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Six times the middle digit of a three digit number is the sum of the other two. if the number is divided by the sum of its digit, the answer is 51 and the remainder is 11. If the digits are reversed, the number becomes 198 lesser. Find the number?

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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