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题库 / Prep2007E2-DS-129
If t is a positive integer and r is the remainder when ~$t^2 + 5t + 6$~ is divided by 7, what is the value of r ?
(1) When t is divided by 7, the remainder is 6.
(2) When t2 is divided by 7, the remainder is 1.
Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
EACH statement ALONE is sufficient.
Statements (1) and (2) TOGETHER are NOT sufficient.
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t^2+5t+6=(t+2)(t+3)
(1) 设t=7x+6（x为倍数）(t+2)(t+3)=(7x+8)(7x+9)=49x^2+49x+63x+72，前4项都是7的倍数，72不是7的倍数，余数为2，所以整个式子余数为2，充分。
(2) 设t^2=7x+1（x为倍数）t=根号7x+1 t^2+5t+6=7x+1+5根号7x+1+6=7x+7+5根号7x+1，前2项都是7的倍数，但5根号7x+1无法确定除以7以后的余数，不充分。
t^2+5t+6=(t+2)(t+3)
(1) 设t=7x+6（x为倍数）(t+2)(t+3)=(7x+8)(7x+9)=49x^2+49x+63x+72，前4项都是7的倍数，72不是7的倍数，余数为2，所以整个式子余数为2，充分。
(2) 设t^2=7x+1（x为倍数）t=根号7x+1 t^2+5t+6=7x+1+5根号7x+1+6=7x+7+5根号7x+1，前2项都是7的倍数，但5根号7x+1无法确定除以7以后的余数，不充分。
（2）假设t=7k+m，发现m=1和6都能使t^2除以7余1，所以并不能确定m等于多少（当时忽视了m=6）
t^2+5t+6=(t+2)(t+3) (1) 设t=7x+6（x为倍数）(t+2)(t+3)=(7x+8)(7x+9)=49x^2+49x+63x+72，前4项都是7的倍数，72不是7的倍数，余数为2，所以整个式子余数为2，充分。 (2) 设t^2=7x+1（x为倍数）t=根号7x+1 t^2+5t+6=7x+1+5根号7x+1+6=7x+7+5根号7x+1，前2项都是7的倍数，但5根号7x+1无法确定除以7以后的余数，不充分。
如果t除以7的余数是6，那么t的平方处以7的余数也是6
你在扯什么犊子？ n mod 7 = 6, n*n mod 7 = 6*6 mod 7 = 1
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t^2可以写成t*t，所以只知道t被7除的情况就可以。发现最后余数为6/7？待确认。
请问什么是t2 是t的平方吗
(1) When t is divided by 7, the remainder is 6 --> t=7q+6t=7q+6 --> (t+2)(t+3)=(7q+8)(7q+9)(t+2)(t+3)=(7q+8)(7q+9). Now, no need to expand and multiply all the terms, just notice that when we expand all terms but the last one, which will be 8*9=72, will have 7 as a factor and 72 yields the remainder of 2 upon division by 7. Sufficient.
The numbers x and y are NOT integers. The value of x is closest to which integer?
(1) 4 is the integer that is closest to x+y --> 3.5