## 2007年9月30日 星期日

### 培正數學邀請賽：經典重溫（二）

#### 7 則留言:

Andy Chan 提到...

Pick's Formula？願聞其詳。

Kahoo 提到...

Andy：

Andy Chan 提到...

Pop 提到...

Let n be the number of votes for Mr Tung, N be the total number of votes

according to the question,
6655/10000 <= n/N < 6665/10000
2000n / 1331 >= N > 2000n / 1333
n + 669n/1331 >= N > n + 667n / 1333

N bounded by [n+ 667n/1333, n +669n/1331)
Now, find min. n such that the integral part of LHS and RHS are diff.

consider only 667n/1333 and 669n/1331

The remainder of 667n/1333
for odd n (2m+1) = 667 + m --(1)
for even n (2m) = m --------(2)

The remainder of 669n/1331
for odd n (2m+1) = 669 + 7m ---(3)
for even n (2m) = 7m ---------(4)

where m = 0, 1, 2, ...

clearly, (3) is growing faster with respect to n and lead to diff. integral part for LHS and RHS

Thus,
669 + 7m > 1331
m > 662 /7 = 94.57
therefore, min m = 95, i.e. min n = 2*95+1=191
(N = 287)

Kahoo 提到...

pop:
Your solution is very nice, and it does not require the observation that 66.6% is close to two-thirds. Would you mind letting your solution be included in a future version of the PCIMC solutions?

Pop 提到...

Kahoo, it's my pleasure.

As i can only get the numerical solution from the net, that s why i work out my own.