This is a popular type of problem that appeared in mathematics textbooks in the 1970s and 1980s. can you find the answer? the su
m of the digits of a two-digit number is 6. if the digits are reversed, the difference between the new number and the original number is 18. find the original number.
A digit is a number in one of the places, so for example the number 54 has two digits; a tens place digit (5) and a ones place digit (4).
Say the mystery number is a two digit number = xy * that's not x times y but two side by side digits.
Info given: <span>the sum of the digits of a two-digit number is 6 x + y = 6 </span> <span>if the digits are reversed, yx the difference between the new number and the original number is 18.
**To obtain the number from digits you must multiply by the place and add the digits up. (Example: 54 = 10(5) + 1(4))
Original number = 10x + y Reversed/New number = 10y + x
Difference: 10y + x - (10x + y) = 18 9y - 9x = 18 9(y - x) = 18 y - x = 18/9 y - x = 2
Now we have two equations in two variables </span>y - x = 2 <span>x + y = 6
Re-write one in terms of one variable for substitution. y = 2 + x sub in to the other equation to combine them. x + (2 + x) = 6 2x + 2 = 6 2x = 6 - 2 2x = 4 x = 2
That's the tens digit for the original number. Plug this value into either of the equations to obtain y, the ones digit.
