Puzzle for May 2, 2021 ( )
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Find the 5-digit number ABCDE by solving the following equations:
A, B, C, D, and E each represent a one-digit non-negative integer.
* "A mod C" equals the remainder of A ÷ C.
** "D mod A" equals the remainder of D ÷ A.
Many thanks to Tom H for sending us this challenging puzzle. Thank you, Tom!
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Hint #1
eq.2 may be written as: eq.2a) B – A = (A + C + D + E) ÷ 4 Multiply both sides of the above equation by 4: 4 × (B – A) = 4 × (A + C + D + E) ÷ 4 which becomes 4×B – 4×A = A + C + D + E Add 4×A to both sides of the above equation: 4×B – 4×A + 4×A = A + C + D + E + 4×A which becomes eq.2b) 4×B = 5×A + C + D + E
Hint #2
Add D and A to both sides of eq.1: C – D + D + A = E – A + D + A which becomes C + A = E + D which may be written as eq.1a) A + C = D + E
Hint #3
In eq.2b, replace D + E with A + C (from eq.1a): 4×B = 5×A + C + A + C which becomes 4×B = 6×A + 2×C Divide both sides of the above equation by 2: 4×B ÷ 2 = (6×A + 2×C) ÷ 2 which becomes 2×B = 3×A + C Subtract 3×A from both sides: 2×B – 3×A = 3×A + C – 3×A which becomes eq.2c) 2×B – 3×A = C
Hint #4
In eq.1a, substitute 2×B – 3×A for C (from eq.2c): A + 2×B – 3×A = D + E which becomes 2×B – 2×A = D + E which is the same as eq.1b) 2×(B – A) = D + E
Hint #5
Subtract A from each side of eq.3: B – A = A + ((D + E) ÷ C) – A which becomes B – A = (D + E) ÷ C Multiply both sides of the above equation by C: C × (B – A) = C × (D + E) ÷ C which becomes eq.3a) C × (B – A) = D + E
Hint #6
In eq.1b, substitute C × (B – A) for D + E (from eq.3a): 2×(B – A) = C × (B – A) Divide each side of the above equation by (B – A): 2×(B – A) ÷ (B – A) = C × (B – A) ÷ (B – A) which simplifies to 2 = C (assumes B – A ≠ 0)
Hint #7
Confirm: B – A ≠ 0 Substituting 0 for B – A in eq.2a would yield: 0 = (A + C + D + E) ÷ 4 which would mean 0 = A + C + D + E Since A, C, D, E are all non-negative, the above equation would make: 0 = A = C = D = E However: C ≠ 0 (from eq.3 and eq.4) and A ≠ 0 (from eq.5) so therefore: B – A ≠ 0
Hint #8
Substitute 2 for C in eq.2c: 2×B – 3×A = 2 Add 3×A to both sides of the equation above: 2×B – 3×A + 3×A = 2 + 3×A which becomes 2×B = 2 + 3×A Divide both sides by 2: 2×B ÷ 2 = (2 + 3×A) ÷ 2 which becomes eq.2d) B = (2 + 3×A) ÷ 2 eq.2d implies that A must be an even integer.
Hint #9
Confirm: A = even integer Substituting an odd integer for A in eq.2d would yield: B = (2 + 3×(odd integer)) ÷ 2 which would become B = (2 + odd integer) ÷ 2 which would mean B = (odd integer) ÷ 2 which would make B = integer + ½ Since B is an integer, then B ≠ integer + ½ which means A ≠ odd integer and makes A = even integer
Hint #10
Substitute 2 for C in eq.5: eq.5a) D mod A = 2 Since D mod A < A then eq.5a makes: A > 2
Hint #11
To make eq.2d true, check several possible even integer values (> 2) for A: If A = 4, then B = (2 + 3×4) ÷ 2 = (2 + 12) ÷ 2 = 14 ÷ 2 = 7 If A = 6, then B = (2 + 3×6) ÷ 2 = (2 + 18) ÷ 2 = 20 ÷ 2 = 10 If A > 6, then B > 10 Since B must be a one-digit integer, then the above equations make: B = 7 and A = 4
Hint #12
Substitute 4 for A, and 2 for C in eq.4: 4 mod 2 = E which makes 0 = E
Solution
Substitute 4 for A, 2 for C, and 0 for E in eq.1a: 4 + 2 = D + 0 which makes 6 = D and makes ABCDE = 47260