(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 (C)6 total solutions, n=400,470,2290,2360,2430,2500.
OR
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≤70k+50 Squaring the second inequality, we get