JEE Main 2023 — Mathematics Algebra

Let the $4-$4−digit pin be $abcd$abcd

Now given sum of first two is equal to last two, so $a+b=c+d$a+b=c+d

And number available are $0,1,2,3,4,5,6,7$0,1,2,3,4,5,6,7

Now Let $a+b=c+d=\lambda$a+b=c+d=λ

Also the greatest digit that he has used is $7$7

So, $a+b=c+d=\lambda \geq 7$a+b=c+d=λ≥7

Now taking all cases starting from $\lambda =7$λ=7 we get,

$\lambda =7,(a,b)\rightarrow (0,7)\text{or}(7,0)$λ=7,(a,b)→(0,7)or(7,0), then $(c,d)\rightarrow (1,6),(2,5),(3,4),(4,3),(5,2),(6,1)$(c,d)→(1,6),(2,5),(3,4),(4,3),(5,2),(6,1) and vice-versa, so $2\times 12=24$2×12=24 numbers

$\lambda =8,(a,b)\rightarrow (1,7)\text{or}(7,1)$λ=8,(a,b)→(1,7)or(7,1), then $(c,d)\rightarrow (2,6),(3,5),(6,2),(5,3)$(c,d)→(2,6),(3,5),(6,2),(5,3) and vice versa so total $2\times 8=16$2×8=16 numbers,

$\lambda =9,(a,b)\rightarrow (2,7)\text{or}(7,2)$λ=9,(a,b)→(2,7)or(7,2) then $(c,d)\rightarrow (3,6),(4,5),(5,4),(6,3)$(c,d)→(3,6),(4,5),(5,4),(6,3) and vice versa, so total $2\times 8=16$2×8=16 numbers,

$\lambda =10,(a,b)\rightarrow (3,7),(7,3),\text{so}(c,d)\rightarrow (6,4),(4,6)\rightarrow 8$λ=10,(a,b)→(3,7),(7,3),so(c,d)→(6,4),(4,6)→8 numbers

$\lambda =11,(a,b)\rightarrow (4,7),(7,4),\text{so}(c,d)\rightarrow (6,5),(5,6)\rightarrow 8$λ=11,(a,b)→(4,7),(7,4),so(c,d)→(6,5),(5,6)→8 numbers

\lambda =12,13&14 are not possible,

So, Total numbers $=24+16+16+8+8=72$=24+16+16+8+8=72