Question

How many ordered four-tuples of integers (a, b, c, d) with 0 < a < b < c < d < 100 satisfy a + d = b + c and bc − ad = 93?​

Answer 1

follow the picture to know the answer of the question . there are 5 four tuples that satisfy the conditions

Answer 2

SOLUTION

TO DETERMINE

The number of ordered four-tuples of integers (a, b, c, d) with 0 < a < b < c < d < 100 satisfy a + d = b + c and bc - ad = 93

EVALUATION

Here it is given that integers (a, b, c, d) with 0 < a < b < c < d < 100 satisfy a + d = b + c

Without any loss of generality we assume that

b = a + m

c = a + m + n

Now a + d = b + c gives

d = b + c - a

⇒ d = a + m + a + m + n - a

⇒ d = a + 2m + n

Again bc − ad = 93 gives

$\sf{(a + m)(a + m + n) - a(a + 2m + n) = 93}$

$\sf{ \implies \: {(a + m)}^{2} + n(a + m ) - a(a + 2m + n) = 93}$

$\sf{ \implies \: {a}^{2} + 2am + {m}^{2} + an + mn - {a}^{2} - 2am - a n = 93}$

$\sf{ \implies \: {m}^{2} + mn = 93}$

$\sf{ \implies \: m(m + n) = 93}$

$\sf{ \implies \: either \: \: m(m + n) = 1 \times 93 \: \: \: or \: \: m(m + n) = 3 \times 31}$

Case : 1

$\sf{m(m + n) = 1 \times 93} \: \: gives \:$

m = 1 and m + n = 93

Consequently m = 1 , n = 92

It is given that

0 < d < 100

⇒ 0 < a + 2m + n < 100

⇒ 0 < a + 2 + 92 < 100

⇒ 0 < a + 94 < 100

⇒ 0 < a < 6 ( Since a > 0 )

So there are 5 values of a

So in this case the number of ordered four-tuples of integers (a, b, c, d) is 5

Case : 2

$\sf{m(m + n) = 3 \times 31} \: \: gives \:$

m = 3 and m + n = 31

Consequently m = 3 , n = 28

It is given that

0 < d < 100

⇒ 0 < a + 2m + n < 100

⇒ 0 < a + 6 + 28 < 100

⇒ 0 < a + 34 < 100

⇒ 0 < a < 66 ( Since a > 0 )

So there are 65 values of a

So in this case the number of ordered four-tuples of integers (a, b, c, d) is 65

Hence the required total number of ordered four-tuples of integers (a, b, c, d)

= 5 + 65

= 70

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