John chooses a 5-digit positive integer and deletes one of its digits to make a 4-digit number. The sum of this 4-digit number and the original 5-digit number is 52713. What is the sum of the digits of the original 5-digit number?

9 answers

Solved it!!

Five options possible are:
1. abcde + abcd
2. abcde + abce
3. abcde + abde
4. abcde + acde
5. abcde + bcde
But as last digit of sum is 3 then only (1) is possible. (e + e – cannot be an odd number)
a must be 4 or 5
Then work out using 4 and 5. ‘a’ has to be 4.

47921
+4792 = 52713
So answer is 23 (4+7+9+2+1)

I think it should be 5+7+9+2+1=24

the answer is 23 confirmed.

I agree with 47921 plus 4792

Ha ha.You are cheating in math kangaroo.You searched up the question didn't you?If you did not then you are a good person if you did it is okay.I did that too because I could not solve it.It is really simple.All you have to do is just work it out and your answer is C(23)LoL everyone else explained it so don't worry.

I have a awesome response

At least I think so.

Have not given complete solution

I think it must be 47921