Apply Euclid"s theorem for $17, 5$ .

17 = 5 \times 3 + 2

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17 = 5 \times 2 + 7

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17 = 5 \times 4 - 3

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None of the above

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Solution

The correct option is A $17 = 5 \times 3 + 2$
According to the definition of Euclid's theorem,
$a = b \times q + r$ where $0 \leq r < b .$
so,
$17 = 5 \times 3 + 2$

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Similar questions

Q. Name the property under multiplication used in each of the following:
(i)

\frac{2}{3} \times \frac{4}{5} = \frac{4}{5} \times \frac{2}{3}

(ii)

(\frac{- 3}{4}) \times 1 = \frac{- 3}{4} = 1 \times (\frac{- 3}{4})

(iii)

(\frac{- 17}{28}) \times (\frac{- 28}{17}) = 1

(iv)

\frac{1}{5} \times (\frac{7}{8} \times \frac{4}{3}) = (\frac{1}{5} \times \frac{7}{8}) \times \frac{4}{3}

(v)

\frac{2}{7} \times (\frac{9}{10} + \frac{2}{5}) = \frac{2}{7} \times \frac{9}{10} + \frac{2}{7} \times \frac{2}{5}

Q. 2, 3, 5, 7, 11, ?, 17

Prove that:
(i) $\frac{1}{3 + \sqrt{7}} + \frac{1}{\sqrt{7} + \sqrt{5}} + \frac{1}{\sqrt{5} + \sqrt{3}} + \frac{1}{\sqrt{3} + 1} = 1$
(ii) $\frac{1}{1 + \sqrt{2}} + \frac{1}{\sqrt{2} + \sqrt{3}} + \frac{1}{\sqrt{3} + \sqrt{4}} + \frac{1}{\sqrt{4} + \sqrt{5}} + \frac{1}{\sqrt{5} + \sqrt{6}} + \frac{1}{\sqrt{6} + \sqrt{7}} + \frac{1}{\sqrt{7} + \sqrt{8}} + \frac{1}{\sqrt{8} + \sqrt{9}} = 2$

Q. (7x+14)/(3-17-3x)=(4x+2)/3-5

$(\frac{5}{3} + \frac{17}{51}) \div 2 - 5 + 4 =$

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Apply Euclid"s theorem for 17,5.

The correct option is A 17=5×3+2According to the definition of Euclid's theorem, a=b×q+r where 0≤r<b.so,17=5×3+2

Apply Euclid"s theorem for $17, 5$ .

The correct option is A $17 = 5 \times 3 + 2$
According to the definition of Euclid's theorem,
$a = b \times q + r$ where $0 \leq r < b .$
so,
$17 = 5 \times 3 + 2$