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20 Aug 2025, 13:53
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
33% (02:28) correct
67%
(03:01)
wrong
based on 57
sessions
History
Date
Time
Result
Not Attempted Yet
A 3-digit integer N is between 200 and 500. The sum of its digits is 10. When N is doubled, the sum of the digits of 2N is 11. How many such integers N are there?
(A) 12
(B) 15
(C) 18
(D) 21
(E) 24
Explanation (brief):
Let N have digits h, t, u with h ∈ {2,3,4} and t + u = 10 − h. When doubling, digit sums satisfy sum(2N) = 2·sum(N) − 9k, where k is the number of carry operations in N + N. Here 11 = 2·10 − 9k ⇒ k = 1, so exactly one carry occurs.
This happens in two disjoint ways:
(A) 12
(B) 15
(C) 18
(D) 21
(E) 24
Explanation (brief):
Let N have digits h, t, u with h ∈ {2,3,4} and t + u = 10 − h. When doubling, digit sums satisfy sum(2N) = 2·sum(N) − 9k, where k is the number of carry operations in N + N. Here 11 = 2·10 − 9k ⇒ k = 1, so exactly one carry occurs.
This happens in two disjoint ways:
- Units-only carry: u ≥ 5 and t ≤ 4.
- Tens-only carry: u ≤ 4 and t ≥ 5 (and h + h + 1 ≤ 9 holds for h = 2,3,4).
- h = 2 ⇒ t + u = 8: 4 pairs in each case ⇒ 8 total.
- h = 3 ⇒ t + u = 7: 3 + 3 = 6 total.
- h = 4 ⇒ t + u = 6: 2 + 2 = 4 total.
20 Aug 2025, 13:57
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In a bit more easier language, if the spoiler was too confusing. We’re looking for 3-digit numbers between 200 and 500 whose digits add up to 10. Also, when we double the number, the digits of the new number must add up to 11.
Step 1: Possible hundreds digit
Since the number is between 200 and 500, the first digit (hundreds place) can only be 2, 3, or 4.
Step 2: Sum of the digits
The three digits must add up to 10. So if the first digit is 2, then the other two must add to 8. If the first digit is 3, the other two must add to 7. If the first digit is 4, the other two must add to 6.
Step 3: What happens when doubling?
When you double a number, sometimes the digits “carry over” (like 7+7=14 means the 4 stays and the 1 carries). Every carry makes the digit sum go down by 9.
- The digit sum starts at 10.
- Doubling should give a sum of 11.
- If there were no carries, the sum would become 20.
- To get from 20 down to 11, exactly one carry must happen (because 20 – 9 = 11).
Step 4: How can one carry happen?
There are only two ways:
- A carry from the units digit only (last digit ≥ 5, but middle digit small enough not to cause another carry).
- A carry from the tens digit only (middle digit ≥ 5, but last digit small enough not to cause another carry).
Step 5: Count them
- If the hundreds digit is 2 → there are 8 numbers.
- If the hundreds digit is 3 → there are 6 numbers.
- If the hundreds digit is 4 → there are 4 numbers.
Total = 18 numbers.
General Discussion
07 Oct 2025, 19:52
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Cannot find another way to solve it.
We need to find 3-digit numbers between 200 and 500 whose digits add up to 10.
When we double the number, the sum of the digits of the new number should be 11.
Step 1: Hundreds digit
Since the number is between 200 and 500, the first digit (hundreds place) can only be 2, 3, or 4.
Step 2: Sum of digits = 10
If the first digit is 2 → the other two digits must add up to 8.
If the first digit is 3 → the other two digits must add up to 7.
If the first digit is 4 → the other two digits must add up to 6.
Step 3: What happens when we double the number
When you double a number, sometimes a carry happens (for example, 7×2 = 14 → the 1 is carried over).
Each carry reduces the total digit sum by 9.
The original sum of digits = 10.
If there were no carries, doubling would make the sum 20.
But we’re told the new sum is 11 → which means one carry happened (because 20 − 9 = 11).
Step 4: When can one carry happen
It can happen in two cases:
Final Answer: 18 three-digit numbers fit all conditions.
We need to find 3-digit numbers between 200 and 500 whose digits add up to 10.
When we double the number, the sum of the digits of the new number should be 11.
Step 1: Hundreds digit
Since the number is between 200 and 500, the first digit (hundreds place) can only be 2, 3, or 4.
Step 2: Sum of digits = 10
If the first digit is 2 → the other two digits must add up to 8.
If the first digit is 3 → the other two digits must add up to 7.
If the first digit is 4 → the other two digits must add up to 6.
Step 3: What happens when we double the number
When you double a number, sometimes a carry happens (for example, 7×2 = 14 → the 1 is carried over).
Each carry reduces the total digit sum by 9.
The original sum of digits = 10.
If there were no carries, doubling would make the sum 20.
But we’re told the new sum is 11 → which means one carry happened (because 20 − 9 = 11).
Step 4: When can one carry happen
It can happen in two cases:
- A carry from the ones place only (last digit ≥ 5, but middle digit small enough to avoid another carry).
- A carry from the tens place only (middle digit ≥ 5, but last digit small enough to avoid another carry).
- If the hundreds digit is 2 → 8 valid numbers
- If the hundreds digit is 3 → 6 valid numbers
- If the hundreds digit is 4 → 4 valid numbers
Final Answer: 18 three-digit numbers fit all conditions.
nafuzzz
