推荐回答(3个)
分析与解:
设开始有x个桃子,我们把x写成(x+4)-4.
第一个猴子来了,吃掉1个,还有桃子
(x+4)-4-1=(x+4)-5,
这时恰好可分成5份,每份的桃子数为
[(x+4)-5]/5=(x+4)/5-1
(x+4)/5必须为整数,所以(x+4)是5的倍数,
第一个猴子藏掉一份后,剩下的桃子为:
(4/5)×[(x+4)-5]=(4/5)×(x+4)-4
同样,第二个猴子来了,一吃一藏之后,剩下的桃子数为
(4/5)×[(4/5)×(x+4)-5]
由于(4/5)×(4/5)×(x+4)是整数,故(x+4)应是5×5=25的倍数,
如此一来五个猴子一吃一藏,恰好剩下
(4/5)×(4/5)× (4/5)×(4/5) ×(4/5) ×(x+4)-5个桃子,
故(x+4)必须是5×5×5×5×5的倍数,
即x+4=5^5
所以: x=3125-4=3121
即开始最少有3121个桃子.
答案是错的!最小满足条件的是 621个桃子!
设最后一只猴子拿走了$n个,桃子总数是¥m个;
公式如下:
for($n=1;$n<10000;$n++){
$m = 5*(5*(5*(5*$n+1)/4+1)/4+1)/4+1;
if(is_int($m)){
echo $n;
echo '
';
echo $m;
echo '
';
break;
}
}
得出当总数为621的时候即可满足条件,最后一只猴子拿走了63个;
以下是一万以内的满足情况!
最后一只猴子拿走的桃子数63 桃子总数为621
最后一只猴子拿走的桃子数127 桃子总数为1246
最后一只猴子拿走的桃子数191 桃子总数为1871
最后一只猴子拿走的桃子数255 桃子总数为2496
最后一只猴子拿走的桃子数319 桃子总数为3121
最后一只猴子拿走的桃子数383 桃子总数为3746
最后一只猴子拿走的桃子数447 桃子总数为4371
最后一只猴子拿走的桃子数511 桃子总数为4996
最后一只猴子拿走的桃子数575 桃子总数为5621
最后一只猴子拿走的桃子数639 桃子总数为6246
最后一只猴子拿走的桃子数703 桃子总数为6871
最后一只猴子拿走的桃子数767 桃子总数为7496
最后一只猴子拿走的桃子数831 桃子总数为8121
最后一只猴子拿走的桃子数895 桃子总数为8746
最后一只猴子拿走的桃子数959 桃子总数为9371
最后一只猴子拿走的桃子数1023 桃子总数为9996
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