Answer:
61
Explanation:
Reversing the digits of a 2-digit number changes its value by a multiple of 9. The multiple is the difference between the digits. Here, the value is changed by 45 = 5·9, so the difference between the two digits is 5. We are told their sum is 7, so the digits must be 1 and 6. The larger digit is in the tens place to make this problem work out right.
The original integer is 61.
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Solution using a single variable
Let t represent the original digit that is in the 10s place. Then 7-t is the units digit and the value of the original number is ...
10t +(7-t) = 9t +7
The value of the number with the digits reversed is ...
10(7-t) +t = 70-9t
The difference between these two numbers is said to be 45, so we have ...
(9t +7) -(70 -9t) = 45
18t -63 = 45
18t = 108
108/18 = t = 6
The tens digit is 6, so the units digit is 7-6=1. The original number is 61.
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Solution using two variables
Let t and u represent the tens and units digits of the number respectively. Then we have ...
t + u = 7
(10t+u) -(10u+t) = 45
The second equation simplifies to ...
9t -9u = 45
9(t -u) = 45 . . . . . . the relationship between the digits described above
t -u = 5 . . . . . . . . . divide by 9
Now, we can add this equation to the one expressing the sum of digits:
(t +u) +(t -u) = (7) +(5)
2t = 12
t = 6
u = 7-t = 1 . . . . . from the first equation (t+u=7)
The original number is 61.