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the sum of the digits of a certain two-digit number is7. reversing its digits increase the number by 9. what is the number?

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Answer: the number is 12.

Explanation:

Let's call the two-digit number "x". We know that the sum of its digits is 7, so we can represent the number as 10a + b, where a and b are the digits of the number and a is the tens digit and b is the units digit.

The statement that "reversing its digits increases the number by 9" means that the number obtained by reversing the digits is x + 9. We can represent this reversed number as 10b + a.

So, we have two equations:

10a + b = x

10b + a = x + 9

Now, we can substitute the value of x from the first equation into the second equation:

10b + a = 10a + b + 9

Expanding both sides, we get:

10b + a = 10a + b + 9

9b = 9a + 9

b = a + 1

We know that 0 < a < 9, since the number is a two-digit number. So, we can conclude that 0 < b < 10.

We can now substitute the value of b into the first equation:

10a + (a + 1) = x

11a + 1 = x

11a = x - 1

a = (x - 1) / 11

Since a is the tens digit of the number, it must be an integer. We can test different values of x to find the value that satisfies this condition.

If x = 20, then a = (20 - 1) / 11 = 1. This gives us b = a + 1 = 2. The original number is 10a + b = 10 + 2 = 12, which satisfies all the conditions.

So, the number is 12.

User Mike Venzke
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