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?

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 agree with 47921 plus 4792

Have not given complete solution

the answer is 23 confirmed.

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.

I think it must be 47921

To find the sum of the digits of the original 5-digit number, we need to figure out the original number itself. Let's break down the problem step by step:

1. Let's assume the original 5-digit number is abcde, where a, b, c, d, and e represent individual digits.

2. When John deletes one of the digits, we need to determine all possible 4-digit numbers that could have been created. To do this, we can consider removing each digit (a, b, c, d, and e) one at a time and calculating the sum for each scenario.

3. Let's consider the case where we remove digit 'a'. In this case, the resulting 4-digit number would be bcde. The sum of the original and 4-digit numbers would be abcde + bcde, which can be written as 10000a + 1000b + 100c + 10d + e + 1000b + 100c + 10d + e = 2000a + 200b + 200c + 20d + 2e.

4. Similarly, for the other digits, we would get the following sums:
- Removing 'b': 1000a + 100c + 10d + e + 1000c + 100d + 10e = 2000a + 200c + 110d + 11e.
- Removing 'c': 1000a + 100b + 10d + e + 1000b + 100d + 10e = 2000a + 200b + 110d + 11e.
- Removing 'd': 1000a + 100b + 100c + e + 1000b + 100c + 10e = 2000a + 200b + 200c + 11e.
- Removing 'e': 1000a + 100b + 100c + 10d + 1000b + 100c + 10d = 2000a + 200b + 200c + 20d.

5. Combining these equations with the given sum of 52713, we have the equation:
2000a + 200b + 200c + 20d + 2e + 2000a + 200c + 110d + 11e + 2000a + 200b + 110d + 11e + 2000a + 200b + 200c + 11e + 2000a + 200b + 200c + 20d = 52713.

6. Simplifying the equation gives:
8000a + 800b + 800c + 42d + 325e = 52713.

7. Now, we can start finding the values of a, b, c, d, and e by trying different combinations.
We know that a < 10, b < 10, c < 10, d < 10, and e < 10. So we can iterate over these values and check which combination satisfies the equation above.

8. Once we have the values of a, b, c, d, and e, we can calculate the sum of the digits of the original 5-digit number by adding them together: a + b + c + d + e.

Following this process, we can find the sum of the digits of the original 5-digit number.

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