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Dmitry [639]
3 years ago
10

Please help answer this question!! 20 points!!!!

Mathematics
1 answer:
Naily [24]3 years ago
7 0

Answer:

Relative to such interrogate proposed, the answer is identified as 5520, if quantities ‘a’ and ‘b’ were to subject to define, given such conditions.

Step-by-step explanation:

Understood:

- Such quantity is a four (4) digit number, as explicitly stated.

- The value must be divisible by three (3). Thus, once the digits are summed, such quantity must be able to divide, and distribute evenly, into three (3).

- Likewise, such number is obligated to be divisible by five (5). Thus, the last digit of such quantity must end in either a ‘5’ or a ‘0’.

We were given the mathematical expression: 5a2b

5 + 2 = 7 + x = a number that is divisible by three, as stated in the problem.

X could be a list of differentiated number(s), although for a simplicity, x could be equated to as 5 to produce 12 (which is a factor of three, hence, divisible by three).

Now that ‘a’ has been substituted, ‘b’ must be substituted, similarly:

With ‘b’, we could appoint any number, so long that is either a 5 or 0 in the last digit, hence producing a value divisible my five.

For simplicity, which choosing either 5 or 0 does not make a difference, I will choose zero.

Thus, the rectified answer to such an interrogate is 5520. The difference between the largest of such number and the smallest is their abilities to become divisible by a certain constant ‘a’ and ‘b’.

*I hope this helps.

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a = \frac{2b+1}{3b-1}

b = 2/3

2b + 1 = 2*\frac{2}{3}+1\\\\=\frac{4}{3}+1\\\\=\frac{4}{3}+\frac{3}{3}\\\\=\frac{7}{3}\\\\3b-1 = 3*\frac{2}{3}-1\\\\=2- 1 = 1\\\\\\

a = \frac{2b+1}{3b-1}\\\\a=\frac{\frac{7}{3}}{1}\\\\=\frac{7}{3}\\\\

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\frac{2b+1}{3b-1}=a\\\\

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2b + 1 = a*(3b - 1)

2b + 1 = a*3b - 3*a

2b + 1 = 3ab- 3a

2b = 3ab - 3a - 1

2b - 3ab = -3a  - 1

b(2 - 3a) = -3a - 1

b =\frac{-3a - 1}{2 - 3a}

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