How many two-digit positive integers

Question

How many two-digit positive integers are increased by 18 when their digits are reversed?

(A) 3

(B) 4

(C) 5

(D) 6

(E) 7

sajjad1995 · Accepted Answer

Let the initial number be xy or 10x+y.
When the digits are reversed, the number becomes yx or 10y+x

Thus, 10x+y+18=10y+x
   y=x+2

So all numbers where tens digit is 2 less than units digit => 13, 24, 35, 46, 57, 68, 79.
Total -  7

Answer is option-  (E) 7

Mamun34 · Answer

Is there anyother simpl mathmatical technique to find out the amount of integer???

Posted from my mobile device

sajjad1995 · Answer

This is the simple way to solve,
But we can solve it directly.

In this type of questions , the difference will be multiple of 9 and sum will be multiple of 11.
Divide given difference or sum by 9 or 11 respectively.
The answer will be the difference between two digits of the number.

In this case 18/9 = 2 , 2 is the difference between two digits of the number.

Charli08 · Answer

Core concept  here is that any two digit (applies for any amount of digits) can be expressed as follows:

10a + b

Where:  a = tens digit s and  b = units digits

So the question says adding 18 to our number equals our number reversed, lets write that out:

10a + b +18 = 10b + a (** notice our number flipped is now 10b + a since the unit digit becomes the tens digit and vice versa!)

18 = 9b - 9a
18 = 9 (b-a)
 2 = b-a

This means our options are the combination of 2 digits which are 2 apart! Our options thus are:

9 & 7
8 & 6
.
.
.
3 & 1

7 combinations

==>  E

ZIX · Answer

Let the number be  10x + y .

10x + y + 18 = 10y + x

9x + 18 = 9y 
​​​​
 x + 2 = y

And we know that since x is a tens digit, therefore it can take values from 1-9 and since y is a units digit, it can take values from 0-9. So only values  x = 1  to  x = 7  would satisfy this equation.

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How many two-digit positive integers

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How many two-digit positive integers

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