Common Multiples Problems in Mathematics, Number Theory Problems in Elementary Mathematics Common Fa

There is 1 person left at the end of a row of 3 people, 5 people at the end of a row of 7 people, and 6 people at the end of a row of 8 people.

It is equivalent to a row of 3 people, a row of 7 people, and a row of 8 people. In the end, there were 2 people left.

Find the least common multiple of -2

3 7 8-2 = 166 (person).

A: There are 166 people in total.

From the meaning of the title, the number of people in the rehearsal activity is 2 fewer than the common multiple, so the minimum common number is 3 * 7 * 8 = 168 The number of people in the rehearsal activity is at least 168-2 = 166 people, if not the minimum number of people, it is 168n-2 people (n means ......

If the number of people is increased by 2, the least common multiple that can be divisible at the same time is: 3*7*8=168168-2=166, so there are at least 166 people in total.

Let this three-digit number be a, and note that (a-2) is a multiple of 3 and 7, i.e. a multiple of 21, so we can let (a-2)=21k get: a=21k+2

20k+(k+2), and because a is divided by 5 and remains, (k+2) is divided by 5 and remains, so k+2=5n+3 gets k=5n+1 and brings this back to a=21k+2 to get a=105n+23

Therefore, the general solution of the number a is a=105n+23 when divided, divided by 5 and 2, divided by 7 and 2

Since 100<=a<=999, 100<=105n+23<=999 solves the maximum value of n as 9

Therefore, the desired is 105*9+23=968

The number of students participating in gymnastics performances in schools is between 60 100. Divide these students into groups of 8 or 12 equally. At least () of the students participated in this performance.

Test Points: Common Factors and Common Multiples Problems.

Analysis: According to the number of people, they are divided into a group of 8 people, or they are divided into a group of 12 people, so the total number of people is a common multiple of 8 and 12, and then the total number of people is between 60 and 100.

Answer: Solution: 8=2 2 2;

The least common multiple of 8 and 12 is: 2 2 2 3=24;

Then the common multiples of 8 and 12 are: 24, 48, 72, 96, ,...

Since the total number is 60,100, the total number is either 72 or 96, with a minimum of 72.

A: At least 72 students participated in this performance.

So the answer is: 72

Comments: This problem uses the common multiple to solve the square limb method, finds the common multiple of 8 and 12, and then uses the range of the total number of people to solve it.

1: Find: 15,20,25 and the greatest common multiple?

2: The greatest common divisor of the two composite numbers is 1, and the least common multiple is 36, find these two numbers.

3: The greatest common divisor of two prime numbers is: The least common multiple is: (fill in the blank) 4: There are three numbers that are mutually primitive, their least common multiple is 4, and the greatest common multiples of the two numbers are: 24, 40, 60, find these three numbers.

5: The least common multiple of the two numbers is 300, and the greatest common divisor is 15.

I follow my ideas.

First of all, note that 10*11=110

110 So 119 is only possible to be a number below 10 multiplied by a number above 10.

Then, the common multiple ends with 9

Then the two prime numbers must be odd, and they cannot be 5

Under 10, only 3 and 7 meet the conditions.

Then break down 119 a bit

I can't divide it with 3.

It's only 7*17=119

ok~~~

Common multiples (common

multiple) means that in two or more natural numbers, if they have the same multiple, the multiple of the shirt is their common multiple. The smallest of these common multiples is called the lowest common multiple of these integers

common

multiple）

Multiples of 4

Multiples of 6 where the common multiple of is

So, there are still 36 teachers left.

b=90

a÷b=2×7

One is 2 b = 180 and the other is 7 b = 630 or one = b = 90 and the other = a = 1260