how many even three-digit integers haver the property that t

三星福田2022-10-04 11:39:541条回答

how many even three-digit integers haver the property that their digits ,read left to right ,are strictly increasing order?

已提交,审核后显示!提交回复

共1条回复
jingui1234 共回答了24个问题 | 采纳率95.8%
有多少偶数的三位数整数有如下性质,从左向右读,每个数字递增,距离246
这个直接分类法再加上点排列组合
最後一位只能是24680,要严格递增,所以0不可能,分类如下,首位有1-6,六种可能,(7,8,9都无法构成严格递增的偶数),中间只能1-7
1(中间不可小於等於1)开头2,3搭配4,6,8;4,5搭配6,8;6,7搭配8
同理继续2开头3搭配4,6,8;4,5搭配6,8;6,7搭配8
3开头4,5搭配6,8;6,7搭配8
4开头5搭配6,8;6,7搭配8
5开头6,7搭配8
6开头7搭配8
最後结果2*3+2*2+2+3+2*2+2+2*2+2+2+2+2+1=34种
我不知道有没有更简单的方法了- -
1年前

相关推荐

美国数学竞赛AMC2012,Two integers have a sum of 26.When two more in
美国数学竞赛AMC2012,
Two integers have a sum of 26.When two more integers are added to the first two integers the sum is 41.Finally when two more integers are added to the sum of the previous four integers the sum is 57.What is the minimum number of even integers among the 6 integers?
A.1 B.2 C.3 D.4
我爱新产品1年前2
wanghaipeng 共回答了21个问题 | 采纳率76.2%
有两个整数的和是26,
又有两个整数加到前面两个整数上,和为41,
再有两个整数加到前面四个整数上,和是57,
问,
这六个数中,偶数的数量,最小是几?(也就是说最少有几个偶数)

根据上面的答案,分析,肯定是
最少有一个偶数,最少有一个奇数.
最多有五个偶数,最多有五个奇数.
英语翻译How many 3-digit positive integers are odd and do not co
英语翻译
How many 3-digit positive integers are odd and do not contain the digit "5"?
honeylr1年前5
andylaw7000 共回答了15个问题 | 采纳率86.7%
问题:
有多少个三位的正整数是奇数且不包含数字5.
首先,三位数从100到999一共是900个,其中一半是奇数,一半是偶数,所以奇数是450个.
这其中百位是5的占了1/9,一共是50个,去掉之后剩400个.
余下的数中十位是5的占了1/10,一共是40个,去掉之后剩360个.
再余下的数中个位是5的占了1/5,一共是72个,去掉之后剩288个.
所以满足题意的这样的三位数一共288个.
There're 288 3-digit positive integers.
Subscript indices must either be real positive integers or l
Subscript indices must either be real positive integers or logicals.
爱ff菜的喽罗1年前1
诡才 共回答了19个问题 | 采纳率89.5%
x=x(i)-xc%j改成i这个错误是说:角标不能是正整数或逻辑变量 由于你的循环变量是i而你下面的循环又用j rx=x(j)-xc这句应当是rx=x(i)-xc %j改成i
1、three positive integers(正整数)have a sum of 28.The greatest
1、three positive integers(正整数)have a sum of 28.The greatest possible product that these integers can have is_______.
2、two numbers are in the ratio 2:3.when 4 is added to each number the ratio changes to 5:7.The sum of two orginal numbers is______.
3、Did you know?A palindrome is a number(回文数)which reads the samn forwards as backwards e.g.(例如)35453.Next year 2002 is an exanple of a palindrome number.What are the difference between 2002 and the number od the previous(以前)palindrome year?
HLZB1年前1
ClaraShumann 共回答了19个问题 | 采纳率100%
1.196=14*14
2.(2a+4):(3a+4)=5:7 a=8
原来的两个数是16 24
3.2002-1991 = 11
The number of triples (a,b,c) of positive integers such that
The number of triples (a,b,c) of positive integers such that 1/a+1/b+1/c=3/4
is
(A) 16 (B) 25 (C) 31 (D) 19 (E) 34
正确答案是b
sun211年前1
Lilsunflower 共回答了19个问题 | 采纳率89.5%
题目:已知a、b、c是三个正整数,满足方程1/a+1/b+1/c=3/4的三元组(a,b,c)的个数有多少?
其实就是找满足该方程的正整数解,有多少组.
不妨设 a≥b≥c,则1/a≤1/b≤1/c,因此3/a ≤ 1/a+1/b+1/c ≤ 3/c 所以有 a≥4,c≤4
当 c=1 时,方程变为 1/a + 1/b = 3/4-1 = -1/4,不符合条件
当 c=2 时,方程变为 1/a + 1/b = 3/4-1/2=1/4,又1/a≤1/b,用和上面同样的方法可以得 a≥8,b≤8,所以b可以取5,6,7,8,对应的a为20,12,28/3,8.所以这种情况下,有三组解
当 c=3 时,方程变为 1/a + 1/b = 3/4-1/3=5/12,又1/a≤1/b,可得 b≤ 24/5,即 b≤4,而b≥c,故b可取3,4,对应的a为12,6.所以这种情况下,有两组解
当 c=4时,方程变为 1/a + 1/b = 3/4-1/4=1/2,又1/a≤1/b,可得 b≤4,而b≥c,故b=4,此时a=4
综上,在满足 a≥b≥c条件下,有6个解,分别是(20,5,2),(12,6,2),(8,8,2),(12,2,2),(6,4,3),(4,4,4)
但是题目并没有指定a,b,c的大小,于是
(20,5,2) 有3x2x1=6种顺序
(12,6,2)有3x2x1=6种顺序
(8,8,2)有(3x2x1)/2=3种顺序
(12,2,2)有(3x2x1)/2=3种顺序
(6,4,3)有3x2x1=6种顺序
(4,4,4)只有一种顺序
所以总共有 6+6+3+3+6+1=25个解答,答案是B
if n is the product of the integers from 1 to 8 ,inclusive ,
if n is the product of the integers from 1 to 8 ,inclusive ,how many different prime factors greater than 1 dose n have?
a/ 4
b/ 5
c/ 6
d/ 7
e/ 8
飛飛藸1年前1
忆云77 共回答了11个问题 | 采纳率90.9%
n是从1到8(包括1和8)的所有整数的乘积,问其有多少个质因子,大于1的.
1*2*3*4*5*6*7*8=2^7*3^2*5*7
所以一共有2,3,5,7四个
c++ 数据结构,帮下忙re-implement BagOfIntegers to use an array of it
c++ 数据结构,帮下忙
re-implement BagOfIntegers to use an array of item counts as the underlying data structure.
For example,if the bag is {1,3,3,3,3,5,7,7,8} then the first nine elements of your count array will be 0,1,0,4,0,1,0,2,1.
#ifndef _BagOfIntegers
#define _BagOfIntegers
#include "BagInterface.h"
class BagOfIntegers :public BagInterface
{
x05private:
x05static const int DEFAULT_BAG_SIZE = 8;
x05int items[DEFAULT_BAG_SIZE];
x05int itemCount;
x05int maxItems;
int getIndexOf(const int& target) const;
public:
x05BagOfIntegers();
x05int getCurrentSize() const;
x05bool isEmpty() const;
x05bool add(const int& newEntry);
x05bool remove(const int& anEntry);
x05void clear();
x05bool contains(const int& anEntry) const;
x05int getFrequencyOf(const int& anEntry) const;
x05vector toVector() const; };
BagOfIntegers::BagOfIntegers() :itemCount(0),maxItems(DEFAULT_BAG_SIZE)
{}
int BagOfIntegers::getCurrentSize() const{
x05return itemCount;}
bool BagOfIntegers::isEmpty() const{
x05return itemCount == 0;}
bool BagOfIntegers::add(const int& newEntry){
x05bool hasRoomToAdd = (itemCount < maxItems);
x05if (hasRoomToAdd){
x05x05items[itemCount] = newEntry;
x05x05itemCount++;}
x05return hasRoomToAdd;}
bool BagOfIntegers::remove(const int& anEntry){
x05int locatedIndex = getIndexOf(anEntry);
x05bool canRemoveItem = isEmpty() && (locatedIndex > -1);
x05if (canRemoveItem){
x05x05itemCount--;
x05x05items[locatedIndex] = items[itemCount];}
x05return canRemoveItem;}
void BagOfIntegers::clear(){
x05itemCount = 0;}
int BagOfIntegers::getFrequencyOf(const int& anEntry) const{
int frequency = 0;
int searchIndex = 0;
while (searchIndex < itemCount) {
if (items[searchIndex] == anEntry) {
frequency++;}
searchIndex++; }
return frequency;}
bool BagOfIntegers::contains(const int& anEntry) const{
x05return getIndexOf(anEntry) > -1;}
vector BagOfIntegers::toVector() const{
x05vector bagContents;
x05for (int i = 0; i < itemCount; i++)
x05x05bagContents.push_back(items[i]);
return bagContents;}
int BagOfIntegers::getIndexOf(const int& target) const{
x05bool found = false;
int result = -1;
int searchIndex = 0;
while found && (searchIndex < itemCount)) {
if (items[searchIndex] == target) {
found = true;
result = searchIndex; }
else
{
searchIndex++;
} } return result;} #endif
感恩kitty1年前1
bayiforever 共回答了12个问题 | 采纳率91.7%
class BagOfIntegers : public BagInterface
{
private:
...
vector CountArray;
public:
...
void UpdateCountArray()
{
CountArray.clear();
for(int i = 0 ; i < itemCount ; i ++)
{
CountArray.push_back(getFrequencyOf(i));
}
}
};
Assume that a,b,c,d are integers,and
Assume that a,b,c,d are integers,and four equations (a-2b)x=1,(b-3c)y=1,(c-4d)z=1,w+100=d have always solutions x,y,z,w of positive numbers respectively,then the minimum of a is ______.
(英汉词典:to assume假设;integer整数;equation方程;solutions (方程的)解;positive正的;respectively分别地;minimum最小值)!
小男孩181年前1
xiaobao130 共回答了18个问题 | 采纳率88.9%
∵四个方程都有正的解,
∴a>2b,b>3c,c>4d,d>100,
∴可得d min =101,
又∵c>4d=404,得出c min =405,
b>3c=1215,得出b min =1216,
a>2b=2432,得出a min =2433.
故答案为:2433.
Find all positive integers x and Y such that x≤y≤2x and 1+x平
Find all positive integers x and Y such that x≤y≤2x and 1+x平方+y平方=3xy
就太聪明了
hh装备1年前4
phonelee 共回答了13个问题 | 采纳率84.6%
由于x≤y≤2x 得1≤y/x≤2 (positive integers x and Y)
由1+x²+y²=3xy得:(y/x-1)*(y/x-2)=1-1/x²≤0
所以x≤1
又x,y都是positive integers
所以x=1
y=1或者2
请教一下112数学难题的第70题,how many positive integers can be expressed
请教一下112数学难题的第70题,how many positive integers can be expressed as a product of two or more of the prime numbers 5,7,11 and 13 if no one product is to include the same prime factor more than once?eightnineteneleventwelve答案选的是D,
逍遥碳1年前1
huie83126 共回答了18个问题 | 采纳率94.4%
product是乘积的意思.问题译文:如果没有一个乘积包括多于一次的同样素数/质数,那么由素数5,7,11,13表达的正整数有多少个?选项是8,9,10,11,12(个) 共11个,如下:5x7,5x11,5x13,7x11,7x13,11x13 5x7x11,5x7x13,5x11x13,7x11x13 5x7x11x13
1.If x and y are integers and xy+x^2 is odd,which of the fol
1.If x and y are integers and xy+x^2 is odd,which of the following statements must be true?
① x is odd
② y is odd
③ x+y is odd
A.①
B.③
C.① and ②
D.① and ③
E.② and ③
2.For how many integers n is (2n+1)(3n-1) a negative number?
A.None
B.One
C.Two
D.Three
E.Four
求详解
百年樵夫1年前3
bassv8 共回答了19个问题 | 采纳率94.7%
1 x(x+y)是奇数,所以x和x+y都是奇数,选D
2 n=0是上式才是负数,其他都是整数,而0不是自然数,所以没有,选A
Of the three-digit integers greater than 700 ,how many have
Of the three-digit integers greater than 700 ,how many have two digits that are equal to each other and the remaining digit different from the other two?
a/ 90
b/ 82
c/ 80
d/ 45
e/ 36
魉魑魍魅1年前1
心灵寄托2 共回答了15个问题 | 采纳率100%
这一题先翻译一下:
在大于700的三位整数中,有多少个整数其中有两个位数上的数字相等而剩余的数字不和这个数字相等,例如:833,838,883……注意,由于有一个数不能相等,所以888不是.
既然理解了题目的意思,那么我们可以将相同的归为3类:百位和十位相等,十位和个位相等,百位和个位相等.
当百位和十位相等时,题目限制了百位至少为7,至多为9,所以一共三种可能性:7,8,9.而中间0~9中每一种可能性都会导致一个数是不允许的,也就是777,888,999.所以一共有3×10-3=27种;
当百位和十位相等时,思考方式同上,也是27种(不懂再追问)
当个位和十位相等时,考虑到百位的存在,当百位是7时不能取7,百位是8时不能取8,百位上是9时不能取9,所以也是3*10-3=27种,总共是27×3=81种,那么,没一个选项是81,怎么回事呢?
原来在第三种的时候,将700这种情况也算进去了,所以要再减1,等于80,至此,回答完毕.
英语翻译The number of integers between 15 and 51 that are square
英语翻译
The number of integers between 15 and 51 that are squares of integers
数学题里的一句话.求翻译,
jnk99991年前1
jz_post 共回答了18个问题 | 采纳率77.8%
15和51之间是某个整数平方的整数的个数,有16、25、36、49,所以答案是4
杭电acm Problem DescriptionNow give you two integers n m,you j
杭电acm
Problem Description
Now give you two integers n m,you just tell me the m-th number after radix point in 1/n,for example n=4,the first numble after point is 2,the second is 5,and all 0 followed
Input
Each line of input will contain a pair of integers for n and m(1
xiaoyan1985361年前1
奥诗卡 共回答了15个问题 | 采纳率73.3%
你想的太简单了
给你代码,你去研究一下吧
#include
//除法的基本运算
void x_mod_y( int x, int y, int *shang, int *yushu )
{
*shang=0;
while ( x >= y )
{
*shang += 1 ;
x-=y;
}
*yushu=x ;
}
自己去加循环
int main()
{
int n,m;
int i=0;
int x,shang,yushu;
scanf("%d %d",&n,&m );
x=1;
for( i=0;x&&i
90n+23p=4523.If n and p are positive integers in the equatio
90n+23p=4523.If n and p are positive integers in the equation above,what is one possible value of n+p?
怎么算?
是51,118和185
514391年前3
flyingzeus 共回答了14个问题 | 采纳率85.7%
n、p 均为正整数
90n+23p=4523
n=4523/90 - 23/90 p
n+p=4523/90 - 23/90 p +p=(4523+67p)/90 属于正整数
所以4523+67p)是90的倍数
所以当p=1时就符合
n+p=(4523+67)/90=51
由于电脑上不便于操作没有用数学符号,你自己写答案的时候一定要记得用数学符号哟!
Sum It UpGiven a specified total t and a list of n integers,
Sum It Up
Given a specified total t and a list of n integers,find all distinct sums using numbers from the list that add up to t.For example,if t = 4,n = 6,and the list is [4,3,2,2,1,1],then there are four different sums that equal 4:4,3+1,2+2,and 2+1+1.(A number can be used within a sum as many times as it appears in the list,and a single number counts as a sum.) Your job is to solve this problem in general.
Input
The input file will contain one or more test cases,one per line.Each test case contains t,the total,followed by n,the number of integers in the list,followed by n integers x 1 ,...,x n .If n = 0 it signals the end of the input; otherwise,t will be a positive integer less than 1000,n will be an integer between 1 and 12 (inclusive),and x 1 ,...,x n will be positive integers less than 100.All numbers will be separated by exactly one space.The numbers in each list appear in nonincreasing order,and there may be repetitions.
Output
For each test case,first output a line containing `Sums of',the total,and a colon.Then output each sum,one per line; if there are no sums,output the line `NONE'.The numbers within each sum must appear in nonincreasing order.A number may be repeated in the sum as many times as it was repeated in the original list.The sums themselves must be sorted in decreasing order based on the numbers appearing in the sum.In other words,the sums must be sorted by their first number; sums with the same first number must be sorted by their second number; sums with the same first two numbers must be sorted by their third number; and so on.Within each test case,all sums must be distinct; the same sum cannot appear twice.
Sample Input
4 6 4 3 2 2 1 1
5 3 2 1 1
400 12 50 50 50 50 50 50 25 25 25 25 25 25
0 0
Sample Output
Sums of 4:
4
3+1
2+2
2+1+1
Sums of 5:
NONE
Sums of 400:
50+50+50+50+50+50+25+25+25+25
50+50+50+50+50+25+25+25+25+25+25
用C/C++写程序,
jing53321年前1
夏野奔流 共回答了22个问题 | 采纳率86.4%
#include
#include
int num[13];
int x[13];
int sum,n;
int cw;
int result[1000][13];
int count;
void backtrack(int t)
{
if(t>n)
{
if(cw==sum)
{
int i;
for(i=1;i
解一道有英语的奥数题:The sum of n different positive integers is less
解一道有英语的奥数题:The sum of n different positive integers is less than 50.
The sum of n different positive integers is less than 50.The greatest possble value of n is( )
A.10 B.7 C.8 D.9
其中positive integers 是正整数的意思.
求奥数和英语都特厉害的大师相救啊!
回答时顺便列一下式子啊!
爱笑的小猪猪1年前3
若水35 共回答了17个问题 | 采纳率82.4%
n个不同的正整数和小于50,那么这个n的最大可能值为9
,这几个数是从1到9,总和是45=(1+9)*9/2
以下是我做的几道题1.Given a,b,c are positive integers,and a,b are pri
以下是我做的几道题
1.Given a,b,c are positive integers,and a,b are prime numbers,(a的3次方+b的c次方+a)=2005 ,then the value of a+b+c is( )
A.14 B.13 C.12 D. 11
(英汉词典 positive integers:正整数.prime numbers:质数)_
2.已知△ABC的三条高的比是3∶4∶5,且三条边的长均为整数,则△ABC的边长可能是( )
A.10 B.12 C.14 D.16
3.某汽车从A地驶向B地,若每分钟行驶a千米,则11点到达,若每分钟行驶 a千米,则11∶20时距离B地还有10千米;如果改变出发时间,若每分钟行驶 a千米,则11点到达,若每分钟行驶a千米,则11∶20时已经超过B地30千米.A、B两地的路程是 千米.
4、若 abc321是一个六位数,其中a,b,c是三个互异的数字,且都不等于0,1,2,3,又M是7的倍数,那么M的最小值是 .
5、分解因式:(x+1)(x+2)(x+3)(x+6)+(x的2次方)=?(因为2次方符号打不出,用文字表达)
6.如果正整数n有以下性质:n的八分之一是平方数,n的九分之一是立方数,它的二十五分之一是五次方数,那么n就称为“希望数”,则最小的希望数是
以下是答案:
1.D 2.B 3.54km 4.468321 5.(x的平方+6x+6)的2次方
6.(2的15次方)×(3的20次方)×(5的12次方)
答案是知道,就是想知道怎么解,知道的拜托了
罗马战火1年前1
版主VIII 共回答了16个问题 | 采纳率93.8%
1.第一题你看看是不是抄掉了东西,我做了几遍都无解,具体思路如下:
由a^3+b^c+a=2005知a^3
how many integers are there from 11 to 111 inclusive?
how many integers are there from 11 to 111 inclusive?
一定要告诉我你是怎么算出来的噢!( 谁去考了SSHAT 得不可以答!)
jellydong1年前2
lshiwbqd 共回答了24个问题 | 采纳率87.5%
111-11+1=101
麻烦做一道gmat数学题if s and t are positive integers such that s/t=6
麻烦做一道gmat数学题
if s and t are positive integers such that s/t=64.12,which of the following could be the remainder when s is divided by
a)2 b)4 c)8 d)20 e)45
eeddine1年前1
ctcp 共回答了9个问题 | 采纳率100%
e
余数部分应该=0.12t,在2,4,8,20,45中,只有45/0.12=整数375
Find all integers x such that 2x^2+x-6 is a positive integer
Find all integers x such that 2x^2+x-6 is a positive integeral power of a prime positive integer.
随处风流X1年前1
hdmmm 共回答了18个问题 | 采纳率88.9%
Let f(x)=2x^2+x-6 = (2x-3)(x+2).
Suppose a positive integer a divides both 2x-3 and x+2.
Then a must also divide 2(x+2)-(2x-3)=7.
Hence,a can either be 1 or 7.
As a result,2x-3=7^n or -7^n for some positive integer n.
We consider the following cases:
(2x-3) = 1.Then x = 2,which yields f(x) = 4,a prime power.
(2x-3) = -1.Then x = 1,which yields f(x) = -3,not a prime power.
(2x-3) = 7.Then x = 5,which yields f(x) = 49,a prime power.
(2x-3) = -7.Then x = -2,which yields f(x) = 0,not a prime power.
(2x-3) = 7^n or -7^n,for n >= 2.Then,since x + 2 = ((2x-3) + 7)/2,we have that x + 2 is divisible by 7
but not by 49.Hence x + 2 = 7 or -7,yielding x = 5,-9.The former has already been considered,
while the latter yields f(x) = 147.
So x can be either 2 or 5.
英语翻译five integers are arranged from least to greatest.if the
英语翻译
five integers are arranged from least to greatest.if the median is 12 and the only mode is 5,what is the least possible range for the 5 numbers
关键是最后一句问什么?答案给的是9
她在我左边1年前2
孤独的海明威 共回答了19个问题 | 采纳率100%
5个整数从小到大排列.如果中值是12且惟一的众数是5,那么这5个数的最小可能范围是什么?
1.Three positive integers have a sum of 28.The greatest poss
1.Three positive integers have a sum of 28.The greatest possible product that these integers can have is?
2.Jack was trying to tessellate regular pentagons.He managed the following figure.The size of angle.a.is?
1图:




请写明译文,能解答更好,
虚拟20011年前1
尘尘203 共回答了18个问题 | 采纳率88.9%
1. 三个正整数和为28,最大可能的积是多少,没说要不同正整数,就是取最相近9,9,10相乘咯,810.
2. 也就是算360度减去三个正五边形的内角,360-(108×3)=36度
一道英语的数学题. The least integer of consecutive integers is -25.I
一道英语的数学题. The least integer of consecutive integers is -25.If the sum of these integers is
The least integer of consecutive integers is -25.If the sum of these integers is 26,how many integers are in this set?
A 25
B 26
C 50
D 51
E 52
翻译一下题目啦~~ 大师们~
demrccrz1年前4
东南风雅 共回答了17个问题 | 采纳率94.1%
有一个连续的整数列,最小的是-25.如果他们的和是26,问有多少个整数在这个数列中?
显然是52个
-25..-1
0
25.1
26
英语翻译The sum of n different positive integers is less than 50
英语翻译
The sum of n different positive integers is less than 50.The greatest possible value of n is( )
ss0071年前2
泥小虾 共回答了14个问题 | 采纳率85.7%
n个不同的正整数的和是小于50.其中n的最大值可以是()
C++麻烦解释一下题目意思Description:A sequence of n (n> 0) integers is
C++麻烦解释一下题目意思
Description:A sequence of n (n> 0) integers is called a jolly jumper if the absolute values of the difference between successive elements take on all the values 1 through n-1. For instance, 1 4 2 3 is a jolly jumper, because the absolutes differences are 3, 2, and 1 respectively. The definition implies that any sequence of a single integer is a jolly jumper. You are to write a program to determine whether or not each of a number of sequences is a jolly jumper.
Input:Each line of input contains an integer n (n< 3000) followed by n integers representing the sequence.
Output:
For each line of input, generate a line of output saying "Jolly" or "Not jolly". Sample Input:4 1 4 2 3
5 1 4 2 -1 6Sample Output:Jolly
Not jolly
孔方芳1年前1
sxk123321 共回答了18个问题 | 采纳率88.9%
描述:一组由n个数字组成的序列,如果序列中所有两个连续数之间差值的绝对值刚好是1到n-1这些数,那么这个序列就称为jolly jumper.比如,1 4 2 3就是一个jolly jumper.因为差值的绝对值分别是3,2和1.这个定义同时告诉我们单独一个数字也是jolly jumper.你需要写一个程序来确定一组序列是否是jolly jumper.
输入:一行数字,包含一个n,和n个数字组成的序列.
输出:对于每一行输入,确定它们是否是Jolly.如果是输出Jolly.不是的话输出Not jolly.
例子输入:4 1 4 2 3
5 1 4 2 -1 6
例子输出:Jolly
Not jolly
SAT 思路是什么If X and Y are two different integers and the produ
SAT 思路是什么
If X and Y are two different integers and the product 35XY is the square of an integer,which of the following could be equal to XY
A.5 B.70 C.105 D.140 E.350
包豁转1年前1
weareboy 共回答了18个问题 | 采纳率94.4%
35XY=5*7*XY
所以XY必须为5*7*n^2的形式
140=5*7*4符合条件
所以答案选D
What is the greatest of 5 consecutive integers if the sum of
What is the greatest of 5 consecutive integers if the sum of these integers equals 185?
unique8301011年前2
鲍鲍六子 共回答了20个问题 | 采纳率95%
39
这五个整数是:35 36 37 38 39

~~~~满意请将此选为“满意答案”,多谢!(*^__^*) ~~~~
看不懂题干.The sum of 4 integers is even.Which one is larger,A or
看不懂题干.
The sum of 4 integers is even.Which one is larger,A or
A The number of these integers that are odd.
B 4
jzzszz1年前3
lltt022 共回答了23个问题 | 采纳率100%
4个整数的和是偶数,以下哪个数更大,A还是B?
A,这些数当中的奇数的个数
B,4
当然选B的吧...
奇数的个数是0、2或者4啊
我怎么算都是只有一个啊for how many integers n is (2n+1)(3n-1) a negativ
我怎么算都是只有一个啊
for how many integers n is (2n+1)(3n-1) a negative number?
答案说有4个,哪来的啊
legna_cc1年前1
说实话的人 共回答了18个问题 | 采纳率83.3%
(2n+1)(3n-1)
英语翻译Consider the set of integers greater that -2 and less th
英语翻译
Consider the set of integers greater that -2 and less than 6.A subset of this set is the positive factors of 5.What is the complement of this subset?
1057137751年前1
lbag 共回答了9个问题 | 采纳率77.8%
翻译:
考虑大于 -2 小于 6 的整数集合,这个集合的一个子集是5的正因数集,那么这个子集的补集是什么?
保证准确,祝好!
Set S consists of n distinct positive integers none of which
Set S consists of n distinct positive integers none of which is greater than 12
补:What is greater possible value of n two integers in S have a common factor greater than 1谁能翻译一下?
wxf_123_4561年前1
numen800 共回答了16个问题 | 采纳率100%
S由n个各不相同的不大于12的正整数组成
sat数学题The sum of four consecutive odd integers w,x,y,and z i
sat数学题
The sum of four consecutive odd integers w,x,y,and z is 24.What is the median of the set{ w,x,y,z,24}?
醍醐1年前4
人造小太阳 共回答了13个问题 | 采纳率92.3%
题目是在问:4个连续奇数w,x,y,z的和是24 ,这一组数{ w,x,y,z,24}的中位数是多少?
median是中位数的意思 当然 y就是这组数中的中位数啦
4个奇数很好求 w+(w+2)+(w+4)+(w+6)=24 得w=3 y=w+4=7
所以 答案是7
An integer c is a common divisor of two integers x and y if
An integer c is a common divisor of two integers x and y if and only if c is a divisor of x and c is a divisor of y.Which of the following sets of integers could possibly be the set of all common divisors of two integers?A.{-6,-2,-1,1,2,6}B.{-6,-2,-1,0,1,2,6}C.{-6,-3,-2,-1,1,2,3,6}D.{-6,-3,-2,-1,0,1,2,3,6}E.{-6,-4,-3,-2,-1,1,2,3,4,6}看懂是问公因数问题 ,但是不知道答案为什么是C...前辈们指导下
zhubier5201年前1
陆拾年代ing爆爆 共回答了21个问题 | 采纳率85.7%
首先不能有0,去掉答案B和D.根据题目给出的答案选项看,如果6是正确的,则分解为±1,2,3,6.A少3,E多4.
SAT数学题求解答How many integers between 100 and 150, inclusive, c
SAT数学题求解答
How many integers between 100 and 150, inclusive, can be evenly divided by neither 3 nor 5?
A 33 B 26 C 24 D 28 E 27
lacoobye1年前2
斌天雪地 共回答了19个问题 | 采纳率89.5%
问题是问100到150(100与150都包括)的整数既不被3整除又不被5整除的数有多少.
首先,100到150中被3整除的数有102,105,...,150,即共(150-102)/3+1=17个.
其次,100到150中被5整除的数有100,105,...,150,即共(150-100)/5+1=11个.
再次,100到150中被3*5=15整除的数有105,120,...,150,即共(150-105)/15+1=4个.
所以被3整除或者被5整除的数共17+11-4=24个,所以既不被3整除又不被5整除的数有51-24=27个,选E.
多项式(英文)How to list the integers which divide into the consta
多项式(英文)
How to list the integers which divide into the constant term?
For example,P(x)=x^3-7x-6
how to find out those integers are
1,-1,2,-2,3,-3,6,-6?
sz8703051年前1
ljllk 共回答了17个问题 | 采纳率94.1%
Put these integers into the equation P(x)=0
If k,which is one of the integers,is a root of the equation,then (x³-7x-6) can be factorized to (x-k)(x²+px+q).
如何证明:prove that there exists a pair of consecutive integers
如何证明:prove that there exists a pair of consecutive integers such that one of there integers is a perfect square and the other is a perfect cube.
月风满楼1年前2
xnxrds 共回答了28个问题 | 采纳率96.4%
(0,1) is a pair of consecutive integers in which 0 is a perfect square and 1 is a perfect cube.
英语翻译原题是:Find three consecutive odd integers such that their
英语翻译
原题是:
Find three consecutive odd integers such that their sum decreased b the second equals 50.
三个连着的奇数的和减少了什么什么的.翻译并且讲解一下翻译,数学题不用做,
nishusheng-1291年前1
xiao45518 共回答了19个问题 | 采纳率100%
SAT GRE GMAT?找出三个连着的奇数 a,b,c ,其总和减去第二个奇数后等于50 啊,a,b,c应该是23,25,27
英语翻译1.Find two consecutive integers whose sum is 383.2.Find
英语翻译
1.Find two consecutive integers whose sum is 383.
2.Find the measure of two complementary angles if the measure of one is 42 degrees more than twice the measure of the other.
因为是数学题,所以语序很重要,不然运算顺序会错.
希望有能的大大能帮忙!
(真的不要翻译器的,)
3.The sum of the measures of the angles in a triangle is 180 The measure of one angle of a triangle is 20 degrees more than the measure of the second angle.The measure of the third angle is 30 degrees more than twice the sum of the measures of the first two angles.Find the measure of each angle in the triangle.
仓促的青春1年前4
yilupingan 共回答了18个问题 | 采纳率88.9%
1.找两个连续的整数,令他们的和为383
2.找两个互补角,其中一个的度数比 另一个的两倍 多42度.
3.三角形中三角之和为180度.第一个角的度数比第二个多20度,第三个角的度数比 前两个角和的两倍 多30度.求出这个三角形的三个角度数.
注:complementary angles 是互补角 不是余角
翻译无误.祝好!
问一道SAT数学题Which of the following integers,when rounded to the
问一道SAT数学题
Which of the following integers,when rounded to the nearest thousand,resaults in 2000
A 2567
B 1499
C 1097
D 1601
E 2700
wyw-ly1年前1
暗暗2303 共回答了12个问题 | 采纳率100%
当四舍五入的时候,下面哪个整数会得到2000?
当然是1601
英语的数学题,求高人指导,急!how is division of integers similar to the di
英语的数学题,求高人指导,急!
how is division of integers similar to the dicision of whole numbers? how is it different? use examples to explain
有些打错了,上面打成dicision的是division
gxbslcd1年前2
boleesoo 共回答了21个问题 | 采纳率76.2%
整数分为正整数、负整数和0,
正整数,比如1,15,37等等
负整数,比如 -3,-45等等
7、The sum of n different positive integers is less than 50.T
7、The sum of n different positive integers is less than 50.The greatest possible value of n is( )
(A)10 (B)9 (C)8 (D)7
叶罗江南1年前2
幽谷遂 共回答了20个问题 | 采纳率90%
没问题
英语翻译The sum of n different positive integers is less than 50
英语翻译
The sum of n different positive integers is less than 50.The greatest possible value of n is( )
cnihc20001年前2
情舞枫阳 共回答了20个问题 | 采纳率95%
n个不同的正整数的和是小于50.其中n的最大值可以是()
英语翻译Properties of Integers Sets simplifying algebraic expres
英语翻译
Properties of Integers
Sets
simplifying algebraic expression
solving linear equations with one unknown
solving two linear equations with two unknowns
solving equations by factoring
solving quadratic equations
intersecting lines and angles
polygons
rectangular solids and cylinders
coordinate geometry
zeee1年前3
lele_2005 共回答了29个问题 | 采纳率89.7%
Properties of Integers 整数的属性Sets 集合simplifying algebraic expression 简化代数式solving linear equations with one unknown 解答一元线性方程式solving two linear equations with two unknowns 解答二元...
sat的2道数学题!1.If j k and n are consecutive integers such that
sat的2道数学题!
1.If j k and n are consecutive integers such that 0
chenyx5251年前4
黎铭 共回答了20个问题 | 采纳率85%
1.如果j,k,n 是连续的三个整数,并且 0
问一到英语数学题Suppose all the integers have been colored red,green
问一到英语数学题
Suppose all the integers have been colored red,green,and blue (one color per integer),such that
the sum of any two green integers (equal or unequal) is blue;
the sum of any two blue integers (equal or unequal) is green;
the negative of any blue integer is green and vice versa.
How many such colorings are there in which 2007 is green?Describe them and prove that there are no others.
rangrang1231年前2
媳妇至尊宝 共回答了20个问题 | 采纳率80%
假如所有的完整事物已经被染色红色、绿色、和蓝色的 (每完整的事物一种颜色),以致于任何的二个绿色的完整事物 (对手或不相等的) 的总数是蓝色的; 任何的二个蓝色的完整事物 (对手或不相等的) 的总数是绿色的; 任何蓝...
SAT数学题求解The integers 1 through 6 appear on the six faces of
SAT数学题求解
The integers 1 through 6 appear on the six faces of a cube, one on each face. If three such cubes are rolled , what is the probability that the sum of the numbers on the top faces is 17 or 18? 答案是1/54,求思考过程和详细解释
connie2481年前3
小猪莎拉 共回答了17个问题 | 采纳率94.1%
三次的和为18只有一种情况,6+6+6=18
三次的和为17的有如下几种情况
6+6+5=17
6+5+6=17
5+6+6=17
所以一共是4种情况下可以得到17或者18
扔一次有6种情况,扔3次的话就是6*6*6=216种情况
所以满足条件的概率就是4÷216=1/54
All the positive integers less than 49 are multiplied togeth
All the positive integers less than 49 are multiplied together.How many zeros will this product end in?
为梦而存1年前1
陈洪湖 共回答了15个问题 | 采纳率86.7%
9个.(nine)
the average of 19 consecutive integers is 99.the largest of
the average of 19 consecutive integers is 99.the largest of these inters is()是什么意思?
42238871年前2
ff资源 共回答了16个问题 | 采纳率93.8%
19个连续的整数的平均数是99,这些整数中,最大的是多少