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garri49 [273]
3 years ago
14

The sum of the square of 3 consecutive positive integers is 110 . What are the numbers

Mathematics
1 answer:
fgiga [73]3 years ago
6 0
Let your 1st integer be X, then the 2nd will be (X+1) and the 3rd will be (X+2). Since we need the sum of the squares to equal 110 ⇒ X² + (X+1)² + (X+2)² = 110 Expanding the left-hand side: X² + X² + 2·X + 1 + X² +4·X + 4 = 110 3·X² + 6·X + 5 = 110 3·X² + 6·X - 105 = 0 Solve using the quadratic formula and you get roots: X = 5, X = -7 Your problem doesn't state that we have to restrict the solution to positive integers only and since we are summing the squares we have 2 solutions that work: 5, 6, 7 and -7, -6, -5
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Since the first digit is a factor of 20, the factors of 20 are 1,2,4,5,10,20. We only need the single digit factors which are 1,2,4 and 5. These 4 numbers can be permuted in 1 way for the first digit, so we have ⁴P₁.
For the second digit, we have 10 digits permuted in 1 way, ¹⁰P₁ and also for the third digit, we have 10 digits permuted in 1 way, ¹⁰P₁ and for the last digit, which is divisible by 5, it is either a 0 or 5, so we have two digits permuted in 1 way, ²P₁.
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