四道五年级奥数20悬赏金,高手即来,一共5道题,跪求!帮我!!!!!!!!!!我一定感激不尽!!!!!

2024-12-02 23:18:01
推荐回答(4个)
回答1:

1)这和的分母是偶数,所以三个数相乘是偶数,也就是说其中一个是2
设另外两个数是b和c,那么
1661/1986 —1/2=334/993
恰好993=331×3, 所以是2,3,331
和为336

2)4875=3×5×5×5×13
所以这两个数只能是从3,5,5,5,13这些数中选一些相乘得到
而这些数不超过64的有3,5,15,25,39,63
和为64,两个数是25和39
恰好25有2个5, 39有1个3和1个13(要看有没有超标,比如各有2个5就超标了)
答案:25 ,39差为14

3)2294=2×31×37
可能的组合有2和31×37, 31和2×37 ,37和2×31 ,还有1和2294(千万别忘了这一个!)
两个数的和被5除不可能余1,楼主搞错啦?
余4的话,有 1145或25
余0的话,有 43或2293

4)设为a,B,c,D,e,F,g,H,i
至少4偶数,必有3个是3的倍数,至少1个是5的倍数,至少1个是7的倍数
质数最多时偶数是B,D,F,H这4个
要质数多那么不是质数那部分尽可能吧份额拿掉就行了(就是倍数什么的,你懂的)
B到H恰好是6个数,让BH是3的倍数,那就可以耗掉3的2个份额了,
剩下那个份额,只能给e了(就是说e是3的倍数)
而5和7在BDeFH中耗掉它们的份额,也是有可能的
所以就最多4个质数了

5)1872=2×2×2×2×3×3×13=16×9×13
(因数太多了,我就把几个乘到一起,比较直观)
约数个位是8并且都是两位数的的组合有 48×39 (含8或18或16×13的都被排除了)
所以应该是45×39=1755

PS:第一题一般一点的做法是应该是
334/993=1/b+1/c=(b+c)/bc
所以334bc=993(b+c)
b=993c/(334c-993)为整数,c是质数
所以334c-993= 993或331或3或1或993c或331c或3c或c
所以c=331或3,对应b分别是3或331
不过第一题也不是碰巧分解出来的
两奇质数倒数相加后分子分母还是不可约的,
因为它们的差与它们自身都不可约
此类题分析奇偶性和分解质因数都是很重要的

第二题主要陷阱还是怕超标

第三题应该错了吧?改回来之后再hi我吧~

第四题很经典,我做了之后去查查是那几个数,结果如下
101至109这九个连续的自然数中,101,103,107,109是质数
主要是假设的地方,假设2,3的话,是倍数的那个,位置变动不大
(额,我说的很抽象,希望你能懂)

第五题是逆推再排除再换算

PPS:
2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,
53,59,61,67,71,73,79,83,89,97,
101,103,107,109,113,127,131,137,139,149,
151,157,163,167,173,179,181,191,193,197,199,
211,223,227,229,233,239,241,
251,257,263,269,271,277,281,283,293,
307,311,313,317,331,337,347,349,
353,359,367,373,379,383,389,397,
401,409,419,421,431,433,439,443,449,
457,461,463,467,479,487,491,499,
503,509,521,523,541,547,
557,563,569,571,577,587,593,599,
601,607,613,617,619,631,641,643,647,
653,659,661,673,677,683,691,
701,709,719,727,733,739,743,
751,757,761,769,773,787,797,
809,811,821,823,827,829,839,
853,857,859,863,877,881,883,887,
907,911,919,929,937,941,947,
953,967,971,977,983,991,997

[总共168个]
以上是1000以内质数

回答2:

1、3个质数的倒数之和是1986分之1661,则这三个质数之和为多少?
把1986分解质因数,1986=2*3*331,那么三个质数分别是2,3,331,和是:2+3+331=336

2、如果两数的和为64,两数的积可以整除4875,那么这两个数的差等于多少?
4875分解质因数,4875=5*5*5*39,那么两个数只能是64=25+39,所以差是39=25=14

3、已知两个数的和被5除余1,他们的积是2294,他们的差等于( )。
2294分解质因数,2294=2*31*37=2*1147=62*37=31*74,依题意,都没有和被5除余1的呀,会不会题目错了?

4、9个连续的自然数,他们都大于80,其中质数最多有多少个?
9个连续的自然数至少有4个偶数,至少有一个是5或者是0在个位的,所以还剩有4个。

5、在做一道两位数乘以两位数的乘法题,玲不小心把一乘数中的数字5看成了8,此得积是1872。那么原来的乘积是( 1755 )。

1872分解因数,1872=2*2*2*2*3*3*13=48*39,原数是45*39=1755

回答3:

感觉五年级的奥数有点像初等数论的例咧!
第一题,把三个数设为x,y,z,则有1/x + 1/y + 1/z = (xy + yz + zx)/xyz=(1661K)/(1986K)
所以xyz =1986K = 2 * 3 * 331K,明显,当且仅当K为1时,这个三个数才为质数,所以三个数的和为2+3+311=316
第二题,设两个数为x,y,且x>y,因为两数的积可以整除4875,所以4875=xyK=3*5*5*5*13,又因为x+y=64,推算一下可得x为3*13=39,y为5*5=25,所以差为39-25=14
第三题,算不出来,2294=2*31*37,但这三个数不管怎么组合都组不了两个数的和被5除余1,看哪个高手能解答。
第四题,9个连续自然数中至多有5个连续的奇数,即第一个数为奇数,最后一个也是奇数,同时可推算第二五八这三个被三整除时其中有二个偶数,所以最多有5-1=4个
第五题,把1872分解为2×2×2×2×3×3×13 ,又因为是两位数与两位数相乘,所以只能为
48*39或78*24然后把8换回5,即可得 45×39=1755或75×24=1800

回答4:

1、336
2、14
3、25
4、4
5、1755

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