Problem 318
2011 nines
Consider the real number √2+√3.
When we calculate the even powers of √2+√3 we get:
(√2+√3)2 = 9.898979485566356…
(√2+√3)4 = 97.98979485566356…
(√2+√3)6 = 969.998969071069263…
(√2+√3)8 = 9601.99989585502907…
(√2+√3)10 = 95049.999989479221…
(√2+√3)12 = 940897.9999989371855…
(√2+√3)14 = 9313929.99999989263…
(√2+√3)16 = 92198401.99999998915…
It looks like that the number of consecutive nines at the beginning of the fractional part of these powers is non-decreasing.
In fact it can be proven that the fractional part of (√2+√3)2n approaches 1 for large n.
Consider all real numbers of the form √p+√q with p and q positive integers and p<q, such that the fractional part of (√p+√q)2n approaches 1 for large n.
Let C(p,q,n) be the number of consecutive nines at the beginning of the fractional part of (√p+√q)2n.
Let N(p,q) be the minimal value of n such that C(p,q,n) ≥ 2011.
Find ∑N(p,q) for p+q ≤ 2011.
2011个9
考虑实数√2+√3。
计算√2+√3的偶数次幂，我们得到：
(√2+√3)2 = 9.898979485566356…
(√2+√3)4 = 97.98979485566356…
(√2+√3)6 = 969.998969071069263…
(√2+√3)8 = 9601.99989585502907…
(√2+√3)10 = 95049.999989479221…
(√2+√3)12 = 940897.9999989371855…
(√2+√3)14 = 9313929.99999989263…
(√2+√3)16 = 92198401.99999998915…
看起来似乎这些幂的小数部分开头连续的9的数目始终不降。
事实上，可以证明(√2+√3)2n的小数部分在n充分大时趋向于1。
考虑所有可以表示为√p+√q的实数，要求p和q均为正整数，p<q，并且(√p+√q)2n的小数部分在n充分大时趋向于1。
记C(p,q,n)为(√p+√q)2n的小数部分开头连续的9的数目。
记N(p,q)为使得C(p,q,n) ≥ 2011的最小n值。
对于所有p+q ≤ 2011，求∑N(p,q)。