Among two consecutive natural number, find the least two numbers that the total of each digit of either number can be divided by seven.
Answer
69999 and 70000
Solution
When the one's digit of the smaller number is except 9, it never happen that the total of each digit of either number can be divided by 7.
There is the case that the one's digit of the smaller number is 9.
Consider the highest digit of the smaller number is X which is not 9 and N pieces of all lower digit are 9 such as X9, X99, X999 ----.
When N is 4, 9 × 4 = 36 and the larger number can be expressed as Y0000.
When both 36+X and Y can be divided by 7, X = 6 and Y = 7.
Therefore numbers to be found are 69999 and 70000.
Answer
69999 and 70000
Solution
When the one's digit of the smaller number is except 9, it never happen that the total of each digit of either number can be divided by 7.
There is the case that the one's digit of the smaller number is 9.
Consider the highest digit of the smaller number is X which is not 9 and N pieces of all lower digit are 9 such as X9, X99, X999 ----.
When N is 4, 9 × 4 = 36 and the larger number can be expressed as Y0000.
When both 36+X and Y can be divided by 7, X = 6 and Y = 7.
Therefore numbers to be found are 69999 and 70000.
