Puzzle for December 4, 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
Add A, F, and C to both sides of eq.5: D – A – F + A + F + C = A + B – C + A + F + C which becomes D + C = 2×A + B + F which is the same as eq.5a) C + D = 2×A + B + F
Hint #2
In eq.4, replace C + D with 2×A + B + F (from eq.5a): E – F = B + 2×A + B + F – E which becomes E – F = 2×A + 2×B + F – E Add F and E to both sides of the above equation: E – F + F + E = 2×A + 2×B + F – E + F + E which becomes 2×E = 2×A + 2×B + 2×F Divide both sides by 2: 2×E ÷ 2 = (2×A + 2×B + 2×F) ÷ 2 which becomes eq.4a) E = A + B + F
Hint #3
In eq.3, substitute A + B + F for E (from eq.4a): A + B + F + F = A + B + C which becomes A + B + 2×F = A + B + C Subtract A and B from each side of the above equation: A + B + 2×F – A – B = A + B + C – A – B which simplifies to eq.3a) 2×F = C
Hint #4
Add F to both sides of eq.4: E – F + F = B + C + D – E + F which becomes E = B + C + D – E + F which may be written as E = B + C + D + F – E Substitute A – C + E for D + F (from eq.2) in the equation above: E = B + C + A – C + E – E which becomes E = B + A which is equivalent to eq.4b) E = A + B
Hint #5
Substitute E for A + B (from eq.4b) into eq.4a: E = E + F Subtract E from each side of the equation above: E – E = E + F – E which makes 0 = F
Hint #6
Substitute 0 for F in eq.3a: 2×0 = C which makes 0 = C
Hint #7
eq.6 may be written as: A = (B + D + E + F) ÷ 4 Multiply both sides of the above equation by 4: 4 × A = 4 × (B + D + E + F) ÷ 4 which becomes 4×A = B + D + E + F which may be written as eq.6a) 4×A = B + D + F + E
Hint #8
Substitute 0 for C in eq.2: D + F = A – 0 + E which becomes eq.2a) D + F = A + E Subtract A from each side of eq.4b: E – A = A + B – A which becomes eq.4c) E – A = B
Hint #9
In eq.6a, substitute E – A for B (from eq.4c), and A + E for D + F (from eq.2a): 4×A = E – A + A + E + E which becomes 4×A = 3×E Divide both sides of the above equation by 4: 4×A ÷ 4 = 3×E ÷ 4 which makes A = ¾×E
Hint #10
Substitute ¾×E for A in eq.4c: E – ¾×E = B which makes ¼×E = B
Hint #11
Substitute 0 for F and C, and ¾×E for A in eq.2: D + 0 = ¾×E – 0 + E which makes D = 1¾×E
Solution
Substitute ¾×E for A, ¼×E for B, 0 for C and F, and 1¾×E for D in eq.1: ¾×E + ¼×E + 0 + 1¾×E + E + 0 = 15 which simplifies to 3¾×E = 15 Divide both sides of the above equation by 3¾: 3¾×E ÷ 3¾ = 15 ÷ 3¾ which means E = 4 making A = ¾×E = ¾ × 4 = 3 B = ¼×E = ¼ × 4 = 1 D = 1¾×E = 1¾ × 4 = 7 and ABCDEF = 310740