1. The title and answer of the divisible problem of Olympic Mathematics in primary school
( 1)2673 135 (2)89906 15496
Solve the problem (1) 2673 135 = 2,673,135, 2+673+ 135 = 8 10.
Because 8 10 can be divisible by 27 but not by 37, 2673 135 can be divisible by 27 but not by 37.
(2)89906 15496=8,990,6 15,496,8+990+6 15+496=2, 109。
2. The title and answer of the divisible problem of Olympic Mathematics in primary school
From left to right, students numbered 1 to 199 1 line up, and from left to right, students numbered 1 to1report. Then the rest of the students count from 1 to1from left to right. If they count to 1 1, the rest of the students leave the team. The remaining students count off for the third time from 1 to 1 1 from left to right. Students who reported to 1 1 stayed behind, and the rest of the students left the team. Then the first person from the left is (). Test center: divisible questions.
Analysis: The original number of students left by the first report is a multiple of 1 1; If these students continue to report, the initial number of remaining students is a multiple of11×1=121,and so on.
Solution: The initial number of students left after the first registration is a multiple of 1 1;
The initial number of students left after the second registration is a multiple of 12 1;
The initial number of students left after the third report is all multiples of 133 1;
So in the end, there is only one classmate left, and his initial number is1331;
A: The starting number of the first person from the left is 133 1.
3. The title and answer of the divisible problem of Olympic Mathematics in primary school
Can you arrange 1 00 natural numbers from1to 100 on the circumference so that at least two of any five connected numbers can be divisible by 3? If the answer is "yes", name an arrangement; If the answer is "no", you need to explain. Test center: the separability of numbers.
Analysis: according to the meaning of the question, the hypothesis method can be used for analysis. Any five natural numbers of 100 are connected and can be divided into 20 groups, so that at least two of any five connected numbers can be divisible by 3, then 40 numbers will be multiples of 3. In fact, only 33 natural numbers from 1 to 100 are multiples of 3.
Solution: suppose that the number 100 can be arranged on the circumference according to the requirements of the topic.
Divide them into five groups according to the order of arrangement, and you can get 20 groups.
There are no * * * same numbers in each group, so at least two of the five numbers in each group are multiples of three.
The divisibility of five-year series in primary schools: Olympiad problem and answer: such a * * * will have at least 40 numbers that are multiples of 3. But in fact, only 33 natural numbers from 1 to 100 are multiples of 3.
Cause conflicts, so you can't.
A: No.
4. The title and answer of the divisible problem of Olympic Mathematics in primary school.
1. Students in Class One and Class Two of a grade are planting trees. The total number of trees planted in the two classes is the same, ranging from 250 to 300. One student in both classes doesn't plant trees and is responsible for delivering water to others. Each of the students who planted trees in Class One planted seven trees, while each of the students who planted trees in Class Two planted 13 trees. How many students are there in two classes? 2. Students in Class One and Class Two of a grade are planting trees. The total number of trees planted in the two classes is the same, ranging from 250 to 300. One student in both classes doesn't plant trees and is responsible for delivering water to others. Each of the students who planted trees in Class One planted seven trees, while each of the students who planted trees in Class Two planted 13 trees. Can the total number of trees in a class be divisible by 7?
Students in Class One and Class Two of a grade are planting trees. The total number of trees planted in the two classes is the same, ranging from 250 to 300. One student in both classes doesn't plant trees and is responsible for delivering water to others. Each of the students who planted trees in Class One planted seven trees, while each of the students who planted trees in Class Two planted 13 trees. Can you tell me how many trees have been planted in a class?
Students in Class One and Class Two of a grade are planting trees. The total number of trees planted in the two classes is the same, ranging from 250 to 300. One student in both classes doesn't plant trees and is responsible for delivering water to others. Each of the students who planted trees in Class One planted seven trees, while each of the students who planted trees in Class Two planted 13 trees. How many students are there in two classes?
5. The title and answer of the divisible problem of Olympic Mathematics in primary school
After 1992, add three numbers to form a seven-digit number, which can be divisible by 2, 3, 5, 1 1 respectively. What is the minimum value of this seven-digit number? Test center: the separability of numbers.
Analysis: let's assume that the three digits composed of three digits are abc, which can be divisible by 2 and 5, indicating that C = 0；; From the fact that seven digits can be divisible by 3, we know that1+9+9+2+A+B+C = 21+A+B+C can be divisible by 1 1, so A+B can be divisible by 3. Seven digits can be divisible by 1 1, which means that (1+9+a+c)-(9+2+b) = a-b-1can be divisible by1; Finally, the seven digits should be the smallest, so take a+b=3 and a-b= 1, so that a=2 and b= 1. Then answer;
Solution: let's assume that the three digits composed of three digits are abc, and the seven digits can be divisible by 2 and 5, that is, C = 0；;
From the fact that seven digits can be divisible by 3, we know that1+9+9+2+A+B+C = 21+A+B+C can be divisible by 1 1, so A+B can be divisible by 3.
From the fact that the seven digits are divisible by 1 1, we can know that (1+9+a+c)-(9+2+b) = a-b-1can be divisible by1;
The seven digits of the composition should be the smallest, so take a+b=3 and a-b= 1, so that a=2 and b= 1.
So the smallest seven digits are 19922 10.
[Note] The usual solution for students is that according to the condition that seven digits can be divisible by 2, 3, 5, 1 1 respectively, seven digits must be the common multiple of 2, 3, 5,1,while 2, 3, 5,1.
In this way,1992000 ÷ 330 = 6036 ...120, so the seven digits that meet the meaning of the question should be (6036+ 1) times, that is,1992000+(330-/)
Comments: The answer to this question should be combined with the meaning of the question, according to the characteristics that numbers can be divisible by 2, 3, 5, 1 1, and then draw a conclusion.