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In two digits number sum of digits 8, reverse number 18 less than the original number, find the original number?

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Answer 1

Step-by-step explanation:

The sum of the digits of a two digit number is 8 and the difference between the number and that formed by reversing the digits is 18. Can you find the number?

Using the J programming language:

Generate all the two-digit integers (10+i.90) and store them in a (a=.). Separate the digits in each integer (10 10#:) & store the separated pairs in s (s=.), Reverse the order of each pair (|.”1 s), and convert the reversed pair into an integer (10#.). Subtract the reversed-digit integers from the original integers in a (a-10#.) take the absolute value of the result of the subtraction (|), and mark all the locations where the result of the subtraction equals 18 (18=). AND that first mark vector (*.) with a second mark vector created by adding the digit pairs i

The number is 10a+ b

a+b=8

10a+b +18 = 10b + a

9a-9b + 18 = 0

a-b = -2

2a = 6

a = 3

b= 5

So the number is 35

Answer 2

Answer:

Let's set up a system with two variables: x = tens place of our answer, y = units place of our answer.

Digit sum of a two digit number is 6:

x+y=6

Reverse the digits and you get 18 less than the original value:

10y+x=10x+y-18

Now let's solve:

y=6-x

10(6-x)+x=10x+(6-x)-18

60-10x+x=10x+(6-x)-18

60-9x=9x-12

72=18x

x=4

y=6-4

y=2

So our original number was 42. Sum of digits is 6. Swap the order of the digits to make 24, and you have a number 18 less than the original number.