If $a b + 4 = c d$ and $b a + 40 = d c$ , where $a b, c d, b a$ and $d c$ are $2$ digit prime numbers. Further $b$ and $d$ are the prime numbers and $a, c$ are neither prime nor composite. Find the value of $\frac{a b + b a}{c d + d c}$ .

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Solution

$a b + 4 = d c . . . . (1)$
$b a + 40 = d c . . . . (2)$
$a, c \to$ neither prime nor composite $a = c = 1$
$b, d$ prime$
from (1)
$b + 10 (a) + 4 = c (10) + d$
$\Rightarrow$ $b + 10 (1) + 4 = 1 (10) + d$
$\Rightarrow$ $b + 4 = d . . . . (1)$
From (2)
$10 (b) + a + 40 = 10 (d) + c$
$\Rightarrow$ $10 b + 1 + 40 = 10 d + 1$
$\Rightarrow$ $40 = 10 (d - b)$
and $b$ and $d$ are prime
$\Rightarrow$ $b = 3$ and $d = 7$ (only possibility)
$\frac{a b + b a}{c d + d c} = \frac{13 + 31}{17 + 71} = \frac{44}{88} = \frac{1}{2}$

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If ab+4=cd and ba+40=dc, where ab,cd,ba and dc are 2 digit prime numbers. Further b and d are the prime numbers and a,c are neither prime nor composite. Find the value of ab+bacd+dc.

ab+4=dc....(1)ba+40=dc....(2)a,c→ neither prime nor composite a=c=1b,d prime$from (1)b+10(a)+4=c(10)+d⇒ b+10(1)+4=1(10)+d⇒ b+4=d....(1)From (2)10(b)+a+40=10(d)+c⇒ 10b+1+40=10d+1⇒ 40=10(d−b)and b and d are prime⇒ b=3 and d=7 (only possibility)ab+bacd+dc=13+3117+71=4488=12

If $a b + 4 = c d$ and $b a + 40 = d c$ , where $a b, c d, b a$ and $d c$ are $2$ digit prime numbers. Further $b$ and $d$ are the prime numbers and $a, c$ are neither prime nor composite. Find the value of $\frac{a b + b a}{c d + d c}$ .