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Word problems on reversing digits of numbers
Problem 1Sum of the digits of a two digit number is 10. When we interchange the digits, it is found that the resulting number is greater than the original number by 36.
Find the number.
Solution
Let a and b are the digits of your number n, so that n = 10a + b.
The number after interchanging digits is 10b + a.
Then you have the system of two equations for the unknowns a and b:
.
Now, simplify it:
.
Simplify it one more time:
.
Solve it.
The solution is b=7, a=3.
Hence, the number is 10a + b = 10*3+7 = 37.
Answer. 37.
Problem 2When a two-digit number is divided by the product of the two digits, the quotient is 2.
If 27 is added to the number, the original number turns into a new number with the digits being swapped around. Find the number.
Solution
Let a and b be the first (the left) and the second (the right) digit of the number respectively, so that the number is 10a + b.
The first condition says that = , or
10a + b = 2ab. (1)
The number with swapped digits is 10b + a, so the second condition says that
10a + b + 27 = 10b + a, or 9a - 9b = -27, or a-b = -3.
So, we have a system of two equations
.
Express b from the last equation, b = a + 3, and substitute it into the previous equation. You will get
10a + a + 3 = 2a*(a+3), or
- - = .
Solve this quadratic equation using the quadratic formula. You will get the roots = and = .
The negative root doesn't suit, because we are looking for the digit, which should be non-negative integer less than 10. The root = 3 is good.
So, the number is 36.
Please check yourself that this number satisfies all conditions of the problem.
Answer. 36.
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