Question

Maximum value of n for 100^8-111112^2 is divisible by #^n

Answer 1

Answer:

8^2 is divisible and n=2

Step-by-step explanation:

=100^8-1,11,112^2

=100^2×100^2×100^2 - 1,11,112^2

=[10,00,000]^2-[1,11,112]^2

= [10,00,000+1,11,112] × [10,00,000-1,11,112]

=(11,11,112)×(8,88,888)

=(8×1,38,889)(8×1,11,111)

=8^2(1,38,889)(1,11,111)

Hence the maximum NUMBER is to divisible exactly is ..8^2

so

n=2