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Sia ? 6 years, 6 months ago
Given integers are 408 and 1032 where 408 < 1032
By applying Euclid’s division lemma, we get 1032 = 408 {tex}\times{/tex} 2 + 216.
Since the remainder ≠ 0, so apply division lemma again on divisor 408 and remainder 216, we get the relation as
408 = 216 {tex}\times{/tex} 1 + 192.
Since the remainder ≠ 0, so apply division lemma again on divisor 216 and remainder 192
216 = 192 {tex}\times{/tex} 1 + 24.
Since the remainder ≠ 0, so apply division lemma again on divisor 192 and remainder 24
192 = 24 × 8 + 0.
Now the remainder has become 0. Therefore, the H.C.F of 408 and 1032 = 24.
Therefore,
24 = 1032m - 408 {tex}\times{/tex} 5
1032m = 24 + 408 {tex}\times{/tex} 5
1032m = 24 + 2040
1032m = 2064
{tex}m = \frac{{2064}}{{1032}}{/tex}
Therefore, m = 2.
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Sia ? 6 years, 6 months ago
Let the ten's digit of the required number be x and the unit's digit be y.
As per given condition the sum of the digits of a two-digit number is 12.
Then, x + y = 12. ...(i)
Required number = (10x + y).
Number obtained on reversing the digits = (10y + x).
As per given condition the number obtained by interchanging its digits exceeds the given number by 18.
{tex}\therefore{/tex} (10y + x) - (10x + y) = 18
{tex}\Rightarrow{/tex}10y + x - 10x - y = 18
{tex}\Rightarrow{/tex} 9y -9x = 18
{tex}\Rightarrow{/tex} y -x = 2. ...(ii)
On adding (i) and (ii), we get
(x + y) + (y -x) = 12 + 2
{tex}\Rightarrow{/tex} 2y = 14
{tex}\Rightarrow{/tex} y = 7.
Putting y = 7 in (i), we get
x + 7 = 12
{tex}\Rightarrow{/tex} x = 12 - 7 = 5
{tex}\therefore{/tex} x = 5 and y = 7.
Hence, the required number is 57.
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Sia ? 6 years, 6 months ago
Let n be any positive integer. Applying Euclids division lemma with divisor = 5, we get
{tex}\style{font-family:Arial}{\begin{array}{l}n=5q+1,5q+2,5q+3\;and\;5q+4\;\\\end{array}}{/tex}
Now (5q)2 = 25q2 = 5m, where m = 5q2, which is an integer;
{tex}\style{font-family:Arial}{\begin{array}{l}(5q\;+\;1)^{\;2}\;=\;25q^2\;+\;10q\;+\;1\;=\;5(5q^2\;+\;2q)\;+\;1\;=\;5m\;+\;1\\where\;m\;=\;5q^2\;+\;2q,\;which\;is\;an\;integer;\\\;(5q\;+\;2)^2\;=\;25q^2\;+\;20q\;+\;4\;=\;5(5q^2\;+\;4q)\;+\;4\;=\;5m\;+\;4,\\\;where\;m\;=\;5q^2\;+\;4q,\;which\;is\;an\;integer;\\\;(5q\;+\;3)^{\;2}\;=\;25q^2\;+\;30q\;+\;9\;=\;5(5q^2\;+\;6q+\;1)\;+\;4\;=\;5m\;+\;4,\\\;where\;m\;=\;5q^2\;+\;6q\;+\;1,\;which\;is\;an\;integer;\\\;(5q\;+\;4)^2\;=\;25q^2\;+\;40q\;+\;16\;=\;5(5q^2\;+\;8q\;+\;3)\;+\;1\;=\;5m\;+\;1,\;\\where\;m\;=\;5q^2\;+\;8q\;+\;3,\;which\;is\;an\;integer\\\end{array}}{/tex}
Thus, the square of any positive integer is of the form 5m, 5m + 1 or 5m + 4 for some integer m.
It follows that the square of any positive integer cannot be of the form 5m + 2 or 5m + 3 for some integer m.
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Aman Preet 7 years, 6 months ago
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