + 3×E = 9 – 2 which becomes 7 + 3×E = 7 Subtract 7 from each side of the above equation: 7 + 3×E – 7 = 7 – 7 which makes 3×E = 0 which means E = 0
Solution
Substitute 7 for C, 3 for D, 0 for E, and 2 for F in eq.5: A + (7 × 3) + 3 – 0 = (3 ^ 3) – 2 which becomes A + 21 + 3 = 27 – 2 which becomes A + 24 = 25 Subtract 24 from both sides of the above equation: A + 24 – 24 = 25 – 24 which makes A = 1 and makes ABCDEF = 117302