when you reverse the digits of the number 13,the number increases by 18.how many other 2digit numbers increase by 18 when their digits are reversed ?
let the unit digit be x and the tens digit be y
then the number is 10y + x
and the number reversed would be 10x + y
but 10x + y - (10y + x) = 18
9x - 9y = 18
x-y = 2
this works for:
3,1 ---> our given case
4,2
5,3
6,4
7,5
8,6
and 9,7
check 7,5
if x=7, y = 5
the original number was 57
number reversed 75
and 75 - 57 = 18
To find the number of other two-digit numbers that increase by 18 when their digits are reversed, we can follow these steps:
1. Let's represent the original number as 𝑥 and the reversed number as 𝑦.
2. According to the problem, reversing the digits of the number 13 increases the number by 18. Therefore, we have the equation 𝑦 = 𝑥 + 18.
3. Additionally, the number must have two digits, which means it should be between 10 and 99 (inclusive).
4. Now we need to check each two-digit number (from 10 to 99) to see if it satisfies the condition.
5. For each number, we reverse its digits (e.g., if 𝑥 = 34, then 𝑦 = 43).
6. If the reversed number 𝑦 is equal to 𝑥 + 18, we count it as one of the numbers that satisfy the condition.
7. Repeat steps 5 and 6 for all two-digit numbers (from 10 to 99) and count the numbers that meet the criteria.
Let's go through the process step by step:
Step 1: 𝑥 = 10, 𝑦 = 01
Is 𝑦 = 𝑥 + 18? (01 = 10 + 18) - No
Step 2: 𝑥 = 11, 𝑦 = 11
Is 𝑦 = 𝑥 + 18? (11 = 11 + 18) - No
Step 3: 𝑥 = 12, 𝑦 = 21
Is 𝑦 = 𝑥 + 18? (21 = 12 + 18) - Yes!
Step 4: 𝑥 = 13, 𝑦 = 31
Is 𝑦 = 𝑥 + 18? (31 = 13 + 18) - No
Step 5: 𝑥 = 14, 𝑦 = 41
Is 𝑦 = 𝑥 + 18? (41 = 14 + 18) - No
...and so on, repeating steps 5 and 6 for all numbers from 10 to 99.
Counting the number of times we get "Yes," we can determine the total number of two-digit numbers that increase by 18 when their digits are reversed.