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A number consisting two digit is four times the sum of its digits and if 27 be added to it the digits are reversed?
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A number consisting two digit is four times the sum of its digits and ...
Given Information:
- Let the two-digit number be represented by 10a + b, where a is the digit in the tens place and b is the digit in the units place.
- The number is four times the sum of its digits: 10a + b = 4(a + b)
- When 27 is added to the number, the digits are reversed: 10a + b + 27 = 10b + a

Equations to solve:
1. 10a + b = 4(a + b)
2. 10a + b + 27 = 10b + a

Solving the equations:
- From equation 1: 10a + b = 4a + 4b
- Rearranging terms: 6a = 3b
- Dividing by 3: 2a = b
- Substituting 2a for b in equation 2:
10a + 2a + 27 = 10(2a) + a
12a + 27 = 20a + a
27 = 19a
a = 27 / 19
a = 1.42

Conclusion:
- Since a should be a whole number, the assumption that the number consists of two digits is incorrect. The number must actually be a single-digit number.
- Therefore, the initial assumption was flawed, and the problem cannot be solved with the given conditions for a two-digit number.
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A number consisting two digit is four times the sum of its digits and if 27 be added to it the digits are reversed?
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