In the first 100 natural numbers, how many can be divisible by 2, 5 or 7
One is 70 2 * 5 * 7 = 70, which is the least common multiple of 257, and the one that can be adjusted within 100 is 70
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- 1. In the first 100 natural numbers, how many can be divisible by 2 or 3 or 5 In the first 100 natural numbers, how many are divisible by 2, 3 and 5
- 2. In the first 100 natural numbers, how many are divisible by 2, 3 or 5
- 3. Use scientific counting method to express that 0.00005 = 43.78 times 10 to the power of - 2 = - 0.002 times 10 to the power of 3 = 4.5 times 10 to the power of 2 times 10 to the power of 3 = 4
- 4. Given the nth power of equation a, can you write out all the qualified integers a and n so that the nth power of equation a = 64 holds? Please have a try As above
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- 7. A. B and C are three prime numbers less than 20, a + B + C = 30
- 8. There are three prime numbers a, B and C, a + B = 20, B + C = 30, a × B × C =? The results I know: 7, 13, 17, the best equation
- 9. a. If a + B + C = 162 and a * B + A * C + b * C = 6279, then a * b * C +?
- 10. ABC is three prime numbers and the product of ABC is five times of the sum of ABC. How much is the sum of a ^ 2 + B ^ 2 + C ^ 2 abc=5(a+b+c) Because a, B, C are all prime numbers, so one of a, B, C is 5, and because the band formula is rotational symmetry, any change of the order of a, B, C does not affect the result Let a = 5 To 5BC = 25 + 5B + 5C Divide both sides by 5 bc=5+b+c bc-b-c+1=6 (b-1)(c-1)=6=1*6=2*3 If it is decomposed into 2 * 3, then B and C are 3 and 4 respectively So (B-1) (C-1) = 1 * 6 Let B-1 = 1 and C-1 = 6 So B = 2, C = 7 a^2+b^2+c^2=5^2+2^2+7^2=78 Why does ABC have a number of 5
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- 12. In the first 100 natural numbers, the sum of all numbers divisible by 3 is
- 13. What is the sum of the first 100 positive natural numbers divisible by 3
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- 16. At 1, 2, 3 Among the 1000 natural numbers of 1000, the natural numbers that can be divisible by 3,5 but cannot be divisible by 15
- 17. In natural numbers less than 1000, the number that can be divisible by 6 but not by 8 Urgent! 11
- 18. How many natural numbers less than 1000 can be divided by 6 but not by 8! There is a process, otherwise I don't understand
- 19. In less than 1000 natural numbers, by 3 integer division, the rest 2, by 7 integer division, the sum of the rest 3 natural numbers? Give me a program
- 20. How many of the 100 natural numbers from 1 to 100 can't be divisible by 3 and 11?