(Recall that ⌊x⌋ is the greatest integer not exceeding x.)
Answer Choices:
A. 2
B. 4
C. 6
D. 30
E. 32 Solution:
We can first consider the equation without a floor function:
70n+1000​=n​
Multiplying both sides by 70 and then squaring:
n2+2000n+1000000=4900n
Moving all terms to the left:
n2−2900n+1000000=0
Now we can determine the factors:
(n−400)(n−2500)=0
This means that for n=400 and n=2500, the equation will hold without the floor function.
Now we can simply check the multiples of 70 around 400 and 2500 in the original equation:
For n=330, left hand side =19 but 182<330<192 so right hand side =18
For n=400, left hand side =20 and right hand side =20
For n=470, left hand side =21 and right hand side =21
For n=540, left hand side =22 but 540>232 so right hand side =23
Now we move to n=2500
For n=2430, left hand side =49 and 492<2430<502 so right hand side =49
For n=2360, left hand side =48 and 482<2360<492 so right hand side =48
For n=2290, left hand side =47 and 472<2360<482 so right hand side =47
For n=2220, left hand side =46 but 472<2220 so right hand side =47
For n=2500, left hand side =50 and right hand side =50
For n=2570, left hand side =51 but 2570<512 so right hand side =50
Therefore we have 6 total solutions, n=400,470,2290,2360,2430,2500=(C)6​
OR
This is my first solution here, so please forgive me for any errors.
We are given that
70n+1000​=⌊n​⌋
⌊n​⌋ must be an integer, which means that n+1000 is divisible by 70 . As 1000≡20(mod70), this means that n≡50(mod70), so we can write n=70k+50 for k∈Z.
Therefore,
70n+1000​=7070k+1050​=k+15=⌊70k+50​⌋
Also, we can say that 70k+50​−1<k+15 and k+15 \leq \sqrt
Squaring the second inequality, we get k2+30k+225≤70k+50⟹k2−40k+175≤0⟹(k−5)(k−35)≤0⟹5≤k≤35.
Similarly solving the first inequality gives us k<19−155​ or k>19+\sqrt
155​ is larger than 12 and smaller than 13 , so instead, we can say k≤6 or k≥32.
Combining this with 5≤k≤35, we get k=5,6,32,33,34,35 are all solutions for k that give a valid solution for n, meaning that our answer is (C)6​ .