Question

Let a sequence have 1000 zeroes. Instep 1, to every position in the sequence we add 2. Instep 2, to every even position in the sequence we add 2.instep 3 , to every position which is a multiple of 3 we add 2 . This is continuous upto 1000 the step . After 1000 th step , what will be the value in the 600 th position a) 64 b )24 c)48 d)124

Answer 1

Answer:

C

Step-by-step explanation:

Let the sequence be a₁,a₂,a₃,a₄,a₅,.........a₁₀₀₀.

Step 1: Every position in the sequence is increased by 2.

Step 2: Every even position in the sequence is increased by 2.

Step 3: Every  3rd term in the sequence is increased by 2.

........

Analysis:

a₁ is added by 2 only in step 1

a₂ is added by 2 in  steps 1 and 2

a₃ is added by 2 in steps 1 and 3

............

Clearly it is evident that any term aₓ is added by 2 in step n ,

if and only if n is the factor of x.

Now, if we factorize 600 we get , 600 =2³×3×5².

Thus total number of factors of 600 are (1+3)×(1+1)×(1+2) = 24.

So,  during each of these 24 steps a₆₀₀ will be added by 2.

Thus a₆₀₀ would be added by 2  24 times , hence a₆₀₀ would be

increased by 24×2 = 48.