Digit Flip!

Let the tenth and unit digits of the two-digit integer be $a$ and $b$ respectively. Then the integer $n = 10a + b$ and that:

$\begin{aligned} \left(1 + \frac{1}{5}\right)(10a+b) & = 10b + a \\ \frac{6}{5}(10a+b) & = 10b+a \\ 12a + \frac{6}{5} b & = 10b + a \\ 11a & = \frac{44}{5} b \\ \Rightarrow a & = \frac{4}{5}b \end{aligned}$

For $a$ to be an integer, $b$ must be divisible by $5$ and as $b$ can only takes on 0 through 9, the only case is $b=5 \implies a = 4$ and $n = \boxed{45}$ .

1. less than 100, hence, either 2 digit or single digit number.
2. digits interchange, hence, minimum 2 digits number required.
3. value increases when digits interchanged, hence, unit digit is greater than tenth's digit.
4. let, the number is 10p+q, where p<q.
5. A/Q, (10q+p) - (10p+q) = (10p+q)/5 => p/q = 4/5 = 4k/5k (say), which is a ratio, now since for k>1 e.g. for k=2, q=5X2=10, which exceeds single digit limit, hence q must be equals 5. so, the required number is 45.

let a two digit whole number= 10 X+Y then according to questions, (10 X+Y) * 1.2 = 10Y+X => 11X= 8.8Y => X/Y = 4/5 So a two digit number is 10*4+5= 45. Answer is 45

If x is original number then reverse will be of value (6 x)/5 . to have integer solution x should be divisible by 5 . then if x is of form (b 10+a). Since a can be be 0 or 5 clearly if a is 0 reversing will decrease the number so a is 5 then for b is less than 5 by so by trial we see only 4 satisfies our above equations.

The original two digit number has to end in 5,as otherwise, 20% increase will not yield a whole number. Then one has to check 15,25,35,45 and need not go beyond that.

Its simple,if we flip any two digit number their difference is always a number divisible by 9. Hence just go for a two digit number whose 20% is 9. That is: 20% of x=9 solving for x we will get 45.

We know the number has to be a multiple of 5. 5, 15, 25, 35, 45. It follows that 45 works.

Let 10x+y → x+10y Then 6/5 * (10x+y)=x+10y >x=4/5 *y x, y are two integers between 0~9 Take x=4 y=5

Make a equation for it.
It is a two digit number so let the digits be A and B
6/5(10a+b) =10b+a.
12a+6/5b=10b+a.
12a-a=10b-6/5b.
11a=44b/5.
55a=44b.
5a=4b.
Now check a= 4 or 5 You will get the answer 45

It is 10 x+y = 1,2 (10*y+x), which we can convert to x/y =1,25. Since 4 * 0,25=1 and x,y must be one digit integers, we can choose y=4 and x=5 to solve the equation. Therefore, the answer is 45.

The method I used is reliant on logic:

When you flip digits, the difference of the numbers are determined by the difference of the digits that make up the number. If the two digits are 1 away from each other, the difference is 9 and if the two digits are 2 away from each other the difference is 18 (it grows by a multiple of 9). However, this question asks for a value within the range of two digits while having a difference equal to 20% greater than the previous value.

(examples to help explain) 32-23=9 (difference of digits are 1) 42-24=18 (difference of digits are 2) 96-69=27 (difference of digits are 3)

We see that a difference of the digits by the value of 2 doesn't work because the difference between the numbers created when swapped is 18. Which is too far (shown below)

x + 18= y 18= x/5 x= 18 x 5 x= 90 y= 108

108 is a number greater than the 2 digit. So you can deduce that the only possible answers are within 1 digit of each other.

If the digits that make up the number are 1 away from each other, then the difference then the difference between the numbers generated is 9.

x + 9 = y 9= x/5 x= 45 y= 54

YAY! It's within 2 digits and it works!

Lets have a number ab which has a and b as digits. When you open in up it has the value of 10a+b, when you flip the digits it becomes ba which is equal to 10b+a. When you take the ratio of ab as 100 percent and ba as 120 percent it becomes something like this. (10a+b)/120=(10b+a)/100 which then becomes 44a=55b. A and b have to be digits so they must be 5 and 4. The numbers are 45 and 54

The difference is always a multiple of 9 . So (9+x)/x =1.2 > 0.2x=9 so X=45 , I chose n=1 because that it is the only multiple which will result a 2 whole digit numbers. For n=2 for example it will result a number which is less than the original number and also only 1 digit whole number which is 9 and which come from X=90

Let "ab" be the number, if the number resulting from switching digits has to be greater than original number, b>a.. take numbers 01-10, 12-21, 23-32 for example.. You can observe that the difference in each case is "9".. This 9 has to be "20%", then 100%= 45, that's how you arrive at the answer..

Let 10x + y be the two digit integer, where x,y < 10. Thus 10y + x is when the digits in the integer are swapped.

This means that 10y + x = (10x + y) + (10x + y)/5, so simplifying it gives us 10y + x = 6(10x + y)/5.

Because the number is an integer, y must be divisible by 5. Therefore, y = 0 or 5.

If y=0, then 10y + x = 6(10x + y)/5 becomes x = 6(10x)/5, then 5x = 60x, so x = 0. This cannot be the answer as 0 is not a two digit integer.

Therefore y = 5, and substituting this value into the equation gives us 10(5) + x = 6(10x + (5))/5. Simplifying this gives us 50 + x = 12x + 6 Then, 44 = 11x, thus x = 4.

So, the two digit integer is 10(4) + 5 = 45.

(1+1/5)(10x+y)=(10y+x) => 6(10x+y)=5(10y+x) => 60x+6y=50y+5x => 55x=44y => x/y=4/5 so the integer is 10*4+5=45

First digit must be smaller than second digit so that interchanged number is greater than initial number. Since it is an integer, one fifth of the number has to be a whole number (the initial number must end in 5). Lastly, the initial number must be less than 100. That leaves us with 05, 15, 25, 35, and 45. Check all of those and see which one satisfies the conditions.

Not sure about that guys above me but these were my conclusions.The numbers tenth place has to be less than that of the ones place to increase by any means, also it must increase by a positive integer.The number reversed MUST be divisible by 6 so I use these criteria and decide that the highest number possible would be 69(96) and then work down from there using only numbers divisible by 3 until I met 45(54).Not algebra but it worked ;).

When the numbers of a two digit integer (made of two consecutive numbers) are interchanged, the difference between the two integers is always 9. Now if this change is 1/5th of the integer then the solution is 9x5=45. Simple 😊

My solution: We are naming the initial two-digit integer A, to which we are adding A/5 to get a number B, a number that can also be formed by interchanging A's digits.

You might already know that applying mod 9 to an integer is equivalent to adding its digits, then adding the digits of the result and so on until there's only one digit left.

So, if A (mod 9)= x B (mod 9)= x, because both numbers have the same digits, and so, B-A (mod 9)= x-x= 0. So, B-A= A/5= 9n. Then, A= 45n. As A < 100, n can only be 1 or 2. If n=1, A=45, B=54, the only solution as, n=2, A=90, B=09, doesn't work.

Let the number be $10x+y$ .According to the condition: $10x+y+\frac{1}{5}(10x+y)=10y+x\\10x+y+2x+\frac{y}{5}=10y+x\\12x+\frac{6}{5}\times y=10y+x\\12x-x=10y-\frac{6}{5}y\\11x=\frac{44}{5}y\\x=\frac{4}{5}y$ In order for $x$ to be an integer, $y$ must be a multiple of 5.The only numbers from 0 to 9 which are factors of 5 are 0 and 5 itself.Putting $y=0$ gives $x=y=0$ ,but since it's given that $10x+y$ is a two digit number,so that case is ruled out.Hence,we are left with $y=5$ which gives $x=4$ and $10x+y=\boxed{45}$

Let number b (10a + b) then interchanged number = (10b + a)

(10b + a) – (10a + b) = (10a + b)/5

solving 45(b – a) = 10a + b

55a = 44 b or 5a = 4b

then a = 4 and b = 5 and number is 45

I simply did 10x+y +1/5 (10x+y)=10y+x Then solved to get x/y=4/5......so 10x+y=4×10+5=45

XY +XY/5=YX 6XY/5=YX number should be multiple of 5 and Y-X=1 Satisfying above condition the final answer going to be 45

Let the original no be 10X+Y and the reversed one 10Y+X.

Now acc to Ques:

1). (10Y+X) - (10X+Y) = (10X+Y)/5.

2). 9Y - 9X = 2X + Y/5.
3). 4Y = 55X.
4). Y:X = 5:4.
5). Or Y=5, X=4.
6). The original no that was (10X+Y) = (10×4+5) = 45.

Let $\overline{AB}$ be the two digit number. This implies that

$\overline{AB} \cdot \frac{6}{5} = \overline{BA}$

$(10A + B)\cdot \frac{6}{5} = 10B + A \implies 5A = 4B \implies A = 4, B = 5$

Therefore, $\boxed{45}$ is the number.

I chose x and y such that 10x + y is an integer with x and y between 0 and 10. Thus, rearranging the digits gives you 10y + x which we want to be greater than the original integer by one fifth. So I set up the equation 1.2 (10x + y) = 10y + x which yields y = 5x/4. Since I was looking for the fifth of the original to add to the original my x and y would have to be added to get the larger number. Knowing that we are working with inregers, I set x = 4 and got y = 5 because they have to be between 0 and 10 and because these are the only integers that fit the conditions. Therefore, adding these integers I got 54. Lastly I check my work (1/5) 45 + 45 = 54. QED.