Question
If β534β = β50β, β758β = β138β,
β3156β = β71, then β4738β = ?Solution
The sum of all the squares of the digits of first number is second number. In 534 = 50, 5Β² + 3Β² + 4Β² = 25 + 9 + 16 = 50 In 758 = 138, 7Β² + 5Β² + 8Β² = 49 + 25 + 64 = 138 In 3156 = 71, 3Β² + 1Β² + 5Β² + 6Β² = 9 + 1 + 25 + 36 = 71 Similarly, in 4738 = 4Β² + 7Β² + 3Β² + 8Β² = 16 + 49 + 9 + 64 = 138
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