\sf Observe \: that \: when \: a \: terminating \: decimal \: is \: converted \: into \: simplest \: fraction,Observethatwhenaterminatingdecimalisconvertedintosimplestfraction,
\textsf{its denominator contains only prime factors of 2 or 5 or both.}its denominator contains only prime factors of 2 or 5 or both.
\textsf{The above 6 inequalities mean: in integers less than 2012,}The above 6 inequalities mean: in integers less than 2012,
\textsf{there are 10 integers with only prime factor 2;}there are 10 integers with only prime factor 2;
\sf there \: are \: 8 \: integers \: with \: only \: prime \: factors \: of\: 2 \:and\: 1\: prime \:factor\: of \:5;thereare8integerswithonlyprimefactorsof2and1primefactorof5;
\sf there \: are \: 6 \: integers \: with \: only \: prime \: factors \: of\: 2 \:and\: 2\: prime \:factors\: of \:5;thereare6integerswithonlyprimefactorsof2and2primefactorsof5;
\sf there \: are \: 4 \: integers \: with \: only \: prime \: factors \: of\: 2 \:and\: 3\: prime \:factors\: of \:5;thereare4integerswithonlyprimefactorsof2and3primefactorsof5;
\textsf{there is 1 integer with 1 prime factor 2 and 4 prime factors of 5;}there is 1 integer with 1 prime factor 2 and 4 prime factors of 5;
\textsf{there are 4 integers with only prime factor 5.}there are 4 integers with only prime factor 5.
Answers & Comments
Answer:
\begin{gathered}\textsf{First 5 terminating decimals}\begin{cases} \sf \frac{1}{2} = 0.5 \\\sf \frac{1}{4} = \frac{1}{2^2} =0.25\\\sf \frac{1}{5} = 0.2\\\sf \frac{1}{8} = \frac{1}{2^3}= 0.125\\\sf \frac{1}{10} = \frac{1}{2 \times 5} =0.1\end{cases}\end{gathered}
First 5 terminating decimals
⎩
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎧
2
1
=0.5
4
1
=
2
2
1
=0.25
5
1
=0.2
8
1
=
2
3
1
=0.125
10
1
=
2×5
1
=0.1
\sf Observe \: that \: when \: a \: terminating \: decimal \: is \: converted \: into \: simplest \: fraction,Observethatwhenaterminatingdecimalisconvertedintosimplestfraction,
\textsf{its denominator contains only prime factors of 2 or 5 or both.}its denominator contains only prime factors of 2 or 5 or both.
\begin{gathered}\begin{array}{cc} \sf 2^{10} < 2012 < 2^{11} & \quad \sf 5^4 < 2012 < 5^5\\\sf 5^2 \times 2^6 < 2012 < 5^2 \times 2^7 & \quad \sf 5^3 \times 2^4 < 2012 < 5^3 \times 2^5\\\sf 5^4 \times 2 < 2012 < 5^4 \times 2^2 & \quad \sf 5 \times 2^{8} < 2012 < 5 \times 2^9\end{array}\end{gathered}
2
10
<2012<2
11
5
2
×2
6
<2012<5
2
×2
7
5
4
×2<2012<5
4
×2
2
5
4
<2012<5
5
5
3
×2
4
<2012<5
3
×2
5
5×2
8
<2012<5×2
9
\textsf{The above 6 inequalities mean: in integers less than 2012,}The above 6 inequalities mean: in integers less than 2012,
\textsf{there are 10 integers with only prime factor 2;}there are 10 integers with only prime factor 2;
\sf there \: are \: 8 \: integers \: with \: only \: prime \: factors \: of\: 2 \:and\: 1\: prime \:factor\: of \:5;thereare8integerswithonlyprimefactorsof2and1primefactorof5;
\sf there \: are \: 6 \: integers \: with \: only \: prime \: factors \: of\: 2 \:and\: 2\: prime \:factors\: of \:5;thereare6integerswithonlyprimefactorsof2and2primefactorsof5;
\sf there \: are \: 4 \: integers \: with \: only \: prime \: factors \: of\: 2 \:and\: 3\: prime \:factors\: of \:5;thereare4integerswithonlyprimefactorsof2and3primefactorsof5;
\textsf{there is 1 integer with 1 prime factor 2 and 4 prime factors of 5;}there is 1 integer with 1 prime factor 2 and 4 prime factors of 5;
\textsf{there are 4 integers with only prime factor 5.}there are 4 integers with only prime factor 5.
\sf Hence, \: there \: are \: 10 + 8 + 6 + 4 + 1 + 4 = 33 \: terminating \: decimals.Hence,thereare10+8+6+4+1+4=33terminatingdecimals.
Explanation:
Answer:
puso
Explanation:
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