FIND all digits a, b, c such that 2008 abc ( a number ) are simultaneously divisible by 5 , 7 and 9
Find all digits a, b, c such that 2008abc( a number) are simultaneously divisible by 5, 7 and 9
sorry i can speak,read and write english
A 6 digit number begins with the digit 8.The number with divisible by 9 and all digits of the number are different.What is the possible smallest value of this number?
Gọi số cần tìm là 8abcde .
8 + a + b + c + d + e = số chia hết cho 9 .
Vì 8abcde có giá trị nhỏ nhất nên a = 0 ; b = 1 ; c = 2 ; d = 3
Vậy bây giờ có :
8 + 0 + 1 + 2 + 3 e = 14 + e = số chia hết cho 9
=> e = 4
Số cần tìm là 801234
gọi số phải tìm là 8abcde
Ta có : 8+a+b+c+d+e phải là số chia hết cho 9
Vì là số bé nhất nên a = 0 ;b =1 ;c = 2 và d = 3
Tổng các chữ số lúc này là : 8+0+1+2+3=14
Ta có : 14+e là số chia hết cho 9
Suy ra e = 4 vì 14+4 = 18 mà 18 chia hết cho 9
Vậy số phải tìm là 801234
If B is not divisible by 24 and B is the greatest 5 - digit number that the sum of all digits is 21 then B= ?
Nếu B là không chia hết cho 24 và B là số 5 chữ số lớn nhất mà tổng của tất cả các chữ số là 21 thì B = ...........
If B is not divisible by 24 and B is the greatest 5 - digit number that the sum of the sum of all digits is 21 then B = ?
Câu 9:
The average mass of a group of children is 47kg. If of the number of children are girls and their average mass is 39kg. Find the average mass of the boys.
Answer: The average mass of the boys is kg
Câu 10:
A 6-digit number begins with the digit 8. The number is divisible by 9 and all digits of the number are different. What is the possible smallest value of this number?
Answer:
Câu 9 : 52 kg
Câu 10 : 801234
đúng 100% nhớ tk 3 nhé
9.Young male average weight is 47 kg + (47-39) = 55 kg
10. The minimum value is 800 001
The average mass of a group of children is 47kg. If 5/13 of the number of children are girls and their average mass is 39kg. Find the average mass of the boys.
Answer: The average mass of the boys is kg
A 4 - digits numbers divisible by all of the numbers from 1 to 10 without any remainder. Find the smallest possible value of the 4-digits number
A 4 - digits numbers divisible by all of the numbers from 1 to 10 without any remainder. Find the smallest possible value of the 4-digits number
Trả lời hộ mik ai làm nhanh mik k cho
Thank you
Exer 1: Given two natural numbers whose sum are 78293. The bigger number where 5 is the units digit and 2 is hundred digit. If we clean these digits then we obtain a number which equals the smaller number. Find two natural numbers.
Exer 2: Prove that: If x, y \(\in\) N and x + 2y divisible by 5 then 3x - 4y divisibles by 5.
Exer 3: Given that 2x + 5y \(⋮\) 7. Prove that 4x + 3y \(⋮\) 7.
Exer 1:
Solution:
Suppose that, the unknown number is: \(\overline{x215}\) (where x \(\in\) N).
When we clean three digits then the smaller number is \(\overline{x}\).
We have: \(\overline{x215}\) + \(\overline{x}\) = 78293
\(\Rightarrow\) 1000. \(\overline{x}\) + 215 + \(\overline{x}\) = 78293
1001. \(\overline{x}\) = 78078
x = 78
Thus, we found two natural number: 78215 and 78.
Exer 2:
Solution:
We have: x + 2y \(⋮\) 5
\(\Rightarrow\) 2x + 4y \(⋮\) 5
(2x + 4y) + (3x - 4y) = 5x \(⋮\) 5
\(\Rightarrow\) 2x + 4y \(⋮\) 5
Deduce 3x - 4y \(⋮\) 5.
Exer 3:
Solution:
We have: 2x + 5y \(⋮\) 7
4x + 10y \(⋮\) 7
(4x + 10y) - (4x + 3y) = 7y \(⋮\) 7
\(\Rightarrow\) 4x + 10y \(⋮\) 7
Deduce 4x + 3y \(⋮\) 7.
A 4 - digits numbers divisible by all of the numbers from 1 to 10 without any remainder. Find the smallest possible value of the 4-digits number