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Elementary four math problems
Solution: prime factorization

30=2×3×5

33=3× 1 1

42=2×3×7

52=2×2× 13

65=5× 13

66=2×3× 1 1

77=7× 1 1

78=2×3× 13

105=3×5×7

These factors include six 2, six 3, three 5, three 7, three 1 1, three 13,

It should be divided into three groups, so each group should have two 2, two 3, one 5, one 7, one 1 1 and one 13.

Therefore, the grouping is as follows:

(42,65,66),(33,52, 105),(30,77,78)