Puzzle for July 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
Help Area
Hint #1
Add C to both sides of eq.3: E + F + C = A + B – C + D + C which becomes E + F + C = A + B + D which may be written as eq.3a) C + E + F = A + B + D Add F to both sides of eq.4: C + E + F = A – B + D – F + F which becomes eq.4a) C + E + F = A – B + D
Hint #2
Subtract the left and right sides of eq.4a from the left and right sides of eq.3a, respectively: C + E + F – (C + E + F) = A + B + D – (A – B + D) which becomes C + E + F – C – E – F = A + B + D – A + B – D which simplifies to 0 = 2×B which means 0 = B
Hint #3
eq.6 may be written as: F = (A + C + D) ÷ 3 Multiply both sides of the equation above by 3: F × 3 = ((A + C + D) ÷ 3) × 3 which becomes eq.6a) 3×F = A + C + D In eq.2, replace B with 0: C + D = A – 0 + F which becomes eq.2a) C + D = A + F
Hint #4
In eq.6a, replace C + D with A + F (from eq.2a): 3×F = A + A + F which becomes 3×F = 2×A + F Subtract F from both sides of the equation above: 3×F – F = 2×A + F – F which becomes 2×F = 2×A Divide both sides by 2: 2×F ÷ 2 = 2×A ÷ 2 which makes F = A
Hint #5
In eq.4, substitute 0 for B, and A for F: C + E = A – 0 + D – A which becomes eq.4b) C + E = D
Hint #6
eq.5 may be written as: E = (B + C + F) ÷ 3 Multiply both sides of the above equation by 3: E × 3 = ((B + C + F) ÷ 3) × 3 which becomes 3×E = B + C + F Substitute 0 for B, and A for F: 3×E = 0 + C + A which becomes eq.5a) 3×E = C + A
Hint #7
Substitute C + E for D (from eq.4b), 0 for B, and A for F in eq.2: C + C + E = A – 0 + A which becomes 2×C + E = 2×A Subtract 2×C from each side of the equation above: 2×C + E – 2×C = 2×A – 2×C which becomes eq.2b) E = 2×A – 2×C
Hint #8
Substitute (2×A – 2×C) for E (from eq.2b) in eq.5a: 3×(2×A – 2×C) = C + A which becomes 6×A – 6×C = C + A In the above equation, add 6×C to both sides, and subtract A from both sides: 6×A – 6×C + 6×C – A = C + A + 6×C – A which becomes eq.5b) 5×A = 7×C
Hint #9
Multiply both sides of eq.5a by 5: 5×(3×E) = 5×(C + A) which becomes 15×E = 5×C + 5×A Substitute 7×C for 5×A (from eq.5b) in the above equation: 15×E = 5×C + 7×C which becomes 15×E = 12×C Divide both sides by 12: 15×E ÷ 12 = 12×C ÷ 12 which makes 1¼×E = C
Hint #10
Substitute 1¼×E for C in eq.4b: 1¼×E + E = D which makes 2¼×E = D
Hint #11
Substitute 1¼×E for C in eq.5a: 3×E = 1¼×E + A Subtract 1¼×E from each side of the above equation: 3×E – 1¼×E = 1¼×E + A – 1¼×E which makes 1¾×E = A and also makes F = A = 1¾×E
Solution
Substitute 1¾×E for A and F, 0 for B, 1¼×E for C, and 2¼×E for D in eq.1: 1¾×E + 0 + 1¼×E + 2¼×E + E + 1¾×E = 32 which simplifies to 8×E = 32 Divide both sides of the above equation by 8: 8×E ÷ 8 = 32 ÷ 8 which means E = 4 making A = F = 1¾×E = 1¾ × 4 = 7 C = 1¼×E = 1¼ × 4 = 5 D = 2¼×E = 2¼ × 4 = 9 and ABCDEF = 705947