a. B, C and D are four different natural numbers. The product of multiplication of these four natural numbers is 1988. Find the maximum and minimum of a + B + C + D

a. B, C and D are four different natural numbers. The product of multiplication of these four natural numbers is 1988. Find the maximum and minimum of a + B + C + D

The number is decomposed into a prime factor of 1 × 2 × 2 × 7 × 71;
Then, the five factors are combined into four factors, in which the sum of the sequence with the smallest difference is the smallest, and the sum of the sequence with the largest difference is the largest
1+2+2+497=502,
2+2+7+71=82

If ABCD is four unequal natural numbers and a × B × C × d = 1988, find the maximum of a + B + C + D

1988=2x2x7x71
The maximum value of a + B + C + D can be obtained when a = 1, B = 2, C = 7 and d = 142
142+1+2+7=152

A multiplied by 3 / 2 = B multiplied by 1 / 20 = C divided by 3 / 2 = D divided by 15 (ABCD is a non-zero natural number)?
All inequalities

1、90A＋3B＞40C；90A＋40C＞4D； 90A＋4D＞3B；90A＋4D＞40C.2、3B＋40C＞4D；3B＋4D＞90A；3B＋4D＞40C.3、40C＋4D＞90A；40C＋4D＞3B.
4、90A＋3B＋40C＞4D；90A＋3B＋4D＞40C；90A＋40C＋4D＞3B；
5、3B＋40C＋4D＞90A；

The sum of natural numbers a, B, C and D is 90. The results of a + 2, B-2, C multiplied by 2 and D divided by 2 are the same. What is B?

A + 2, B-2, C multiplied by 2, D divided by 2 are the same
A+2=B-2=2C=D/2
A = B-4, C = B / 2-1, d = 2b-4 are obtained
And a + B + C + D = 90
B = 22