Problem 24 · 2013 Math Kangaroo
Stretch
Number Theory
Algebra & Patterns
place-valuedigit-sum
Robert chose a five-digit positive number. He removed one of its digits, leaving a four-digit number. The sum of this four-digit number and the original five-digit number is 52713. What is the digit sum of the original five-digit number?
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Answer: C — 23
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Hint 1 of 2
The five-digit number plus the four-digit number is 52713; estimate the five-digit number's size.
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Hint 2 of 2
Removing a digit relates the two numbers through place value — set up and solve.
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Approach: place-value relation between N and the trimmed number
- Let N be the five-digit number and M the four-digit number; N + M = 52713.
- Since M is N with one digit removed, N is a little under 52713, around 47921.
- Solving gives N = 47921 (and M = 4792), whose digits sum to 4+7+9+2+1 = 23.
- So the digit sum of the original number is 23.
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