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Sagot :
Answer:
725
Step-by-step explanation:
The three digit number can be represented by
[tex]100x +10 y + z[/tex]
x, y, and z are from the set {0,1,2,...,9}. This is an important domain since it ensures us that the sum 100x+10y+z is three digits. x is our hundreds-digit, y is the tens-digit, and z is the ones-digit.
For example, if x = 1, y = 4, and z = 5, then our three-digit number is
[tex]100x +10 y + z\\100(1) + 10(4)+ 5 = 145[/tex]
Our three digit in this example is 145.
Let us look at the problem and try to find relationships between x, y, and z.
Six times the middle digit of a three digit number is the sum of the other two.
This means that:
[tex]6y=x+z\\\\x-6y+z = 0[/tex]
Equation 1 is [tex]x-6y+z = 0[/tex]
if the number is divided by the sum of its digit, the answer is 51 and the remainder is 11.
When a number is divided and there is a remainder, we can treat the remainder as a fraction, whose numerator is the remainder, and denominator is the divisor.
For example, if we divide 20 by 3 we get:
[tex]20 / 3 = 6 rem. 2\\\\\frac{20}{3}= 6 + \frac{2}{3}[/tex]
Going back to the problem, the number is divided by the sum of its digits, this can be represented as
[tex]\frac{100x+10y+z}{x+y+z}[/tex]
then its results is 51 remainder 11, so the result is
[tex]51 + \frac{11}{x+y+z}[/tex]
These 2 are equal, so
[tex]\frac{100x+10y+z}{x+y+z}= 51 + \frac{11}{x+y+z}[/tex]
We can multiply both sides by [tex](x+y+z)[/tex] to give us
[tex]100x+10y+z = 51(x+y+z)+ 11[/tex]
Simplifying gives us:
[tex]100x+10y+z = 51x+51y+51z + 11\\49x-41y-50z = 11[/tex]
Equation 2 is [tex]49x-41y-50z = 11[/tex].
If the digits are reversed, the number becomes 198 lesser
This means that the hundreds-digit and ones-digit are swapped. Our number becomes
[tex]100z+10y+x[/tex]
Since it becomes 198 lesser, it can be interpreted
[tex]100z+10y+x = 100x+10y+z - 198[/tex]
Simplifying gives us
[tex]100z+10y+x = 100x+10y+z - 198\\99x-99z = 198\\99(x-z) = 99(2)\\x-z= 2\\x = z+2[/tex]
Equation 3 is [tex]x =z+2[/tex].
We have a system of 3 variables.
[tex]x-6y+z = 0\\49x-41y-50z = 11\\x = z+2\\[/tex]
We can substitute the value of x given by the third equation to equation 1 and 2.
[tex]x-6y+z = 0\\x = z+2\\\\z+2 -6y + z = 0\\2z -6y + 2 = 0\\z-3y = -1\\-3y+z=-1\\3y-z=1[/tex]
[tex]x = z+2\\49x-41y-50z = 11\\\\49(z+2) - 41y-50z=11\\49z+98-41y-50z=11\\-41y-z=11-98\\-41y-z=-87\\41y+z = 87\\[/tex]
We now have a system of linear equations.
[tex]\left \{ {3y-z=1} \atop {41y+z = 87}} \right.[/tex]
We can add them together to eliminate z.
[tex]3y-z + 41y+z = 1+87\\44y = 88\\y = 2[/tex]
We now have y = 2. The tens digit of the number is 2.
We can substitute y = 2 back to the system to get z.
[tex]y=2\\3y-z=1\\3(2)-z = 1\\6-z = 1\\6-1=z\\5= z[/tex]
We have z = 5. The ones digit of the number is 5.
We can substitute z = 5 to our earlier systems of 3 variables
[tex]x = z+2\\z = 5\\x = 5+2\\x=7[/tex]
We have x = 7. The hundreds digit of the number is 7.
The number is 725.
If we check, we can verify the relationships.
- Six times the middle digit of a three digit number is the sum of the other two
[tex]6*2 = (7+5)\\12 = 12[/tex]
- if the number is divided by the sum of its digit, the answer is 51 and the remainder is 11
[tex]\frac{725}{7+2+5} = 51+\frac{11}{7+2+5}\\\\\frac{725}{14} = 51+\frac{11}{14}[/tex]
- If the digits are reversed, the number becomes 198 lesser
[tex]527 = 725-198\\527 = 527[/tex]
To know more about these kinds of word problems, click here
https://brainly.ph/question/2464235
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