Given, 52n−48n+k is divisible by 24 Let ‌‌E(n)=52n−48n+k and ‌‌E(n)≡0(bmod‌24),∀n∈N ⇒‌‌52n−48n+k≡0(bmod‌24) ⇒‌‌k=48n−52n(bmod‌24) For n=1, k‌=48(1)−52×1(bmod‌24) ‌=48−25(bmod‌24) ‌=23(bmod‌24) For n=2 ‌k=48(2)−51(bmod‌24) ‌=96−625(bmod‌24) ‌=529(bmod‌24)=23(bmod‌24) Thus, divisibility for 24 to hold for all n, we must have k≡23(bmod‌24) Thus, the least positive integral value of k is 23 .