The sum of the digits of a two-digit positive integer is seven. subtracting 45 from this integer yields another two-digit integer with the same digits, but in reversed order. what is the original integer?

61

Step-by-step 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.

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.

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.

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step-by-step explanation: