Puzzle for April 24, 2021 ( )
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, and DE are 2-digit number (not A×B, C×D, or D×E).
Scratchpad
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Hint #1
eq.5 may be written as: E – F = D – (B + C) In the above equation, replace B + C with D + E (from eq.3): E – F = D – (D + E) which is equivalent to E – F = D – D – E which becomes E – F = –E Add F and E to both sides: E – F + F + E = –E + F + E which makes eq.5a) 2×E = F
Hint #2
Subtract C and E from both sides of eq.3: B + C – C – E = D + E – C – E which becomes B – E = D – C In the equation above, replace D – C with A – B – E (from eq.2): B – E = A – B – E Add E and B to both sides: B – E + E + B = A – B – E + E + B which makes eq.3a) 2×B = A
Hint #3
In eq.4, substitute 2×B for A, and 2×E for F: 2×B – C + 2×E = B + C – E In the above equation, add C and E to both sides, and subtract B from both sides: 2×B – C + 2×E + C + E – B = B + C – E + C + E – B which simplifies to eq.4a) B + 3×E = 2×C
Hint #4
eq.6 may be written as: 10×C + D + 10×D + E + F = 10×A + B + C which becomes 10×C + 11×D + E + F = 10×A + B + C Subtract C from both sides of the equation above: 10×C + 11×D + E + F – C = 10×A + B + C – C which becomes eq.6a) 9×C + 11×D + E + F = 10×A + B In eq.4, add C and E to both sides, and subtract A from both sides: B + C – E + C + E – A = A – C + F + C + E – A which becomes B + 2×C – A = F + E which may be written as eq.4b) B + 2×C – A = E + F
Hint #5
In eq.6a, substitute B + 2×C – A for E + F (from eq.4b): 9×C + 11×D + B + 2×C – A = 10×A + B which becomes 11×C + 11×D + B – A = 10×A + B In the equation above, subtract B from both sides, and add A to both sides: 11×C + 11×D + B – A – B + A = 10×A + B – B + A which becomes 11×C + 11×D = 11×A Divide both sides by 11: (11×C + 11×D) ÷ 11 = 11×A ÷ 11 which becomes eq.6b) C + D = A
Hint #6
Substitute C + D for A (from eq.6b) in eq.2: D – C = C + D – B – E In the above equation, add C to each side, and subtract D from each side: D – C + C – D = C + D – B – E + C – D which becomes eq.2a) 0 = 2×C – B – E
Hint #7
Substitute B + 3×E for 2×C (from eq.4a) in eq.2a: 0 = B + 3×E – B – E which becomes 0 = 2×E which means 0 = E
Hint #8
Substitute 0 for E in eq.5a: 2×0 = F which means 0 = F
Hint #9
Substitute 0 for E in eq.4a: B + 3×0 = 2×C which makes B = 2×C
Hint #10
Substitute (2×C) for B in eq.3a: 2×(2×C) = A which makes 4×C = A
Hint #11
Substitute 4×C for A in eq.6b: C + D = 4×C Subtract C from each side of the above equation: C + D – C = 4×C – C which becomes D = 3×C
Solution
Substitute 4×C for A, 2×C for B, 3×C for D, and 0 for E and F in eq.1: 4×C + 2×C + C + 3×C + 0 + 0 = 10 which simplifies to 10×C = 10 Divide both sides of the above equation by 10: 10×C ÷ 10 = 10 ÷ 10 which means C = 1 making A = 4×C = 4 × 1 = 4 B = 2×C = 2 × 1 = 2 D = 3×C = 3 × 1 = 3 and ABCDEF = 421300