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The sum of all natural numbers within 100 divided by 5 and 1 is______ ．
Solution: all natural numbers within 100 divided by 5 and remaining 1 form an arithmetic sequence with tolerance of 5: 1, 6, 11 The sum of them is 1 + 6 + 11 + 96 = (1 + 96) × [(96-1) △ 5 + 1] △ 2 = 97 × 20 △ 2, = 970
Which natural number in 1-100 can be divided by 3 and 5, and the remaining 1 can be divided by 7
3X5=15,
Therefore, the natural numbers in 1-100 can be divided by 3 and 5, and the remaining 1 is 16, 31, 46, 61, 76, 91
It can be divided by 7 to 91
A natural number less than 50 can be divided by 3 and is a multiple of 5, but there is no factor 2. How many such numbers are there
I need method, the answer doesn't matter
A:
If there is no factor 2, then the natural number is odd
If it is a multiple of 5, the single digit is 5
Is a multiple of 3, then the sum of the digits is a multiple of 3
The natural numbers less than 50 satisfy the following conditions:
15、45
The minimum natural number that can be divided by 8 and 12 at the same time is ()
24
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