JEE Main 2023 — Mathematics Algebra

Given,

$${25}^{190}-{19}^{190}-{8}^{190}+{2}^{190}$$25190−19190−8190+2190

Now $({25}^{190}-{19}^{190}),{8}^{190},{2}^{190}$(25190−19190),8190,2190 are divisible by $2$2, as ${25}^{190}$25190 will give remainder as $1$1 and ${19}^{190}$19190 will give remainder as $1$1 when divided by $2$2, so combined ${25}^{190}-{19}^{190}$25190−19190 will give remainder $0$0

Now, ({25}^{190}-{8}^{190})&(-{19}^{190}+{2}^{190}) are divisible by $17$17, ${a}^{n}-{b}^{n}$an−bn is divisible by $a-b$a−b if $n$n is even,

So given number is divisible by $34$34 as it is divisible by 2&17 both,

Now,${25}^{190}={(7\times 3+4)}^{190}$25190=(7×3+4)190, dividing by $7$7 we get,

$${(7\times 3+4)}^{190}={4}^{190}={16}^{95}={2}^{95}=4\cdot {8}^{31}$$(7×3+4)190=4190=1695=295=4⋅831

Now when $4\cdot {(7+1)}^{31}$4⋅(7+1)31 divided by $7$7 will give remainder as $4\cdot 1=4$4⋅1=4

And ${19}^{190}={(7\times 3-2)}^{190}$19190=(7×3−2)190, when divided by $7$7 we get,

$${(7\times 3-2)}^{190}={(-2)}^{190}={2}^{190}=2\times {8}^{63}=2\times {(7+1)}^{63}$$(7×3−2)190=(−2)190=2190=2×863=2×(7+1)63

So, when $2\times {(7+1)}^{63}$2×(7+1)63 divided by $7$7 will give remainder as $2$2

Similarly ${8}^{190}={(7+1)}^{190}$8190=(7+1)190 will give $1$1 remainder

And ${2}^{190}=2{(7+1)}^{63}$2190=2(7+1)63 will give $1$1 remainder,

So combining all we get,${25}^{190}-{19}^{190}-{8}^{190}+{2}^{190}=4-2-1+1=5$25190−19190−8190+2190=4−2−1+1=5

$\Rightarrow$⇒Not divisible by $7$7, so it will not be divisible by $14$14