Solution: Let's say that $n$ be any arbitrary positive odd integer. Upon dividing $n$ by 6, let $m$ be the quotient and $r$ be the remainder. Therefore, by Euclid's division lemma, we get $n=6 m+r$,...
Solution: (i) When applying Euclid's algorithm, that is dividing $2520$ by $405$, we obtain, Quotient $=6$, Remainder $=90$ $\therefore 2520=405 \times 6+90$ Again upon applying Euclid's algorithm,...
Solution: It is stated by the Euclid's division algorithm that for any two positive integers $a$ and $b$, there exist unique integers $q$ and $r$, such that $a=b q+r$. where $0 \leq r \leq b$.
![\[43.\overline{123456789}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-817e128760c54fb2ce3df345c7c4a60a_l3.png "Rendered by QuickLaTeX.com")
(i) 43.123456789 Since it has a terminating decimal expansion, it is a rational number in the form of p/q and q has factors of 2 and 5 only. (ii) 0.120120012000120000. . . Since, it has...
Solution: According to Euclid’s division Lemma, Let n be the positive integer And b equals to 3 Where q is the quotient and r is the remainder $n\text{ }=3q+r$, 0<r<3 implies remainders may be...
Solution: Where q is an integer and r = 0, 1, 2, 3, 4, 5, then 6q + r is a positive integer. The positive integers are then of the form: 6q, 6q+1, 6q+2, 6q+3, 6q+4, and 6q+5. Taking cube on L.H.S...
Solution: Let x and y be 2k + 1 and 2p + 1 the two odd positive numbers, respectively Therefore, ${{x}^{2}}~+\text{ }{{y}^{2}}~=\text{ }{{\left( 2k+\text{ }1 \right)}^{2}}~+{{\left( 2p\text{ }+1...
Solution: Any odd positive integer n can be written in the form 4q + 1 or 4q + 3 as we know. Now, according to the question, For $n~=4q~+1$, Therefore ${{n}^{2}}~-1\text{ }={{\left( 4q~+\text{ }1...
Solution: Let b=4 and a be any odd integer. According to Euclid’s algorithm, For some integer $m\ge ~0,\text{ }a=4m+r$ And r = 0,1,2,3 as 0 ≤ r < 4. As a result we have, a = 4m or 4m +...
Solution: Let a be the positive integer. According to Euclid’s division algorithm, $a=6q+r$, where the value of 0 ≤ r < 6 ${{a}^{2}}~={{\left( 6q+r \right)}^{2}}~=36{{q}^{2}}~+\text{...
Solution: Let ‘a’ be the positive integer According to Euclid’s division lemma, $a=bm+r$ The value of b is 5 according to the question, $a=5m+r$ So, r= 0, 1, 2, 3, 4 For r = 0,$a=5m$. For r =...
Solution: Let any positive integer be ‘a’ and the value b is 4. According to Euclid Division Lemma, $a=bq+r$, where, 0 ≤ r < b $a=3q+r$, where, 0 ≤ r < 4 The possible values of ‘r’ according...
Solution: According to Euclid’s division lemma, $~a=bq+r$ According to the question, For b = 4. $a=4k+r$, 0 < r < 4 For r = 0, we obtain, $a=4k$...
Solution: No. Justification: Consider 3q + 1as the positive integer, where q is a natural number. (3q + 1)2 = 9q2 + 6q + 1 = 3(3q2 + 2q) + 1 = 3m + 1, (where m being an integer equals to 3q2 +...
Solution: No, the square of any positive integer cannot be of the form 3m + 2 where m is a natural number Justification: Applying the Euclid’s division lemma, The positive integer ‘a' can be written...
Solution: Yes, the statement is true. Justification: Consider 2, 3, 4 as the three consecutive numbers \[\left( 2\text{ }\times \text{ }3\text{ }\times \text{ }4 \right)/6\text{ }=\text{ }24/6\text{...
Solution: Yes, the statement is true. Justification: Let ‘a’ and ‘a + 1’ be the two consecutive positive integers Applying the Euclid’s division lemma, We have, a = bq + r, where 0 ≤ r < b For...
Solution: No, every positive integer, where q is an integer cannot be of the form 4q + 2. Justification: All numbers in the form 4q + 2, where ‘q' is an integer, are even and not divisible by 4. For...
The largest number which divides 70 and 125, leaving remainders 5 and 8, respectively, is (A) 13 (B) 65 (C) 875 (D) 1750 Solution: Option(A) 13 Explanation: According to the question, Find the...
If the HCF of 65 and 117 is expressible in the form 65m – 117, then the value of m is: (A) 4 (B) 2 (C) 1 (D) 3 Solution: Option(B) 2 Explanation: Let's calculate the HCF of 65 and 117, 117 = 1×65 +...
n2 – 1 is divisible by 8, if n is: (A) an integer (B) a natural number (C) an odd integer (D) an even integer Solution: Option(C) an odd integer Explanation: Let x = n2 – 1 In the equation...
For some integer q, every odd integer is of the form: (A) q (B) q +1 (C) 2g (D) 2q +1 Solution: Option(D) i.e. 2q+1 Explanation: Those integers which are not divisible by 2 are odd integers. As a...
For some integer m, every even integer is of the form: (A) m (B) m +1 (C) 2m (D) 2m+1 Solution: Option(C) 2m Explanation: Those integers which are divisible by 2 are even integers. As a result, it...
Solution: Let the positive integer = a According to Euclid’s division algorithm, a = 6q + r, where 0 ≤ r < 6 \[{{a}^{2}}={{\left( 6q+r \right)}^{2}}=36{{q}^{2}}+{{r}^{2}}+12qr\],\[\because...
Let ‘a’ be any positive integer. Then from Euclid’s division lemma, a = bq+r; where 0 < r < b Putting b=6 we get, ⇒ a = 6q + r, 0 < r < 6 For r = 0, we get\[a=6q=2\left( 3q...
Let, n be any positive integer. And since any positive integer can be of the form 6q, or 6q+1, or 6q+2, or 6q+3, or 6q+4, or 6q+5. (From Euclid’s division lemma for b= 6) We have n3 – n =...
Let n be any positive integer. Thus, the three consecutive positive integers are n, n+1 and n+2. We know that any positive integer can be of the form 6q, or...
(i) \[\mathbf{43}.\mathbf{123456789}\] (ii) \[\mathbf{0}.\mathbf{120120012000120000}\ldots \ldots \] $\left( III \right)$ $43.\overline{123456789}$ (i) \[\mathbf{43}.\mathbf{123456789}\] Since ,it...
(i) ${}^{13}/{}_{3125}$ (ii) ${}^{17}/{}_{8}$ (iii) ${}^{64}/{}_{455}$ (iv) ${}^{15}/{}_{1600}$ (v) ${}^{29}/{}_{343}$ (vi)$23/{{2}^{3}}{{5}^{2}}$ (vii)...
$\left( 1 \right)$ ${}^{1}/{}_{\sqrt{2}}$ $(2)7\sqrt{5}$ $(3)6+\sqrt{2}$ $\left( 1 \right)$ Let us assume, ${}^{1}/{}_{\sqrt{2}}$ is rational. Then, we can find co-prime x and y$(y\ne 0)$ such that...
is irrational ?Let us assume $3+2\sqrt{5}$ is rational. Then we can find co-prime x and y($y\ne 0$ ) such that $3+2\sqrt{5}=\frac{x}{y}$ Rearranging , we get $2\sqrt{5}=\frac{x}{y}-3$...
is irrational number?Let, $\sqrt{5}$ is rational number. Therefore ,$\sqrt{5}$ can be written as ${}^{x}/{}_{y}$ (where,x & y are coprime) $y\sqrt{5}=x$ Squaring both sides ${{(y\sqrt{5})}^{2}}={{x}^{2}}$ $$...
Since, Both Sonia and Ravi move in the same direction and at the same time, The method to find the time when they will be meeting again at the starting point is LCM(Lowest Common Multiple) of 18...
and 
are composite numbers?Numbers are of two types –prime numbers(two factors) & composite numbers(more then 2 factors). $(7\times 11\times 13)+13$ Taking 13 as common factor =$13[(7\times 11)+1]$ =$13(77+1)$...
can end with the digit 0 for any natural number n?If last digit of any number is 0,it should be divisible by 10. It also divisible by 2 and 5 as $10=2\times 5$ Prime factorisation of ${{6}^{n}}={{(2\times 3)}^{n}}$ By observation we can say...
As we know that, HCF×LCM=Product of the two given numbers(Multiplication of two given number) Since, HCF(Highest Common Factor) is given 9 of two numbers 306 & 657 Therefore using formula...
(i) 140 (ii) 156 (iii) 3825 (iv) 5005 (v) 7429 i) 140 By Taking the LCM(Lowest Common Multiple) of 140, we will get the product of its prime factor. Therefore, $140=2\times 2\times 5\times...
Given, Number of army contingent members=616 Number of army band members = 32 When we find the HCF(616,32) then we get maximum number of columns in which they can march. We use Euclid’s...
Let a be any positive integer and b=6.Then,by Euclid’s algorithm, $a=6q+r$ for some integer $q\ge 0$ and r=0,1,2,3,4,5 because $0\le r<6$ . Therefore, $a=6q$ or $6q+1$ or $6q+2$ or $6q+3$...
(1)We have 225>135 So, we apply the division lemma to 225 & 135 to obtain 225=135×1+90 Here remainder 90≠0, we apply the division lemma again to 135 and 90 to obtain 135=90×1+45 We consider...