# 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 halflast

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​

## This Post Has 8 Comments

1. jetblackcap says:

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;

$x^{2}-xy+y^{2}$=21

substituting value of x we have

$(9-y)^{2} -(9-y)y+y^{2} =21$

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

2. morbidodyssey says:

4774?

i think

Step-by-step explanation:

3. haleybug6 says:

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

∵ 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

4. costel8532 says:

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.

5. lexysmith1722 says:

(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.

6. ionutvasiliev4237 says:

8493

Step-by-step explanation:

7. dakshshberry says:

a pues no lo se wey po no ma

Step-by-step explanation:buesca en braily.com

8. elliekuprisek2122 says:

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."$x+y = 9$"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"$x^2 - xy + y^2 = 21$this is a system of equations: \displaystyle\left \{ \begin{aligned} x+y & = 9 \\ x^2 - xy + y^2 & = 21 \end{array} \right.can do  $x+y = 9 \leftrightarrow y = 9 - x$. substituting 9 - y for x in the other equation: \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}using x+y = 9, if x = 4 then y = 5. if x = 5 then y = 4. so the numbers are 4 and 5.