Five times the sum of the digits of a two-digit number is 13 less than the original number. If you reverse the digits in the two
-digit number, four times the sum of its two digits is 21 less than the reversed two-digit number. (Hint: You can use variables to represent the digits of a number. If a two-digit number has the digit x in tens place and y in one’s place, the number will be 10x + y. Reversing the order of the digits will change their place value and the reversed number will 10y + x.) The difference of the original two-digit number and the number with reversed digits is .
Let the original 2-digit number be xy. Because 5 times the sum of the digits is 13 less than the number, therefore 5(x + y) = 10x + y - 13 5x + 5y = 10x + y - 13 -5x + 4y = - 13 (1)
The number with reversed digits is yx. Because 4 times the sum of the digits is 21 less than the reversed 2-digit number, therefore 4(x + y) = 10y + x - 21 4x + 4y = 10y + x - 21 3x - 6y = -21 x - 2y = -7 x = 2y - 7 (2)
Answers: The original 2-digit number is 98 The reversed 2-digit number is 89 The difference between the original and the reversed 2-digit numbers is 98 - 89 = 9
