Find the two digit no.?

Question

Find the two digit no. the addation of the two digit is 16. the difference between the no and the no interchanging the original digit place is 18.

Answer 1

Let the two digit number be represented by the digits a and b.

So the number is:
10a + b

Interchanging the digits:
10b + a

Subtracting the two you get:
10a + b - 10b - a = +/-18
9a - 9b = +/-18
9 (a - b) = +/-18
a - b = +/-2

Also given:
a + b = 16

now
a + b = 16 & a - b = 2 solving we get, a = 9 & b =7
again
a + b = 16 & a - b = -2 solving we get, a = 7 & b =9
so the no is 97 or 79

Answer 2

The sum of the two digits is 16; so they can be both 8, or 9 and 7. This is because digits can only be from 0 to 9. but the number is not 88, because when you interchange the digits, you get 88 again, and the difference between the two is 0. And easily, 97 -79 = 18, which satisfies the second condition.

The answer is 97 or 79.

In fact, the difference between a two digit number, and a number formed by interchanging its digits, is always a multiple of 9; further, it is equal to 9 times the difference between the two digits!

Examples:

44 - 44 = 0 = 0 x 9;
54 - 45 = 9 = 1 x 9;
97 - 79 = 18 = 2 x 9;
73 - 37 = 36 = 4 x 9;

So, to solve the original problem, the difference between the digits must be 2, and sum is known to be16, they have to be 7 and 9!

Answer 3

Let the two digit number be represented by the digits a and b.

So the number is:
10a + b

Interchanging the digits:
10b + a

Subtracting the two you get:
10a + b - 10b - a = 18
9a - 9b = 18
9 (a - b) = 18
a - b = 2

Also given:
a + b = 16

Add these two together:
a - b + a + b = 18
2a = 18

So:
a = 9
b = 7

*79 could technically be an answer except I interpret "the difference between the no and the interchanged number" to mean the first number is bigger. 79 - 97 would be -18, not 18.

The sum of the digits is 16 and the difference between 97 and 79 is 18.

The number is 97.

Answer 4

Let digits be x and y
x+y = 16 (given) ...1
the number = 10x+y
revresing the digits number = 10y+x
Let x > y

so (10x+y)-(10y+x) = 18
or 9x - 9y = 18
so x-y = 2.... 2
add 1 and 2 2x = 18 or x = 9
so y = 7

so number was 97.
as difference is mentioned so it could be 79
so 97 or 79.
if 97 it becomes 79 smaller by 18 , if 79 then becomes 97 greater by 18

Answer 5

the no is 97, sum of whose digits is 16 and diff between the no and the no interchanging the original digit place is 18.
i.e. 97-79=18

Answer 6

Let x = the tens digit
16 - x = the ones digit

We have:
10x + (16 - x) = the original number
10(16 - x) + x = the number with the digits reversed

[10x + (16 - x)] - [10(16 - x) + x] = 18
9x + 16 - 160 + 9x = 18
18x = 162
x = 9 (tens digit)
16 - x = 7 (ones digit)

Thus, the original number is 97.

^_^

Answer 7

97
79

Answer 8

97 or 79

Answer 9

97 or 79

Answer 10

let the once digit be x and tens digit be y
(x) + (y) = 16 --------- (1)
18 = 10(y) + (x) - 10(x) - (y)
therefore, 9(y)- 9(x) = 18
(y) - (x) = 2 --------(2)
from 1 & 2
(y) = 9
(x) = 7

hence answer is 79 or 97