Puzzle for April 30, 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.
Scratchpad
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Hint #1
Add A and B to both sides of eq.4: B – A + A + B = D – B + A + B which becomes eq.4a) 2×B = D + A Add B to both sides of eq.5: D + F – B + B = B + E – A + B which becomes eq.5a) D + F = 2×B + E – A
Hint #2
In eq.5a, replace 2×B with D + A (from eq.4a): D + F = D + A + E – A which becomes D + F = D + E Subtract D from each side of the above equation: D + F – D = D + E – D which makes F = E
Hint #3
eq.6 may be written as: C = (B + D + F) ÷ 3 Multiply both sides of the above equation by 3: 3 × C = 3 × (B + D + F) ÷ 3 which becomes eq.6a) 3×C = B + D + F which becomes
Hint #4
In eq.6a, replace B + D with C + F (from eq.3): 3×C = C + F + F which becomes 3×C = C + 2×F Subtract C from each side of the equation above: 3×C – C = C + 2×F – C which makes 2×C = 2×F Divide both sides by 2: 2×C ÷ 2 = 2×F ÷ 2 which makes C = F
Hint #5
Add A to both sides of eq.2: D – A + A = A + E + A which becomes eq.2a) D = 2×A + E In eq.4a, substitute 2×A + E for D (from eq.2a): 2×B = 2×A + E + A which becomes eq.4b) 2×B = 3×A + E
Hint #6
Add B to both sides of eq.4a: 2×B + B = D + A + B which is the same as 3×B = B + D + A Substitute C + F for B + D (from eq.3) in the above equation: eq.4c) 3×B = C + F + A
Hint #7
Substitute E for C and F in eq.4c: 3×B = E + E + A which becomes 3×B = 2×E + A Subtract 2×E fromm both sides of the above equation: 3×B – 2×E = 2×E + A – 2×E which becomes eq.4d) 3×B – 2×E = A
Hint #8
Substitute 3×B – 2×E for A (from eq.4d) into eq.4b: 2×B = 3×(3×B – 2×E) + E which becomes 2×B = 9×B – 6×E + E which becomes 2×B = 9×B – 5×E In the equation above, add 5×E to both sides, and subtract 2×B from both sides: 2×B + 5×E – 2×B = 9×B – 5×E + 5×E – 2×B which makes eq.4e) 5×E = 7×B
Hint #9
Multiply both sides of eq.4d by 5: 5 × (3×B – 2×E) = 5 × A which becomes 15×B – 10×E = 5×A which may be written as 15×B – 2×(5×E) = 5×A Substitute 7×B for 5×E (from eq.4e) into the above equation: 15×B – 2×(7×B) = 5×A which becomes 15×B – 14×B = 5×A which makes B = 5×A
Hint #10
Substitute 5×A for B in eq.4e: 5×E = 7×(5×A) Divide both sides of the above equation by 5: 5×E ÷ 5 = 7×(5×A) ÷ 5 which makes E = 7×A and also makes C = F = E = 7×A
Hint #11
Substitute 7×A for E in eq.2a: D = 2×A + 7×A which makes D = 9×A
Solution
Substitute 5×A for B, 7×A for C and E and F, and 9×A for D in eq.1: A + 5×A + 7×A + 9×A + 7×A + 7×A = 36 which simplifies to 36×A = 36 Divide both sides of the above equation by 36: 36×A ÷ 36 = 36 ÷ 36 which makes A = 1 making B = 5×A = 5 × 1 = 5 C = E = F = 7×A = 7 × 1 = 7 D = 9×A = 9 × 1 = 9 and ABCDEF = 157977