Determine all pairs (m, n) of a two-digit natural number m and a single-digit natural number n satisfying the following conditions.
1) If the number n is given as the number between the two digits of the number m, then a three-digit number is obtained which is 11 times the number m.
2) If the number n is used as a number before the decimal number of the number m, then a three-digit number is obtained that is 21 times as large as the number n.
If I understand this correctly, then if the digits of m are a and b, we have
m = 10a+b
and so
100a+10n+b = 11(10a+b)
100n+10a+b = 21n
But this comes out with negative solutions. Maybe you can fix my logic and take it where it's supposed to go.
To determine all pairs (m, n) satisfying the given conditions, we can start by breaking down the conditions and finding possible values for m and n.
Condition 1:
"If the number n is given as the number between the two digits of the number m, then a three-digit number is obtained which is 11 times the number m."
Let's represent m as x and n as y. According to the condition, if n is placed between the two digits of m, we obtain a three-digit number 11 times m. Therefore, we can write this as an equation:
(100x + 10y + x) = 11x
Simplifying the equation, we get:
110x + 10y = 11x
Subtracting 11x from both sides gives us:
99x + 10y = 0
This equation tells us that 99x is a multiple of 10, which means x is a multiple of 10. Since we are looking for two-digit natural numbers, x can be 10 or 20.
Condition 2:
"If the number n is used as a number before the decimal number of the number m, then a three-digit number is obtained that is 21 times as large as the number n."
Using the same variables x and y, we can set up the equation for condition 2:
(10y + x) = 21y
Simplifying this equation, we get:
x = 20y
This equation tells us that x is always 20 times y. Since y is a single-digit natural number, it can be 1, 2, 3, 4, 5, 6, 7, 8, or 9.
Now that we have the possible values for x and y, we need to check if they satisfy both conditions simultaneously.
For x = 10:
From condition 2: y = 1/2
This is not a valid solution since y should be a natural number.
For x = 20:
From condition 2: y = 1
From condition 1: (100*20 + 10*1 + 20) = 11*20
So, the pair (m, n) = (20, 1) satisfies both conditions.
Therefore, the only pair that satisfies both conditions is (m, n) = (20, 1).