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Setler79 [48]
3 years ago
10

A, b, c and d are positive integers, such that a+b+ ab = 76, c+d+ cd = 54. Find (a+b+c+d)·a·b·c·d.

Mathematics
1 answer:
lutik1710 [3]3 years ago
6 0

Notice that

(1 + <em>x</em>)(1 + <em>y</em>) = 1 + <em>x</em> + <em>y</em> + <em>x y</em>

So we can add 1 to both sides of both equations, and we use the property above to get

<em>a</em> + <em>b</em> + <em>a b</em> = 76  ==>  (1 + <em>a</em>)(1 + <em>b</em>) = 77

and

<em>c</em> + <em>d</em> + <em>c d</em> = 54  ==>  (1 + <em>c</em>)(1 + <em>d</em>) = 55

Now, 77 = 7*11 and 55 = 5*11, so we get

<em>a</em> + 1 = 7  ==>  <em>a</em> = 6

<em>b</em> + 1 = 11  ==>  <em>b</em> = 10

(or the other way around, since the given relations are symmetric)

and

<em>c</em> + 1 = 5  ==>  <em>c</em> = 4

<em>d</em> + 1 = 11  ==>  <em>d</em> = 10

Now substitute these values into the desired quantity:

(<em>a</em> + <em>b</em> + <em>c</em> + <em>d</em>) <em>a</em> <em>b</em> <em>c</em> <em>d</em> = 72,000

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