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Find a two-digit number which exceeds by 12 the sum of the squ

Question

Find a two-digit number which exceeds by $12$ the sum of the squares of its digits and by $16$ the doubled product of its digits.

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Find a two-digit number which exceeds by

12

the sum of the squares of its digits and by

16

the doubled product of its digits.

12 mins ago

Discuss this question LIVE

12 mins ago

Text solutionVerified

let the number be

10 a + b

From the given conditions

10 a + b = a^{2} + b^{2} + 12...... (1)

10 a + b = 2 a b + 16.......... (2)

(1) - (2) ⟹ (a - b)^{2} = 4 ⟹ a - b = 2

⟹ 10 a + (a - 2) = 2 (a) (a - 2) + 16

⟹ 2 a^{2} - 4 a + 16 - 10 a - a + 2 = 0

⟹ a = 6

⟹ b = 4

So the number is

64

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Question Text	Find a two-digit number which exceeds by $12$ the sum of the squares of its digits and by $16$ the doubled product of its digits.
Answer Type	Text solution:1
Upvotes	150