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Sia ? 6 years, 5 months ago
196 and 38220
We have 38220 > 196,
So, we apply the division lemma to 38220 and 196 to obtain
38220 = 196 × 195 + 0
Since we get the remainder is zero, the process stops.
The divisor at this stage is 196,
Therefore, HCF of 196 and 38220 is 196.
Posted by Dhananjay Yaduvansi 6 years, 5 months ago
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Sia ? 6 years, 5 months ago
The required number is the HCF of (245 - 5) and (1029 - 5) i.e., 240 and 1024.
{tex}1024 = 240 \times 4 + 64{/tex}
{tex}240 = 64 \times 3 + 48{/tex}
{tex}64 = 48 \times 1 + 16{/tex}
{tex}48 = 16 \times 3 + 0{/tex}
{tex}\therefore H C F \text { is } 16{/tex}.
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Sia ? 6 years, 6 months ago
{tex}3x - 4y = 7\ and\ 5x + 2y = 3{/tex}
The given system of linear equation is {tex}3x - 4y = 7\ and\ 5x + 2y = 3{/tex}
Now, {tex}3x - 4y = 7{/tex}
{tex}y = \frac { 3 x - 7 } { 4 }{/tex}
When x = 1 then, y = -1
When x = -3 then y = -4
| x | 1 | -3 |
| y | -1 | -4 |
Now, 5x + 2y = 3
{tex}y = \frac { 3 - 5 x } { 2 }{/tex}
When x = 1 then, y = -1
When x = 3 then y = -6
Thus, we have the following table
| x | 1 | 3 |
| y | -1 | -6 |
Graph of the given system of equations are
Clearly the two lines intersect at A(1, -1)
Hence, x = 1 and y = -1 is the solution of the given system of equations.
Posted by Deepanshu Thakue 7 years, 8 months ago
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Sia ? 6 years, 6 months ago
Let a be the positive integer and b = 4.
Then, by Euclid’s algorithm, a = 4q + r for some integer q ≥ 0 and r = 0, 1, 2, 3 because 0 ≤ r < 4.
So, a = 4q or 4q + 1 or 4q + 2 or 4q + 3.
{tex}(4q)^3\;=\;64q^3\;=\;4(16q^3){/tex}
= 4m, where m is some integer.
{tex}(4q+1)^3\;=\;64q^3+48q^2+12q+1=4(\;16q^3+12q^2+3q)+1{/tex}
= 4m + 1, where m is some integer.
{tex}(4q+2)^3\;=\;64q^3+96q^2+48q+8=4(\;16q^3+24q^2+12q+2){/tex}
= 4m, where m is some integer.
{tex}(4q+3)^3\;=\;64q^3+144q^2+108q+27{/tex}
=4×(16q3+36q2+27q+6)+3
= 4m + 3, where m is some integer.
Hence, The cube of any positive integer is of the form 4m, 4m + 1 or 4m + 3 for some integer m.
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