I am a four-digit number. My first digit is a square number.
My last digit is five less than my first. My second digit is half
last digit. My third digit is three times my second digit.
my
I am a four-digit number. My first digit is a square number.
My last digit is five less than my first. My second digit is half
last digit. My third digit is three times my second digit.
my
d.45,54
Step-by-step explanation:
let the two numbers be x and y
x+y=9
y+x=9
let x=9-y
the square of either number minus the product of both digits plus the sqaure of the other digit is expressed as;
[tex]x^{2}-xy+y^{2}[/tex]=21
substituting value of x we have
[tex](9-y)^{2} -(9-y)y+y^{2} =21[/tex]
we have;
81 - 18y + y^2 - 9y + y^2 + y^2 = 21
3y^2 - 27y + 60 = 0, dividing through by 3 we get
y^2 - 9y + 20 = 0 using factorization
(y-5)(y-4) = 0
Two solutions
y = 5, then x = 4
y = 4, then x = 5
:
The two numbers would be 54, and 45
4774?
i think
Step-by-step explanation:
You are number 34
Step-by-step explanation:
* Lets explain how to solve the problem
- The number is less than 100
∴ The number is 2-digit number
- The number is the product of 2 prime numbers
∴ The two prime numbers could be two of these numbers:
2 , 3 , 5 , 7 , 11 , 13 , 17 , 19 , 23 , 29 , 31 , 37 , 41 , 43
∵ The number is less than 100
∴ The two prime numbers are 1-digit numbers OR one of them
is 1-digit and the other is 2-digit
- The sum of its digits is a 1-digit
∴ The sum of its digits is less than 10
- One of its digit is a square number
∴ One of its digit is 1 or 4 or 9 because they are square numbers
* Look to the prime number above the number and chose two of
them try to find the answer
- Start with the product of 1-digit prime numbers
∵ 2 and 7 are prime numbers
∵ 2 × 7 = 14 and the sum of its digits is less than 10 (1 + 4 = 5)
∵ 14 < 100
∵ The reversed of its digits obtained 41 which is a prime number
∵ 4 and 1 are a square number (1² = 1 or 2² = 4)
∴ The number is not 14
- Chose one of them 1-digit and the other 2-digit
∵ 2 and 17 are prime numbers
∵ 2 × 17 = 34 and the sum of its digits is less than 10 (3 + 4 = 7)
∵ 34 < 100
∵ The reversed of its digits obtained 43 which is a prime number
∵ 4 is a square number (2² = 4)
∴ The number is 34
* You are number 34
479
Step-by-step explanation:
The number you want is between 100 and 999At least two of the digits are perfect squares 1 4 9 are the perfect squares.Units > tens; units are 2 to 9 ; tens are 1 to 81 even in the result; the other two are in 1 3 5 7 or 9The units digit is larger than the hundreds digit. The units digit is odd.t = unit - 2. the two odd digits are the tens and the units.t - h= 3one of the digits is 7
The units digit cannot be 1 That would make the 10s digit - 1. The tens digit cannot be 9. That would make the units digits 11.
Let's try x79 If that is correct, the hundreds digit is 4 because the hundreds digit is three less than the tens.
479
479 is between 100 and 999. 4 and 9 are perfect squares. u>t. 4 is the only even number.
(d.) 45, 54
Step-by-step explanation:
Let the first digit = y
Let the second digit = z
y +z = 9 (1)
y²- yz +z² = 21(2)
From equation (2),
z = 9-y(3)
Substitute equation (3) into (2):
y²- y(9-y) +(9-y)² = 21
y²-9y+y²+y²-18y+81 = 21
3y²-27y+ 81 = 21
3y²-27y+ 81-21= 0
3y²-27y+ 60= 0
y²- 9y +20= 0
(y -5) (y-4) =0
y= 5 or y =4
z = 4 or 5 (substituting into (3))
So the numbers are 54 or 45.
8493
Step-by-step explanation:
a pues no lo se wey po no ma
Step-by-step explanation:buesca en braily.com
Let x be the first digit; let y be the second digit."the sum of both digits, of either of two two-digit numbers, in whatever order the digits are written, is 9."[tex]x+y = 9[/tex]"the square of either of the digits of either number, minus the product of both digits, plus the square of the other digit is the number 21"[tex]x^2 - xy + y^2 = 21[/tex]this is a system of equations: [tex]\displaystyle\left \{ \begin{aligned} x+y & = 9 \\ x^2 - xy + y^2 & = 21 \end{array} \right.[/tex]can do [tex]x+y = 9 \leftrightarrow y = 9 - x[/tex]. substituting 9 - y for x in the other equation: [tex]\begin{aligned} x^2 - x(9-x) + (9-x)^2 & = 21 \\ x^2 - 9x + x^2 + (9-x)^2 & = 21 \\ 2x^2 - 9x + \left(81-18x-x^2\right) & = 21 \\ 3x^2 - 27x + 81 & = 21 \\ 3x^2 - 27x + 60 & =0 \\ 3(x^2 - 9x + 20) & =0\\ 3(x-5)(x-4) & =0\\ x& =4,\, 5 \end{aligned}[/tex]using x+y = 9, if x = 4 then y = 5. if x = 5 then y = 4. so the numbers are 4 and 5.