Puzzle for March 21, 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.
Our thanks to Abby S (age 10) for sending us this puzzle. Thank you, Abby!
Scratchpad
Help Area
Hint #1
eq.6 may be written as: E = (B + C + D) ÷ 3 Multiply both sides of the above equation by 3: 3 × E = 3 × ((B + C + D) ÷ 3) which becomes eq.6a) 3×E = B + C + D
Hint #2
eq.5 may be written as: B = (A + D + E) ÷ 3 Multiply both sides of the above equation by 3: 3 × B = 3 × ((A + D + E) ÷ 3) which becomes eq.5a) 3×B = A + D + E
Hint #3
In eq.1, replace B + C + D with 3×E (from eq.6a): A + 3×E + E + F = 31 which becomes eq.1a) A + 4×E + F = 31
Hint #4
eq.1 may be re-written as: A + D + E + B + C + F = 31 In the equation above, replace A + D + E with 3×B (from eq.5a): 3×B + B + C + F = 31 which becomes eq.1b) 4×B + C + F = 31
Hint #5
In eq.1a, substitute 4×B + C + F for 31 (from eq.1b): A + 4×E + F = 4×B + C + F Subtract F from both sides of the above equation: A + 4×E + F – F = 4×B + C + F – F which becomes eq.1c) A + 4×E = 4×B + C
Hint #6
Add C to both sides of eq.5a: 3×B + C = A + D + E + C which may be written as eq.5b) 3×B + C = A + C + D + E Add 2×A and C to both sides of eq.4: D + E – A + 2×A + C = A + B – C + 2×A + C which becomes D + E + C + A = 3×A + B which may be written as eq.4a) A + C + D + E = 3×A + B
Hint #7
Substitute 3×B + C for A + C + D + E (from eq.5b) into eq.4a: 3×B + C = 3×A + B Subtract 3×B from each side of the equation above: 3×B + C – 3×B = 3×A + B – 3×B which becomes eq.4b) C = 3×A – 2×B
Hint #8
Substitute 3×A – 2×B for C (from eq.4b) in eq.1c: A + 4×E = 4×B + 3×A – 2×B which becomes A + 4×E = 2×B + 3×A Subtract A from both sides of the above equation: A + 4×E – A = 2×B + 3×A – A which becomes 4×E = 2×B + 2×A Divide both sides by 2: 4×E ÷ 2 = (2×B + 2×A) ÷ 2 which becomes 2×E = B + A which may be written as eq.1d) 2×E = A + B
Hint #9
Substitute 3×A – 2×B for C (from eq.4b) in eq.3: B + 3×A – 2×B + F = A + D which becomes 3×A – B + F = A + D In the equation above, add B to both sides, and subtract A from both sides: 3×A – B + F + B – A = A + D + B – A which becomes eq.3a) 2×A + F = D + B
Hint #10
Add E and F to both sides of eq.2: F – E + E + F = D – F + E + F which becomes eq.2a) 2×F = D + E Multiply both sides of eq.2a by 2: 2×(2×F) = 2×(D + E) which becomes eq.2b) 2×(2×F) = 2×D + 2×E
Hint #11
Substitute 2×F for D + E (from eq.2a) in eq.5a: 3×B = A + 2×F Subtract A from both sides of the above equation: 3×B – A = A + 2×F – A which becomes eq.5c) 3×B – A = 2×F Divide both sides of eq.5c by 2: (3×B – A) ÷ 2 = 2×F ÷ 2 which becomes eq.5d) 1½×B – ½×A = F
Hint #12
Substitute 3×B – A for 2×F (from eq.5c), and A + B for 2×E (from eq.1d) in eq.2b: 2×(3×B – A) = 2×D + A + B which becomes 6×B – 2×A = 2×D + A + B Subtract A and B from both sides of the above equation: 6×B – 2×A – A – B = 2×D + A + B – A – B which becomes eq.2c) 5×B – 3×A = 2×D
Hint #13
Substitute 1½×B – ½×A for F (from eq.5d) in eq.3a: 2×A + 1½×B – ½×A = D + B which becomes 1½×A + 1½×B = D + B Subtract B from both sides of the above equation: 1½×A + 1½×B – B = D + B – B which becomes eq.3b) 1½×A + ½×B = D
Hint #14
Substitute (1½×A + ½×B) for D (from eq.3b) in eq.2c: 5×B – 3×A = 2×(1½×A + ½×B) which becomes 5×B – 3×A = 3×A + B In the above equation, add 3×A to both sides, and subtract B from both sides: 5×B – 3×A + 3×A – B = 3×A + B + 3×A – B which becomes 4×B = 6×A Divide both sides by 4: 4×B ÷ 4 = 6×A ÷ 4 which makes B = 1½×A
Hint #15
Substitute (1½×A) for B in eq.3b: 1½×A + ½×(1½×A) = D which becomes 1½×A + ¾×A = D which makes 2¼×A = D
Hint #16
Substitute (1½×A) for B in eq.5d: 1½×(1½×A) – ½×A = F which becomes 2¼×A – ½×A = F which makes 1¾×A = F
Hint #17
Substitute (1½×A) for B in eq.4b: C = 3×A – 2×(1½×A) which becomes C = 3×A – 3×A which makes C = 0
Hint #18
Substitute 1½×A for B in eq.1d: 2×E = A + 1½×A which makes 2×E = 2½×A Divide by both sides of the above equation by 2: 2×E ÷ 2 = 2½×A ÷ 2 which makes E = 1¼×A
Solution
Substitute 1½×A for B, 0 for C, 2¼×A for D, 1¼×A for E, and 1¾×A for F in eq.1: A + 1½×A + 0 + 2¼×A + 1¼×A + 1¾×A = 31 which simplifies to 7¾×A = 31 Divide by both sides of the above equation by 7¾: 7¾×A ÷ 7¾ = 31 ÷ 7¾ which means A = 4 making B = 1½×A = 1½ × 4 = 6 D = 2¼×A = 2¼ × 4 = 9 E = 1¼×A = 1¼ × 4 = 5 F = 1¾×A = 1¾ × 4 = 7 and ABCDEF = 460957