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If [log10 1] + [log10 2] + [log10 4] +...+ [log10 n] = n where [x] denotes the greatest integer less than or equal to x, then
(2016)
  • a)
    96 ≤ n < 104
  • b)
    104 < - n < 107
  • c)
    107 ≤ n < 111
  • d)
    111 ≤ n < 116
Correct answer is option 'C'. Can you explain this answer?
Most Upvoted Answer
If[log10 1] + [log10 2] + [log10 4] +...+ [log10 n] = n where [x] deno...
[log10 x] = 0, for any value of x ∈ {1, 2, ......9), ...(1)
Similarly [log10 x] = 1, for x ∈ {10, 11, 12 ... 99}...(2)
and [log10x] = 2, for
x ∈ {100, 101, 102, ... 999} ...(3)
Now consider, 1 ≤ n ≤ 9, then
[log101] + [log102] + [log103] ... [log10 n] = 0 (i.e., ≠ n)
Hence the expression given in the question cannot be satisfied.
Now consider, 10 ≤ n ≤ 99, then [log101] + [log102] ... [log10n]
From (1) and (2), the above expression becomes (0 + 0 ... 9 times) + (1 + 1 + ... (n – 9) times) = n – 9
Using the same approach, for
100 ≤  n ≤ 999, [log 101] + [log102] ... [log 10n] = 90 + 2(n – 99)
If can be seen that, only for the third case i.e., 100 ≤  n ≤  999, can the expression given in the question be satisfied.
Hence 90 + 2(n – 99) =  n
⇒ n = 198 – 90 = 108
∴ 107 ≤ n < 111.
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Community Answer
If[log10 1] + [log10 2] + [log10 4] +...+ [log10 n] = n where [x] deno...
Notice that $n$ must be greater than or equal to $2$ since $[log10 1]$ equals $0.$

If $2\leq n < 10,$="" then="" $[log10="" 1]="" +="" [log10="" 2]="" +="" \cdots="" +="" [log10="" n]="" \leq="" 0="" +="" 0="" +="" \cdots="" +="" 0="" +="" 1="1,$" which="" is="" less="" than="" />

If $10\leq n < 100,$="" then="" $[log10="" 1]="" +="" [log10="" 2]="" +="" \cdots="" +="" [log10="" n]="" \leq="" 0="" +="" 0="" +="" \cdots="" +="" 0="" +="" 1="" +="" 2="3,$" which="" is="" less="" than="" />

If $100\leq n < 1000,$="" then="" $[log10="" 1]="" +="" [log10="" 2]="" +="" \cdots="" +="" [log10="" n]="" \leq="" 0="" +="" 0="" +="" \cdots="" +="" 0="" +="" 1="" +="" 2="" +="" 3="6,$" which="" is="" less="" than="" />

If $1000\leq n < 10000,$="" then="" $[log10="" 1]="" +="" [log10="" 2]="" +="" \cdots="" +="" [log10="" n]="" \leq="" 0="" +="" 0="" +="" \cdots="" +="" 0="" +="" 1="" +="" 2="" +="" 3="" +="" 4="10,$" which="" is="" less="" than="" />

Since $2016$ is between $1000$ and $10000,$ we know that $[log10 1] + [log10 2] + \cdots + [log10 2016] < 2016.$="" therefore,="" the="" equation="" $[log10="" 1]="" +="" [log10="" 2]="" +="" \cdots="" +="" [log10="" n]="n$" is="" never="" satisfied="" for="" $n="" />

Thus, the answer is $\boxed{\text{none of the above}}.$
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If[log10 1] + [log10 2] + [log10 4] +...+ [log10 n] = n where [x] denotes the greatest integer less than or equal to x, then(2016)a)96 ≤ n < 104b)104 < - n < 107c)107 ≤ n < 111d)111 ≤ n < 116Correct answer is option 'C'. Can you explain this answer?
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