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Sia ? 6 years, 6 months ago
Let three consecutive numbers be x, (x + 1) and (x + 2)
Let x = 6q + r 0 {tex}\leq r < 6{/tex}
{tex}\therefore x = 6 q , 6 q + 1,6 q + 2,6 q + 3,6 q + 4,6 q + 5{/tex}
{tex}\text { product of } x ( x + 1 ) ( x + 2 ) = 6 q ( 6 q + 1 ) ( 6 q + 2 ){/tex}
if x = 6q then which is divisible by 6
{tex}\text { if } x = 6 q + 1{/tex}
{tex}= ( 6 q + 1 ) ( 6 q + 2 ) ( 6 q + 3 ){/tex}
{tex}= 2 ( 3 q + 1 ) .3 ( 2 q + 1 ) ( 6 q + 1 ){/tex}
{tex}= 6 ( 3 q + 1 ) \cdot ( 2 q + 1 ) ( 6 q + 1 ){/tex}
which is divisible by 6
if x = 6q + 2
{tex}= ( 6 q + 2 ) ( 6 q + 3 ) ( 6 q + 4 ){/tex}
{tex}= 3 ( 2 q + 1 ) .2 ( 3 q + 1 ) ( 6 q + 4 ){/tex}
{tex}= 6 ( 2 q + 1 ) \cdot ( 3 q + 1 ) ( 6 q + 1 ){/tex}
Which is divisible by 6
{tex}\text { if } x = 6 q + 3{/tex}
{tex}= ( 6 q + 3 ) ( 6 q + 4 ) ( 6 q + 5 ){/tex}
{tex}= 6 ( 2 q + 1 ) ( 3 q + 2 ) ( 6 q + 5 ){/tex}
which is divisible by 6
{tex}\text { if } x = 6 q + 4{/tex}
{tex}= ( 6 q + 4 ) ( 6 q + 5 ) ( 6 q + 6 ){/tex}
{tex}= 6 ( 6 q + 4 ) ( 6 q + 5 ) ( q + 1 ){/tex}
which is divisible by 6
if x = 6q + 5
{tex}= ( 6 q + 5 ) ( 6 q + 6 ) ( 6 q + 7 ){/tex}
{tex}= 6 ( 6 q + 5 ) ( q + 1 ) ( 6 q + 7 ){/tex}
which is divisible by 6
{tex}\therefore {/tex} the product of any three natural numbers is divisible by 6.
Posted by Dinar Hingorani 6 years, 6 months ago
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Sia ? 6 years, 6 months ago
Let my age is x years and my son's age is y years.
{tex}\Rightarrow x = 3y{/tex} .............(i)
Five years later, my age will be {tex}(x + 5)years{/tex} and my son's age will be {tex}(y + 5) years{/tex}.
{tex}\therefore{/tex}{tex}x + 5 = \frac { 5 } { 2 } ( y + 5 ){/tex} [Given]
{tex}\Rightarrow{/tex} {tex}2x - 5y -15 =0{/tex} .............(ii)
Putting x =3y in equation (ii), we get
{tex}6y - 5y -15 =0{/tex} {tex}\Rightarrow{/tex} {tex}y =15{/tex}
Putting {tex}y =15{/tex} in equation (i), we get
{tex}x = 45{/tex}
My present age is 45 years and my son's present age is 15 years.
Posted by Tanvi Singh 6 years, 5 months ago
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Sia ? 6 years, 5 months ago
The given equations are:
{tex}\frac { x + y - 8 } { 2 } = \frac { x + 2 y - 14 } { 3 } = \frac { 3 x + y - 12 } { 11 }{/tex}
Therefore, we have
{tex}\frac { x + y - 8 } { 2 } = \frac { 3 x + y - 12 } { 11 }{/tex}
By cross multiplication,we get
{tex}11x + 11y - 88 = 6x + 2y - 24{/tex}
{tex}11x - 6x + 11y -2y = -24 + 88{/tex}
{tex}5x + 9y = 64 {/tex}..........(i)
and {tex}\frac { x + 2 y - 14 } { 3 } = \frac { 3 x + y - 12 } { 11 }{/tex}
By cross multiplication,we get
{tex}11x + 22y -154 = 9x + 3y - 36{/tex}
{tex}11x - 9x + 22y -3y = -36 + 154{/tex}
{tex}2x + 19y = 118{/tex}.........(ii)
Multiplying (i) by 19 and (ii) by 9, we get
{tex}95x + 171y = 1216{/tex} .......(iii)
{tex}18x + 171y = 1062{/tex} .......(iv)
Subtracting (iv) from (iii),we get
{tex}77 x = 154 \Rightarrow x = 2{/tex}
Substituting x = 2 in (i),we get
{tex}5 \times 2 + 9 y = 64 \Rightarrow 9 y - 54{/tex}
{tex}\therefore{/tex} Solution is x = 2, y = 6
Posted by Shiv Mishra 7 years, 8 months ago
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Posted by Aditya Rai 6 years, 6 months ago
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Sia ? 6 years, 6 months ago
On applying the division lemma to 225 and 135, we get
225 = 135 × 1 + 90
135 = 90 × 1 + 45
90 = 45 × 2 + 0
Hence, HCF(225, 135) = 45
Posted by Rohith Rohith 7 years, 8 months ago
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Posted by Prince Singh 6 years, 6 months ago
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Sia ? 6 years, 6 months ago
Let a be any positive integer and b = 6
∴ by Euclid’s division lemma
a = bq + r, 0≤ r and q be any integer q ≥ 0
∴ a = 6q + r,
where, r = 0, 1, 2, 3, 4, 5
If a is even then then remainder by division of 6 is 0,2 or 4
Hence r = 0, 2,or 4
or A is of form 6q,6q+2,6q+4
As, a = 6q = 2(3q), or
a = 6q + 2 = 2(3q + 1), or
a = 6q + 4 = 2(3q + 2).
If these 3 cases a is an even integer.
but if the remainder is 1,3 or 5 then r=1,3 or 5
or A is of form 6q+1,,6q+3 or,6q+5
Case 1:a = 6q + 1 = 2(3q) + 1 = 2n + 1,
Case 2: a = 6q + 3 = 6q + 2 + 1,
= 2(3q + 1) + 1 = 2n + 1,
Case 3: a = 6q + 5 = 6q + 4 + 1
= 2(3q + 2) + 1 = 2n + 1
This shows that odd integer is of the form 6q + 1, or 6q + 3, or 6q + 5, where q is some integer.
Posted by Tanya Jha 7 years, 8 months ago
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Simar Simar 7 years, 8 months ago
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