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【OG20-P286-380题】
If the positive integer n is added to each of the integers 69, 94, and 121,what is the value of n ?
(1) 69+ n and 94 + n are the squares of two consecutive integers.
(2) 94 + n and 121 + n are the squares of two consecutive integers.
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条件1:69+n=x^2,94+n=(x+1)^2解得x=12,n=75,充分
条件2:94+n=y^2,121+n=(y+1)^2解得y=13,n=75,充分
题目讨论 (4条评论)

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373961gg
are the squares of two consecutive integers--> 为两个连续整数各自的平方
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0 回复 2020-07-17 12:04:17
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kikig
From statement one, we can say that for some positive integer x, 69 + n = x^2 and 94 + n = (x + 1)^2. Let’s subtract the first equation from the second equation: (94 + n) - (69 + n) = (x + 1)^2 - x^2
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0 回复 2019-12-18 16:58:49
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phylliskong38
2 unknown 2 equations ---> solvable
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0 回复 2018-08-28 22:46:00
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Adam.W
1)69+n=a^2 94+n=(a+1)^2 上下相减,a=12, n=75 充分 2) 同样的方法 94+n=b^2 121+n=(b+1)^2 上下相减, b=13, n=75 充分
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0 回复 2018-02-25 21:53:03