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Urgent! Math problem! Math problem!
1. The two archers agree that the number of rings obtained by each arrow is one of the integers from 0 to 1 1. Each of them shot five targets, and the product of the number of rings each person got was exactly 1764, but the total number of rings of Party A was four less than that of Party B, so find the total number of rings of each person.

1764 = 2 * 2 * 3 * 7 * 7. Specify that the number of cycles obtained by each arrow is one of the integers from 0 to 10.

So the 7-ring must be separated, so there are 5 combinations of the 36-ring.

Total of 2,2,9 13

The total number of 2,3,6 is 1 1.

Total of 3,3,4 10

1, 4,9 total 14

1, 6,6 total 13

The total number of rings of A is 4 less than that of B.

So A is 3, 3, 4, 7, 7, with 24 rings, and B is 1, 4, 7, 7, 9, with 28 rings.

2. Given that the greatest common factor of two numbers is 16 and the least common multiple is 160, then these two numbers are ().

160= 16*2*5

So these two numbers can be 16*2*5 and 16, or 16*2 and 16*5.

Namely 160 and 16, or 32 and 80.

3. The sum of three times of a prime number and twice of another prime number is 100, and the product of these two prime numbers is ().

The sum of three times of a prime number A and twice of another prime number B is 100.

Because only the prime number 2 is even, then A is 2, so B=47.

The product is 94

Xiao Li, Xiao Jie and Xiao Ming go to the orphanage as volunteers regularly. Xiao Ming goes every seven days, Xiao Li every five days and Xiao Jie every 1 1 day. They went there once together on April 65, 2008. When will the three of them go together next time?

Xiao Ming goes every 7 days, Xiao Li every 5 days, and Xiao Jie every 1 1 day, asking to go together next time, that is, to find the least common multiple of 8,6, 12, that is, 24.

So the next time we go together is May 12.