Question

Please help me! The sum of the digits of a two-digit numeral is 11. If the digits were reversed, the resulting number would be 45 greater than the original number. What is the original number?

Answer 1

\(ab\) could be your number like \(23\)
then you know that \(a+b=11\) and also that \(10b+a=10a+b+45\)

Answer 2

would u like me to work that?

Answer 3

let consider the problem in simplest way
if sum of 2 number is 11, then we have a couple of pairs: (2,9),(3,8),(4,7), (5,6)
and if we write it back ward, the new one is bigger than the old one. Therefore, in the original one, the first number is less than the second one. I mean with (2,9) you have 29, not 92
so, let see if we write back ward, which couple satisfy the condition that the new one is 45 unit bigger than the original one
(2,9) : old one 29, new one 92 and 92-29 = 63 \(\neq 45\) so, (2,9) is not the answer
(3,8): 83-38= 45 Bingo, I am so lucky to get it after 2 attempts.
so, the number you need to find out is 38
which 3+8 =11
write backward 83 and 83 -38 =45

Answer 4

to satellite73 method. It is the general method
Let me "translate" it to you.
2digit numeral is some number like 23, 34, 45, 88...
pick one of them to say. 23 means 20 +3 = 2*10 +3
Let say your required number is ab (like 23) , so a is a*10 +b
backward is ba , so to this backward, ba = b*10+a
this backward number (ba) = original number (ab) +45