1 A 1 B 1 15 A and B are both natural numbers, find the maximum value of A and B

1/a+1/b=1/15

1/a=1/15-1/b=(b-15)/15ba=15b/(b-15)=(15b-225+225)/(b-15)=15+225/(b-15)

Similarly: b=15+225 (a-15).

A and B are natural numbers, so both A-15 and B-15 are divisors of 225.

The larger the b, the smaller the a; The smaller the b, the larger the a.

The minimum value of a is 16, a-15 = 1, and the maximum value of b is 240, and vice versa.

When the difference between a and b is the largest, that is, when the two numbers are 16 and 240 respectively, a+b has a maximum value of 256a and b has the smallest difference, that is, when both numbers are 30, a+b has a minimum value of 60

Obviously, if both a and b are greater than 15, then a=16, then b=240, a+b=256, this is the maximum.

From 1 a+1 b=1 15, it is easy to obtain a>15, b>15

1/a=1/15-1/b=(b-15)/15b

a=15b/(b-15)=15+225/(b-15)

Because a and b are natural numbers.

There are 9 approximations of 225: 1, 3, 5, 9, 15, 25, 45, 75, 225

When b-15 takes 1, 3, 5, 9, 15, 25, 45, 75, 225, b=16, 18, 20, 24, 30, 40, 60, 90, 240

a=240,90, 60, 40, 30, 24, 20, 18, 16

Therefore, the maximum value of a+b is 256 and the minimum value is 60

Summary. Hello! We know that a+17=b, 14a+b=122, bring a+17=b into 14a+b=122, 14a+a+17=12215a+17=12215a=12215a=122-1715a=105a=105 15=7, so, b=5+17=22<>

Knowing that a and b are both natural numbers, and a+17=b, 14a+b=122, find the values of a and b. (Full form!) )

Hello! We know that with a mask, a+17=b, 14a+b=122, bring a+17=b into 14a+b=122, 14a+a+17=122a+17=12215a+17=12215a=122-1715a=105a=105 15=7, so, b=5+17=22<>

This method is to bring in the elimination method, pro-<>

I would like to ask the teacher.

How did this 15a come about.

14a +a =（14+1）a=15a<>

From 1 Roll Hunger Filial Piety a+1 b=1 15 easy to get a> large draft 15,b>151 a=1 15-1 b=(b-15) 15ba=15b (b-15)=15+225 (b-15) because the limb a and b are natural numbers 225 has 9 approximations: 1, 3, 5, 9, 15, 25, 45, 75, 225 When b-15 takes 1, 3, 5, 9, 15, 25, 45, 75, 225 respectively, b=16, 18, 20....

The divisor of 14 is 140, and their common finger branch filial piety multiples have 140, that is, 1 14 = 10 140, and 1 + 2 + 7 = 10, so 1 140 + 2 140 + 7 140 = 10 140, after the reduction, we get: 1 140 + 1 70 + 1 20 = 1 14

Expand your thinking: 1 14 must have been obtained through the reduction. Then the denominator must be a multiple of 14.

So the denominator and the numerator of the fraction are multiplied by a number that is not 0 at the same time, and the size of the fraction remains the same, and because a, b, and c are different natural numbers, the sum of 1 a+1 b+1 c cannot be 70, so I directly thought that the sum is 6 84, then 1 84 + 2 84 + 3 84 = 6 84, and after the reduction, he obtains: 1 84 + 1 42 + 1 28 = 1 14, and after getting another set of solutions, I began to look for the first.

3. The fourth set of solutions, after calculation, obtains: 1 126 + 1 63 + 1 21 = 1 14;1 168 + 1 56 + 1 21 = 1 14;1/56+1/40+1/35=1/14……

a+17=b,①

14a+b=122,②

Substituting b=a+17 into the formula, obtains: imaginary volt 14a+a+17=122, and solution: a=7 substituting into the differential state with closed sails.

Get: b=a+17=7+17=24

a=7b=24