Puzzle for July 12, 2019 ( )
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.
* AB and CD are 2-digit numbers (not A×B or C×D).
Scratchpad
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Hint #1
In eq.2, replace B + D with C (from eq.3): C + E + F = A + C Subtract C from each side of the above equation: C + E + F - C = A + C - C which becomes eq.2a) E + F = A
Hint #2
In eq.4, replace A with E + F (from eq.2a): D + E - F = E + F In the above equation, add F to each side, and subtract E from each side: D + E - F + F - E = E + F + F - E which simplifies to D = 2×F
Hint #3
In eq.5, substitute B + D for C (from eq.3): A + B = B + D + D + F Subtract B from each side of the above equation: A + B - B = B + D + D + F - B which simplifies to A = 2×D + F Substitute (2×F) for D: A = 2×(2×F) + F which is the same as A = 4×F + F which makes A = 5×F
Hint #4
Substitute 5×F for A in eq.2a: E + F = 5×F Subtract F from each side of the above equation: E + F - F = 5×F - F which means E = 4×F
Hint #5
Substitute 2×F for D in eq.3: B + 2×F = C Subtract B from each side of the equation above: B + 2×F - B = C - B which becomes eq.3a) 2×F = C - B
Hint #6
eq.6 may be written as: 10×C + D + F = 10×A + B Substitute 2×F for D, and (5×F) for A in the above equation: 10×C + 2×F + F = 10×(5×F) + B which becomes 10×C + 3×F = 50×F + B Subtract both B and 3×F from each side: 10×C + 3×F - B - 3×F = 50×F + B - B - 3×F which becomes 10×C - B = 47×F which may be written as eq.6a) 9×C + C - B = 47×F
Hint #7
Substitute 2×F for C - B (from eq.3a) in eq.6a: 9×C + 2×F = 47×F Subtract 2×F from each side: 9×C + 2×F - 2×F = 47×F - 2×F which means 9×C = 45×F Divide both sides by 9: 9×C ÷ 9 = 45×F ÷ 9 which makes C = 5×F
Hint #8
Substitute 5×F for C in eq.3a: 2×F = 5×F - B In the above equation, add B to each side, and subtract 2×F from each side: 2×F + B - 2×F = 5×F - B + B - 2×F which simplifies to B = 3×F
Solution
Substitute 5×F for A and C, 3×F for B, 2×F for D, and 4×F for E in eq.1: 5×F + 3×F + 5×F + 2×F + 4×F + F = 20 which simplifies to 20×F = 20 Divide both sides by 20: 20×F ÷ 20 = 20 ÷ 20 which means F = 1 making A = C = 5×F = 5 × 1 = 5 B = 3×F = 3 × 1 = 3 D = 2×F = 2 × 1 = 2 E = 4×F = 4 × 1 = 4 and ABCDEF = 535241