Question
Explain why 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 are composite numbers.
✓
Answer
Explain: Why 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 are composite numbers.
We can see that both the numbers have common factor 7 and 1.
7 × 11 × 13 + 13 = (77 + 1) × 13
= 78 × 13
7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 = (7 × 6 × 4 × 3 × 2 × 1) × 2
= 1008 × 5
And we know that composite numbers are those numbers which have at least one more factor other than 1.
Hence after simplification we see that both numbers are even and therefore the given two numbers are composite numbers.
We can see that both the numbers have common factor 7 and 1.
7 × 11 × 13 + 13 = (77 + 1) × 13
= 78 × 13
7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 = (7 × 6 × 4 × 3 × 2 × 1) × 2
= 1008 × 5
And we know that composite numbers are those numbers which have at least one more factor other than 1.
Hence after simplification we see that both numbers are even and therefore the given two numbers are composite numbers.
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→Complete the following proof by filling in the blanks.
Proof: Draw seg OD.
$\angle ACB =\square \ldots 90^{\circ} \ldots \ldots \ldots . . \text { angle inscribed in semicircle }$
$\angle DCB =\square \ldots . . .45^{\circ} \ldots CD \text { is the bisector of } \angle C$
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$\angle DOB =\square \ldots . . . .90^{\circ} \text { definition of measure of an arc (I) }$
$\operatorname{seg} OA \cong \operatorname{seg} OB \ldots \ldots . . \text { radii of the circle... } \square \text { (II) }$
$\therefore \text { line } OD \text { is } \square \text { of seg } AB . \ldots . . \text { bisector.... From (I) and (II) }$
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