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Sagot :
{12402, 12420, 12438, 12456, 12474, 12492}
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To solve the problem, let's understand the conditions and constraints:
- We have a 5-digit number with specific digits.
- The tens and ones digits are unknown.
- The number should be divisible by both 6 and 9.
Given:
- Ten thousand's place = 10000
- Thousand's place = 2000
- Hundred's place = 400
- Tens and one's places are unknown.
The number can be represented as (10000 + 2000 + 400 + 10a + b), where (a) is the digit in the tens place and (b) is the digit in the one's place. This simplifies the number to:
- 12400 + 10a + b
Conditions for divisibility:
- A number is divisible by 6 if it is divisible by both 2 and 3.
- A number is divisible by 9 if the sum of its digits is divisible by 9.
Given these conditions, we must find the values of (a) and (b) such that the number (12400 + 10a + b) satisfies both conditions. We list the resulting numbers in increasing order.
1. 12402 (a=0, b=2)
- Divisible by 2 (last digit is even)
- Sum of digits: (1 + 2 + 4 + 0 + 2 = 9) (divisible by 3 and 9)
2. 12420 (a=2, b=0)
- Divisible by 2 (last digit is even)
- Sum of digits: (1 + 2 + 4 + 2 + 0 = 9) (divisible by 3 and 9)
3. 12438 (a=3, b=8)
- Divisible by 2 (last digit is even)
- Sum of digits: (1 + 2 + 4 + 3 + 8 = 18) (divisible by 3 and 9)
4. 12456 (a=5, b=6)
- Divisible by 2 (last digit is even)
- Sum of digits: (1 + 2 + 4 + 5 + 6 = 18) (divisible by 3 and 9)
5. 12474 (a=7, b=4)
- Divisible by 2 (last digit is even)
- Sum of digits: (1 + 2 + 4 + 7 + 4 = 18) (divisible by 3 and 9)
6. 12492 (a=9, b=2)
- Divisible by 2 (last digit is even)
- Sum of digits: (1 + 2 + 4 + 9 + 2 = 18) (divisible by 3 and 9)
So, the numbers that Amy could have written, in increasing order, are:
- {12402, 12420, 12438, 12456, 12474, 12492}
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