求教一道奥数题,答得快又对的,追加分 请教一道小学奥数题,时间要求是5分钟,求解题思路和答案,谢谢...
\u4e00\u9053\u5965\u6570\u9898\uff0c\u4f30\u7b97\u3002\u8bf7\u5927\u54e5\u5927\u59d0\u5927\u4fa0\u5927\u4ed9\u4eec\u6765\u89e3\u7b54\u3002\u8981\u6c42\uff1a\u5feb\u72e0\u51c6\uff01\u7b54\u7684\u597d\u52a0\u5206\u300297-73=24\u5206\uff0c\u800397\u5206\u7684\u60c5\u51b5\u4e0b\u6bd4\u800373\u5206\u7684\u60c5\u51b5\u4e0b\u603b\u5206\u8981\u9ad824\u5206\uff0c\u5e73\u574790\u5206\u6bd4\u5e73\u574787\u5206\u6bcf\u95e8\u8bfe\u7a0b\u8981\u9ad890-87=3\u5206\uff0c\u6240\u4ee5\u9700\u8981\uff1a 24\u9664\u4ee53=8\u95e8\u8bfe\u7a0b\u624d\u80fd\u6ee1\u8db3\u8981\u6c42\u3002
B \u00d7 D= -- D \u7684\u53ea\u6709\uff1a6 \u00d7 2=12 3 \u00d75=15 6 \u00d74=14 7 \u00d75 =35 9 \u00d75=45
\u548c1 \u00d7 \u4efb\u4f55\u6570
\u4e00 \u4e00\u7b97\u6765\uff1a
6 \u00d7 2=12 \uff0c D=2 \uff0cB=6\uff0cC=1 222\u00f712 \u9664\u4e0d\u5c3d\uff0c\u4e0d\u884c
3 \u00d75=15 \uff0c D=5 \uff0cB=3\uff0cC=1\u30012\u30013 555\u00f715 =37 B=7 \u4e86\u4e0d\u884c
6 \u00d74=14 \uff0c D=4 \uff0cB=6\uff0cC=1 444\u00f714 \u9664\u4e0d\u5c3d\uff0c\u4e0d\u884c
7 \u00d75 =35\uff0c D=5\uff0cB=7\uff0cC=1 555\u00f715 =37 \u53ef\u4ee5
B=1 D=1 C=1---8\uff0c 111\u00f711 -----111\u00f781 \u5747\u4e0d\u884c
D=2 C=1--8\uff0c 222\u00f712 -----222\u00f782 \u5747\u4e0d\u884c
D=3, C=1-8 , 333\u00f713 333\u00f783 \u4e0d\u884c
D=4 C=1 --8 444\u00f714 ... 444\u00f784
....... 666\u00f716....666\u00f786
777\u00f717 777\u00f787 \u53ea\u6709777\u00f737=21
888\u00f718... 888\u00f788
999\u00f719 ...999 \u00f799 \u5747\u4e0d\u884c\uff0c
\u6240\u4ee5\uff0c\u7b54\u6848\u4e3a\uff1a 37 \u00d7 15= 555 21\u00d737=777
解:注意到1010.....101×11=1111....111
而N×111=111.....11(共15个)
于是,取k=15,则有:
10101.....101(共15个1)=1111.....111(30个1)/11=N×111×(1000...(当中共14个0)..001/11)
而括号中显然是整数,即10101...101(共15个1)为N的倍数
下面证明15即最小值
假设m为k的最小值,即:101.....101(m个1)为N的倍数且m<15
则1010....101(15个1)-101....101(m个1)=101...101(15-m个1)×10^2m也是A的倍数
因为m是满足条件的最小值,所以m<15-m
m<8
N本身即为13位数,则m只能为7或8
经过计算则可知7和8均不满足条件
所以k=15即为满足条件的最小值
1010101010101/1001001001001=0.1
101010101010101/1001001001001=1
解:设M = 1010101…01 ,其中数字1出现k次,N = 1001001001001。试求出最小的k值使得M能被N整除
解:注意到1010.....101×11=1111....111
而N×111=111.....11(共15个)
于是,取k=15,则有:
10101.....101(共15个1)=1111.....111(30个1)/11=N×111×(1000...(当中共14个0)..001/11)
而括号中显然是整数,即10101...101(共15个1)为N的倍数
下面证明15即最小值
假设m为k的最小值,即:101.....101(m个1)为N的倍数且m<15
则1010....101(15个1)-101....101(m个1)=101...101(15-m个1)×10^2m也是A的倍数
因为m是满足条件的最小值,所以m<15-m
m<8
N本身即为13位数,则m只能为7或8
经过计算则可知7和8均不满足条件
所以k=15即为满足条件的最小值
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