Answer:
87
Step-by-step explanation:
You want the 2-digit number that is 9 more than its value when the digits are reversed, and that has a sum of digits of 15.
Let the tens digit be represented by t. Then the ones digit is (15 -t) and the value of the number is 10t +(15-t) = 9t+15.
When the digits are reversed, the value of the number is 10(15-t) +t.
The first value is 9 more than the last value:
9t +15 = 9 +(10(15 -t) +t)
9t +15 = 9 +150 -10t +t
18t = 144
t = 8
15-t = 7
The number is 87.
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Additional comment
Reversing the digits of a 2-digit number always makes a number that differs by a multiple of 9. That multiplier is the difference between the digits. Knowing this, you know the difference of the digits is 9/9 = 1.
If the sum of digits is 15 and the difference of digits is 1, the two digits are (15±1)/2 = {8, 7}. We want the larger number, so that will be 87, not 78.
Answer:
87
Step-by-step explanation:
Let the original number have the form ab.
a is the tens digit, and b is the ones digit.
The value of the number is 10a + b.
Now reverse the digits.
The new number has the form ba.
The b is the tens digit, and the a is the ones digit.
The value of the send number is 10b + a.
"The sum of two digits of a certain two-digit number is 15."
a + b = 15 Eq. 1
"If the digits are reversed, the number formed is nine less than the original number."
(10a + b) - (10b + a) = 9
10a + b - 10b - a = 9
9a - 9b = 9
a - b = 1 Eq. 2
Solve Eq. 1 and Eq. 2 as simultaneous equations.
a + b = 15
a - b = 1
Add the equations to eliminate b.
2a = 16
a = 8
a + b = 15
8 + b = 15
b = 7
Answer: 87
