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 continues up to 1000th step. After 1000th step, what will be the value in the 600th position

Answer 1

Answer:

48

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.