Puzzle for January 31, 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, A, and B from both sides of eq.2: E + F – E – A – B = A + B + C – E – A – B which becomes eq.2a) F – A – B = C – E In eq.3, subtract E from both sides, and add D to both sides: C – D – E + D = E – F – E + D which becomes C – E = –F + D which may be written as eq.3a) C – E = D – F
Hint #2
In eq.3a, replace C – E with F – A – B (from eq.2a): F – A – B = D – F Add A, B, and F to both sides of the equation above: F – A – B + A + B + F = D – F + A + B + F which becomes 2×F = D + A + B which may be written as eq.3b) 2×F = A + B + D
Hint #3
Add E and B to both sides of eq.4: D – E + E + B = A – B + E + B which becomes D + B = A + E which may be written as B + D = A + E In eq.3b, substitute A + E for B + D: 2×F = A + A + E which becomes 2×F = 2×A + E Subtract E from each side of the equation above: 2×F – E = 2×A + E – E which becomes eq.3c) 2×F – E = 2×A
Hint #4
Add A to both sides of eq.6: C + D – A + A = A + F + A which becomes C + D = 2×A + F Substitute 2×F – E for 2×A (from eq.3c) in the above equation: C + D = 2×F – E + F which becomes C + D = 3×F – E Add E to each side: C + D + E = 3×F – E + E which becomes eq.6a) C + D + E = 3×F
Hint #5
Multiply both sides of eq.3a by 3: 3 × (C – E) = 3 × (D – F) which is equivalent to 3×C – 3×E = 3×D – 3×F Substitute (C + D + E) for 3×F (from eq.6a) in the above equation: 3×C – 3×E = 3×D – (C + D + E) which is equivalent to 3×C – 3×E = 3×D – C – D – E which becomes 3×C – 3×E = 2×D – C – E Add 3×E and C to each side: 3×C – 3×E + 3×E + C = 2×D – C – E + 3×E + C which simplifies to 4×C = 2×D + 2×E Divide both sides by 2: 4×C ÷ 2 = (2×D + 2×E) ÷ 2 which becomes eq.3d) 2×C = D + E
Hint #6
Substitute 2×C for D + E (from eq.3d) in eq.6a: C + 2×C = 3×F which makes 3×C = 3×F Divide each side of the equation above by 3: 3×C ÷ 3 = 3×F ÷ 3 which makes C = F
Hint #7
Substitute C for F in eq.5: C + B = C – B In the equation above, subtract C from both sides, and add B to both sides: C + B – C + B = C – B – C + B which makes 2×B = 0 which means B = 0
Hint #8
Substitute C for F, and 0 for B in eq.2: E + C = A + 0 + C Subtract C from both sides of the equation above: E + C – C = A + 0 + C – C which makes E = A
Hint #9
Substitute A for E in eq.3c: 2×F – A = 2×A Add A to each side of the equation above: 2×F – A + A = 2×A + A which becomes 2×F = 3×A Divide both sides by 2: 2×F ÷ 2 = 3×A ÷ 2 which makes F = 1½×A and also makes C = F = 1½×A
Hint #10
Substitute A for E, and 0 for B in eq.4: D – A = A – 0 Add A to each side of the equation above: D – A + A = A – 0 + A which makes D = 2×A
Solution
Substitute 0 for B, 1½×A for C and F, 2×A for D, and A for E in eq.1: A + 0 + 1½×A + 2×A + A + 1½×A = 28 which simplifies to 7×A = 28 Divide both sides of the above equation by 7: 7×A ÷ 7 = 28 ÷ 7 which means A = 4 making C = F = 1½×A = 1½ × 4 = 6 D = 2×A = 2 × 4 = 8 E = A = 4 and ABCDEF = 406846