Puzzle for May 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.
* BC, CD, and EF are 2-digit numbers (not B×C, C×D, or E×F).
Scratchpad
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Hint #1
In eq.2, replace C with E + F (from eq.5): E + F + F = B + E Subtract E from each side of the above equation: E + F + F - E = B + E - E which becomes 2×F = B
Hint #2
Subtract both C and E from each side of eq.2: C + F - C - E = B + E - C - E which becomes eq.2a) F - E = B - C Subtract E from each side of eq.4: A - E = B + C + E - E which becomes eq.4a) A - E = B + C
Hint #3
Subtract the left and right sides of eq.2a from the left and right sides of eq.4a, respectively: A - E - (F - E) = B + C - (B - C) which is the same as A - E - F + E = B + C - B + C which becomes eq.4b) A - F = 2×C
Hint #4
eq.6 may be written as: 10×B + C + 10×E + F = A + 10×C + D Subtract both C and F from each side of the above equation: 10×B + C + 10×E + F - C - F = A + 10×C + D - C - F which becomes 10×B + 10×E = A + 9×C + D - F which may be written as 10×B + 10×E = A - F + 9×C + D In the equation above, substitute 2×C for A - F (from eq.4b): 10×B + 10×E = 2×C + 9×C + D which may be written as eq.6a) 10×(B + E) = 11×C + D
Hint #5
Substitute C + F for B + E (from eq.2) in eq.6a: 10×(C + F) = 11×C + D which is equivalent to 10×C + 10×F = 11×C + D Subtract 10×C from each side of the above equation: 10×C + 10×F - 10×C = 11×C + D - 10×C which becomes 10×F = C + D Substitute E + F for C (from eq.5): 10×F = E + F + D Subtract F from each side: 10×F - F = E + F + D - F which becomes eq.6b) 9×F = E + D
Hint #6
Subtract B from both sides of eq.4: A - B = B + C + E - B which becomes A - B = C + E In eq.3, substitute C + E for A - B: D - E = C + E Add E to both sides of the above equation: D - E + E = C + E + E which becomes D = C + 2×E Substitute E + F for C (from eq.5): D = E + F + 2×E which becomes eq.4c) D = 3×E + F
Hint #7
Substitute 3×E + F for D (from eq.4c) in eq.6b: 9×F = E + 3×E + F Subtract F from each side: 9×F - F = E + 3×E + F - F which simplifies to 8×F = 4×E Divide both sides by 4: 8×F ÷ 4 = 4×E ÷ 4 which means 2×F = E
Hint #8
Substitute 2×F for E in eq.5: 2×F + F = C which makes 3×F = C
Hint #9
Substitute 2×F for E in eq.6b: 9×F = 2×F + D Subtract 2×F from each side of the above equation: 9×F - 2×F = 2×F + D - 2×F which means 7×F = D
Hint #10
Substitute (3×F) for C in eq.4b: A - F = 2×3×F Add F to both sides: A - F + F = 2×3×F + F which makes A = 7×F
Solution
Substitute 7×F for A and D, 2×F for B and E, and 3×F for C in eq.1: 7×F + 2×F + 3×F + 7×F + 2×F + F = 22 which simplifies to 22×F = 22 Divide both sides of the above equation by 22: 22×F ÷ 22 = 22 ÷ 22 which means F = 1 making A = D = 7×F = 7 × 1 = 7 B = E = 2×F = 2 × 1 = 2 C = 3×F = 3 × 1 = 3 and ABCDEF = 723721