Respuesta :
If a number multiplied by itself is [tex]6889[/tex], the number is [tex]\sqrt{6889}=83[/tex].
If we didn't know that, consider our two digit number whose square ends in 9. Its ones digits must be 3 or 7, because those are the ones whose squares will end in 9. So we try [tex]83^2[/tex] and [tex]47^2[/tex], and 83 works.
If we didn't know that, consider our two digit number whose square ends in 9. Its ones digits must be 3 or 7, because those are the ones whose squares will end in 9. So we try [tex]83^2[/tex] and [tex]47^2[/tex], and 83 works.
ANSWER
83
EXPLANATION
This method makes use of a system of equations
Let x be the tens-place digit
Let y be the ones-place digit
(therefore, x and y are positive integers between 0 and 9 inclusive as they represent digits)
Then this two digit number is 10x + y
"Multiplying it by itself gives 6889" implies
(10x + y)(10x + y) = 6889
or just
(10x + y)² = 6889
Square root both sides.
Note that √(10x + y)² = 10x + y since x and y are positive and √6889 is 83
10x + y = 83
Now use the other information to make another equation
The sum of its digits is 11 so
x + y = 11
We have a system of equations as x and y must satisfy both equations for the question conditions to be met:
10x + y = 83 ... (I)
x + y = 11 ... (II)
This can be solved using elimination. Multiplying (II) by -1 to aim to get rid of y, we get
10x + y = 83 ... (I)
- x - y = -11 ... (II)
Adding the two equations, we get
(10x - x) + (y - y) = (83 - 11)
which is
9x = 72
Divide both sides by 9 to get
x = 8
This is the tens digit.
Can use x + y = 11 ... (II) to figure out value of y,
x + y = 11
⇒ y = 11 - x
⇒ y = 11 - 8
⇒ y = 3
x = 8 and y = 3
x is the tens-place digit
y is the ones-place digit
The number is 83
83
EXPLANATION
This method makes use of a system of equations
Let x be the tens-place digit
Let y be the ones-place digit
(therefore, x and y are positive integers between 0 and 9 inclusive as they represent digits)
Then this two digit number is 10x + y
"Multiplying it by itself gives 6889" implies
(10x + y)(10x + y) = 6889
or just
(10x + y)² = 6889
Square root both sides.
Note that √(10x + y)² = 10x + y since x and y are positive and √6889 is 83
10x + y = 83
Now use the other information to make another equation
The sum of its digits is 11 so
x + y = 11
We have a system of equations as x and y must satisfy both equations for the question conditions to be met:
10x + y = 83 ... (I)
x + y = 11 ... (II)
This can be solved using elimination. Multiplying (II) by -1 to aim to get rid of y, we get
10x + y = 83 ... (I)
- x - y = -11 ... (II)
Adding the two equations, we get
(10x - x) + (y - y) = (83 - 11)
which is
9x = 72
Divide both sides by 9 to get
x = 8
This is the tens digit.
Can use x + y = 11 ... (II) to figure out value of y,
x + y = 11
⇒ y = 11 - x
⇒ y = 11 - 8
⇒ y = 3
x = 8 and y = 3
x is the tens-place digit
y is the ones-place digit
The number is 83
