Puzzle for May 13, 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.
Scratchpad
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Hint #1
Subtract E from both sides of eq.2: B + C – D – A – E = D + E – E which becomes B + C – D – A – E = D which may be written as B – E + C – D – A = D In the above equation, replace B – E with A – C + D (from eq.3): A – C + D + C – D – A = D which simplifies to 0 = D
Hint #2
In eq.4, replace D with 0: 0 = B – C Add C to both sides of the above equation: 0 + C = B – C + C which makes C = B
Hint #3
In eq.3, substitute B for C, and 0 for D: A – B + 0 = B – E Add B and E to both sides of the above equation: A – B + 0 + B + E = B – E + B + E which becomes eq.3a) A + E = 2×B
Hint #4
Add A and F to both sides of eq.6: F – A + A + F = E – F + A + F which becomes 2×F = E + A which is the same as 2×F = A + E Substitute 2×F for A + E in eq.3a: 2×F = 2×B Divide both sides by 2: 2×F ÷ 2 = 2×B ÷ 2 which makes F = B
Hint #5
Substitute B for C and F in eq.5: E – A = A + B + B – E + B which becomes E – A = A + 3×B – E Add E and A to both sides of the equation above: E – A + E + A = A + 3×B – E + E + A which becomes 2×E = 2×A + 3×B Divide both sides by 2: 2×E ÷ 2 = (2×A + 3×B) ÷ 2 which makes eq.5a) E = A + 1½×B
Hint #6
Substitute A + 1½×B for E (from eq.5a) in eq.3a: A + A + 1½×B = 2×B which becomes 2×A + 1½×B = 2×B Subtract 1½×B from both sides of the above equation: 2×A + 1½×B – 1½×B = 2×B – 1½×B which makes 2×A = ½×B Multiply both sides by 2: 2 × 2×A = 2 × ½×B which makes 4×A = B and also makes F = B = C = 4×A
Hint #7
Substitute (4×A) for B in eq.5a: E = A + 1½×(4×A) which becomes E = A + 6×A which makes E = 7×A
Solution
Substitute 4×A for B and C and F, 0 for D, and 7×A for E in eq.1: A + 4×A + 4×A + 0 + 7×A + 4×A = 20 which simplifies to 20×A = 20 Divide both sides of the above equation by 20: 20×A ÷ 20 = 20 ÷ 20 which means A = 1 making B = C = F = 4×A = 4 × 1 = 4 E = 7×A = 7 × 1 = 7 and ABCDEF = 144074