The following sequence is generated by dividing 50 by natural numbers :
\n0, 0, 2, 0, 2, x, 2, 5, …
\nWhat is x?<\/p>\n
[amp_mcq option1=”1″ option2=”2″ option3=”3″ option4=”4″ correct=”option2″]<\/p>\n
Let’s analyse the relationship between n and f(n).
\nNotice the outputs 0, 0, 2, 0, 2, ?, 2, 5.
\nThe inputs are 1, 2, 3, 4, 5, 6, 7, 8.
\nf(1)=0, f(2)=0, f(4)=0. These inputs (1, 2, 4) are powers of 2 ($1=2^0$, $2=2^1$, $4=2^2$).
\nf(3)=2, f(5)=2, f(7)=2. These inputs (3, 5, 7) are primes greater than 2.
\nf(8)=5. This input (8) is $2^3$.
\nf(6)=x. This input (6) is not a power of 2 and not a prime.<\/p>\n
Let’s hypothesize a rule based on the observed pattern:
\n– If n is a power of 2, $n=2^k$: the output depends on k? f(1)=f($2^0$)=0, f(2)=f($2^1$)=0, f(4)=f($2^2$)=0, f(8)=f($2^3$)=5.
\n– If n is a prime greater than 2: f(n)=2. (Matches f(3), f(5), f(7)).
\n– If n is composite and not a power of 2: f(n)=? (Only n=6 in the given range).<\/p>\n
Let’s refine the rule based on inputs being powers of 2 vs not powers of 2.
\n– If n is a power of 2: $n=2^k$. f(1)=0, f(2)=0, f(4)=0, f(8)=5. This still doesn’t follow a simple rule like k, k+c, k^2, etc.
\n– If n is NOT a power of 2: f(3)=2, f(5)=2, f(7)=2. This strongly suggests f(n)=2 for n which are not powers of 2, with a potential exception for n=8.<\/p>\n
Let’s assume the rule is:
\nf(n) = 2 if n is not a power of 2.
\nf(n) = 0 if n is a power of 2, except for n=8.
\nf(8) = 5.<\/p>\n
Let’s test this rule:
\nf(1) = f($2^0$). Power of 2, not 8. Rule says 0. Matches.
\nf(2) = f($2^1$). Power of 2, not 8. Rule says 0. Matches.
\nf(3). Not a power of 2. Rule says 2. Matches.
\nf(4) = f($2^2$). Power of 2, not 8. Rule says 0. Matches.
\nf(5). Not a power of 2. Rule says 2. Matches.
\nf(6). Not a power of 2 ($6 = 2 \\times 3$). Rule says 2. So x=2.
\nf(7). Not a power of 2. Rule says 2. Matches.
\nf(8) = f($2^3$). This is the special case $n=8$. Rule says 5. Matches.<\/p>\n