It is less than 100,000. it's an even number. it is divisible by 2,3, and 6. None of the digits is less than 4. None of the digits is used more than once. the sum of the digits is 30. The digit in the hundreds place is 2 less than the digit in the thousands place. The digit in the ten-thousands place is one more than the digit in the thousands place. The digit in the tens place is one less than the digit in the hundreds place. If the number were increased by 6, it would be divisible by 2,3,4,6, and 9. Find the number.
Number is abcde
a+b+c+d+e = 30
c=b-2
a=b+1
d=c-1
b+1 + b + b-2 + b-3 + e = 30
4b + e = 34
e is 4,6,8
4b = 30,28,26
so, b=7
Number is a7cd6 = 87546 = 2*3*14591
Adding 6, we get 87552 = 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 3 * 3 * 19
68
To find the number that satisfies all the given conditions, we can go step by step. Let's break down the information provided:
1. It is less than 100,000 and an even number.
Since it is an even number, the last digit must be either 0, 2, 4, 6, or 8. However, since none of the digits is less than 4, the last digit cannot be 0 or 2. Therefore, the last digit must be either 4, 6, or 8.
2. It is divisible by 2, 3, and 6.
A number that is divisible by 2 must have an even last digit. Since we have narrowed down the possibilities to 4, 6, or 8, the last digit of the number must be 6.
3. The sum of the digits is 30.
Since the last digit is 6, we need to find three more digits such that their sum is 30 - 6 = 24. To ensure that none of the digits is less than 4, the only possibilities are 7, 8, and 9. Since none of the digits can be used more than once, we will assign the digits one by one and check the conditions further.
4. The digit in the hundreds place is 2 less than the digit in the thousands place.
Let's assume the digit in the thousands place is 9. Therefore, the digit in the hundreds place would be 7 (two less than 9).
5. The digit in the ten-thousands place is one more than the digit in the thousands place.
Since the digit in the thousands place is 9, the digit in the ten-thousands place would be 9 + 1 = 10, which is not a valid digit since it has to be less than 9. Let's consider the digit in the thousands place to be 8 instead.
6. The digit in the tens place is one less than the digit in the hundreds place.
Based on the previous deductions, the digit in the hundreds place is 7. Therefore, the digit in the tens place would be 7 - 1 = 6.
By assigning the digits as mentioned above, we get the number 89,764. This number satisfies all the given conditions.
7. If the number were increased by 6, it would be divisible by 2, 3, 4, 6, and 9.
Adding 6 to the number 89,764 gives us 89,770. To confirm if it satisfies the additional condition, we can check if it is divisible by 4 and 9.
The divisibility rule for 4 is that the last two digits of a number should be divisible by 4. In this case, the last two digits are 70, which is divisible by 4.
The divisibility rule for 9 is that the sum of the digits of a number should be divisible by 9. In this case, 8 + 9 + 7 + 7 + 0 = 31, which is not divisible by 9.
Therefore, the number 89,764 does not satisfy the condition when increased by 6. Please note that there is no number that satisfies all the given conditions while also satisfying the additional condition.