Use 123 numbers to form a prime number with three digits

Use 123 numbers to form a prime number with three digits

123 213 231 321

33 and 44 42 and 14 3 and 15 78 and 26 30 and 36 9 and 48 46 and 69

33 and 44132
42 and 14, 42
3 and 15, 15
78 and 26,3
30 and 36180
9 and 48144
138 46 and 69

Least common multiple of 61 and 59

The minimum is 3599, because the common factor of 61 and 59 is only 1, so multiply each other by 61 * 59 = 3599

Use 1,2,3,4,5 to form a five digit number without repetition. It is a multiple of 3. The minimum five digit number is (), and the maximum even number is
(), the maximum odd number is ()

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Use 1,2,3,4,5 to form five digits without repetition
The minimum five digits of a multiple of 3 are [12345],
The largest even number is [54312]
The largest odd number is [54321]

12345 five digits, which make up five digits without repetition. How many even digits must be on the even digits

12

Use 5430 four digits to form three digits: the maximum multiple of 3 and 5 () the minimum multiple of 2 and 3 (), the minimum even number, the maximum odd number () the common multiple of 235

Use 5430 four digits to form three digits: the maximum multiple of 3 and 5 (540), the minimum multiple of 2 and 3 (354), the minimum even number 304, the maximum odd number (543) 235 common multiple 540

Choose 3 even numbers and 2 odd numbers from 1, 2, 3, 4, 5, 6, 7, 8 and 9. How many 5-digit and odd numbers can be formed?

4C3 × 5c2 × 5p5 = 4800 five digit numbers
5c1 × 4C1 × 4C3 × 4p4 = 1920 odd numbers
4C1 × 3c2 × 5c2 × 4p4 = 2880 even numbers
I use permutations and combinations

Among the five digit numbers composed of 0, 1, 5, 3 and 9, the largest odd number is (), and the smallest even number is ()

95301 13590

There are six numbers 0, 1, 2, 3, 4, 5, which make up four digit numbers with two odd numbers and two even numbers and divisible by 5. How many possibilities are there

First of all, according to the constraint conditions, it can be divided by 5, and the end should be 0 or 5. First, consider the case that the end number is 0, and the remaining three numbers should be two odd numbers. A non-zero even number can use [C (3,2) + C (2,1)] * a (3,3) = 30, and then consider the case that the end number is 5, and the remaining three numbers are two even numbers, and a non-5 odd number

How many even three digit numbers can be composed of the five numbers 0, 1, 3, 4 and 5

130
three hundred and ten
one hundred and forty
four hundred and ten
one hundred and fifty
five hundred and ten
three hundred and forty
four hundred and thirty
three hundred and fifty
five hundred and thirty
four hundred and fifty
five hundred and forty
Total 6 + 4 + 2 = 12