Q: Use Euclids division lemma to show that the cube of any positive integer is of the form \$9m, 9m + 1\$ or \$9m + 8.\$

Using Euclid division algorithm, we know that \$a = bq + r,\$ \$0 \leq r \leq b\$ \$----(1)\$Let a be any positive integer, and \$b = 3.\$Substitute \$b = 3\$ in equation \$(1)\$\$a = 3q + r\$ where \$0 \leq r \leq 3,\$ \$r = 0, 1, 2\$If \$r = 0, a = 3q\$Cube the value, we get \$a^3 = 27q^3\$\$a^3 = 9(3q^3)\$, where m = \$3q^3\$ \$--- (2)\$If \$r = 1, a = 3q + 1\$Cube the value, we get \$a^3 = (3q + 1)^3\$\$a^3 = (27q^3 + 27q^2+9q+1)\$\$a^3 = 9(3q^3 + 3q^2 + 1) + 1\$, where m = \$3q^3 + 3q^2 + q\$ \$---(3)\$If \$r = 2, a = 3q + 2\$Cube the value, we get\$a^3 = (3q+2)^3\$\$a^3 = (27q^3+53q^2+36q+8)\$\$a^3 = 9(3q^3 + 6q^2 + 4q) + 8\$, where m = \$3q^3 + 6q^2 + 4q\$ \$---- (4)\$From equation \$2, 3\$ and \$4,\$The cube of any positive integer is of the form \$9m, 9m + 1\$ or \$9m + 8.\$

Q: Use Euclid's division lemma to show that the square of any positive integer is either of the form \$3m\$ or \$3m+1\$ for some integer m, but not of the form \$3m+2\$.

Let \$a\$ be the positive integer and \$b=3\$.We know \$a=bq+r\$, \$0 ≤ r < b\$Now, \$a=3q+r\$, \$0 ≤ r < 3\$The possibilities of remainder is \$0,1,\$ or \$2\$.Case 1 : When \$a=3q\$\$a^2 = (3q)^2=9q^2=3q\times 3q=3m\$ where \$m = 3q^2\$Case 2 : When \$a = 3q+1\$\$a^{ 2 }=(3q+1)^{ 2 }=(3q)^{ 2 }+(2×3q×1)+(1)^{ 2 }=3q(3q+2)+1=3m+1\$ where \$m= q(3q+2)\$Case 3: When \$a= 3q+2\$\$a^{ 2 }=(3q+2)^2=(3q)^{ 2 }+(2×3q×2)+(2)^{ 2 }=9q^{ 2 }+12q+4=9q^{ 2 }+12q+3+1=3(3q^{ 2 }+4q+1)+1=3m+1\$ where \$m= 3q^2+4q+1\$Hence, from all the above cases, it is clear that square of any positive integer is of the form \$3m\$ or \$3m+1\$.

Q: The length and breadth of a rectangular field is 110 m and 30 m respectively. Calculate the length of the longest rod which can measure the length and breath of the field exactly.

Given that the length and breadth of a rectangular field is \$110\$ m and \$30\$ m respectively. The length of the longest rod that can measure both the length and breadth of the field exactly will be the highest common factor \$(HCF)\$ of the length and breadth.Therefore, factorise \$110\$ as follows:\$110=2×5×11\$Now, factorise \$30\$ as follows:\$30=2×3×5\$We know that \$HCF\$ is the highest common factor.Since the common factor between the numbers \$110\$ and \$30\$ is \$2\times 5=10\$.Hence, the length of the longest rod which can measure the length and breadth of the field exactly, is \$10\ m\$

Q: Show that any positive even integer is of the form \$4q\$ or \$4q+2\$, where \$q\$ is a whole number.

Let \$a\$ be any positive integer and \$b=4\$, then by Euclid's algorithm,\$a=4q+r\$, for some integer \$q\ge 0\$ and \$r=0,1,2,3\$So, \$a=4q\$ or \$4q+1\$ or \$4q+2\$ or \$4q+3\$ because \$0 ≤ r < 4\$.Now, \$4q\$ that is \$(2\times 2q)\$ is an even number.Therefore, \$4q+1\$ is an odd number.Now, \$4q+2\$ that is \$2(2q+1)\$ which is also an even number.Therefore, \$(4q+2)+1=4q+3\$ is an odd number.Hence, we can say that any even integer can be written in the form of \$4q\$ or \$4q+2\$ where \$q\$ is a whole number.

Q: There are 75 rose and 45 lily flowers. These are to be made into bouquets containing both the flowers. All the bouquets should contain the same number of flowers. Find the number of bouquets with maximum number of flowers that can be formed and the number of flowers in them.

Since it is given that there are \$75\$ rose and \$45\$ lily flowers. Therefore,Factorise \$75\$ as follows:\$75=3×5×5\$Now, factorise \$45\$ as follows:\$45=3×3×5\$We know that \$HCF\$ is the highest common factor.Since the common factor between the numbers \$75\$ and \$45\$ is \$3\times 5=15\$.Hence, the number of banquets that can be formed is \$15\$ and total number of flowers are \$5+3=8\$.

Q: Express \$3825\$ as a product of prime factors

Factorise \$3825\$ as follows:\$3825=3×3×5×5×17=3^{ 2 }×5^{ 2 }×17\$Since, \$3,5\$ and \$17\$ are prime numbers as these are the numbers that are only divisible by themselves. Hence, \$3825=3^{ 2 }×5^{ 2 }×17\$

Q: Use Euclids division algorithm to find the HCF of:(i) \$135\$ and \$225\$(ii) \$196\$ and \$38220\$(iii) \$867\$ and \$225\$

(i) \$135\$ and \$225\$Step 1: First find which integer is larger.\$225 > 135\$Step 2: Then apply the Euclid's division algorithm to \$225\$ and \$135\$ to obtain \$225 = 135 \times 1 + 90\$Repeat the above step until you will get remainder as zero.Step 3: Now consider the divisor \$135\$ and the remainder \$90,\$ and apply the division lemma to get\$135 = 90 \times 1 + 45\$\$90 = 2 \times 45 + 0\$Since the remainder is zero, we cannot proceed further.Step 4: Hence, the divisor at the last process is \$45\$So, the H.C.F. of \$135\$ and \$225\$ is \$45.\$(ii) \$196\$ and \$38220\$ Step 1: First find which integer is larger.\$38220 > 196\$Step 2: Then apply the Euclid's division algorithm to 38220 and 196 to obtain\$38220 = 196 \times 195 + 0\$Since the remainder is zero, we cannot proceed further.Step 3: Hence, the divisor at the last process is 196.So, the H.C.F. of \$196\$ and \$38220\$ is \$196.\$(iii) \$867\$ and \$225\$Step 1: First find which integer is larger.\$867 > 255\$Step 2: Then apply the Euclid's division algorithm to \$867\$ and \$255\$ to obtain\$867 = 255 \times 3 + 102\$Repeat the above step until you will get remainder as zero.Step 3: Now consider the divisor \$225\$ and the remainder \$102,\$ and apply the division lemma to get\$255 = 102 \times 2 + 51\$\$102 = 51 \times 2 = 0\$Since the remainder is zero, we cannot proceed further.Step 4: Hence the divisor at the last process is 51.So, the H.C.F. of \$867\$ and \$255\$ is \$51.\$

Q: Prove that the product of three consecutive positive integers is divisible by 6.

Let the three consecutive positive integers be \$n\$, \$n+1\$ and \$n+2\$.Whenever a number is divided by \$3\$, the remainder obtained is either \$0,1\$ or \$2\$.Therefore, \$n=3p\$ or \$3p+1\$ or \$3p+2\$, where \$p\$ is some integer.If \$n=3p\$, then \$n\$ is divisible by \$3\$.If \$n=3p+1\$, then \$n+2=3p+1+2=3p+3=3(p+1)\$ is divisible by \$3\$.If \$n=3p+2\$, then \$n+1=3p+2+1=3p+3=3(p+1)\$ is divisible by \$3\$.So, we can say that one of the numbers among \$n, n+1\$ and \$n+2\$ is always divisible by \$3\$ that is:\$n(n+1)(n+2)\$ is divisible by \$3\$.Similarly, whenever a number is divided by \$2\$, the remainder obtained is either \$0\$ or \$1\$.Therefore, \$n=2q\$ or \$2q+1\$, where \$q\$ is some integer.If \$n=2q\$, then \$n\$ and \$n+2=2q+2=2(q+1)\$ is divisible by \$2\$.If \$n=2q+1\$, then \$n+1=2q+1+1=2q+2=2(q+1)\$ is divisible by \$2\$.So, we can say that one of the numbers among \$n\$, \$n+1\$ and \$n+2\$ is always divisible by \$2\$.Since, \$n(n+1)(n+2)\$ is divisible by \$2\$ and \$3\$.Hence, \$n(n+1)(n+2)\$ is divisible by \$6\$.

Q: Show that any positive odd integer is of the form \$6q + 1\$, or \$6q + 3\$, or \$6q + 5\$, where \$q\$ is some integer.

Using Euclid division algorithm, we know that \$a = bq + r,\$ \$0 \leq r \leq b\$ ----(1)Let \$a\$ be any positive integer and \$b = 6\$.Then, by Euclid’s algorithm, \$a = 6q + r\$ for some integer \$q\geq 0\$, and \$r = 0, 1, 2, 3, 4, 5\$ ,\$\:or\$ \$0 ≤ r < 6\$.Therefore, \$a = 6q \:or\: 6q + 1 \:or\: 6q + 2 \:or\: 6q + 3 \:or\: 6q + 4 \:or\: 6q + 5 \$\$6q+0: 6\$ is divisible by \$2\$, so it is an even number.\$6q+1: 6\$ is divisible by \$2\$, but \$1\$ is not divisible by \$2\$ so it is an odd number.\$6q+2: 6\$ is divisible by \$2\$, and \$2\$ is divisible by \$2\$ so it is an even number.\$6q+3: 6\$ is divisible by \$2\$, but \$3\$ is not divisible by \$2\$ so it is an odd number.\$6q+4: 6\$ is divisible by \$2\$, and \$4\$ is divisible by \$2\$ so it is an even number.\$6q+5: 6\$ is divisible by \$2\$, but \$5\$ is not divisible by \$2\$ so it is an odd number.And therefore, any odd integer can be expressed in the form \$6q + 1\:or\: 6q + 3\; or\: 6q + 5\$

Question 1

Use Euclids division lemma to show that the cube of any positive integer is of the form $9m, 9m + 1$ or $9m + 8.$

Accepted Answer

Using Euclid division algorithm, we know that $a = bq + r,$    $0 \leq r \leq b$   $----(1)$Let a be any positive integer, and $b = 3.$Substitute $b = 3$ in equation $(1)$$a = 3q + r$ where $0 \leq r \leq 3,$ $r = 0, 1, 2$If $r = 0, a = 3q$Cube the value, we get $a^3 = 27q^3$$a^3 = 9(3q^3)$, where m = $3q^3$ $--- (2)$If $r = 1, a = 3q + 1$Cube the value, we get $a^3 = (3q + 1)^3$$a^3 = (27q^3 + 27q^2+9q+1)$$a^3 = 9(3q^3 + 3q^2 + 1) + 1$, where m = $3q^3 + 3q^2 + q$ $---(3)$If $r = 2, a = 3q + 2$Cube the value, we get$a^3 = (3q+2)^3$$a^3 = (27q^3+53q^2+36q+8)$$a^3 = 9(3q^3 + 6q^2 + 4q) + 8$, where m = $3q^3 + 6q^2 + 4q$ $---- (4)$From equation $2, 3$ and $4,$The cube of any positive integer is of the form $9m, 9m + 1$ or $9m + 8.$

Question 2

Use Euclid's division lemma to show that the square of any positive integer is either of the form $3m$ or $3m+1$ for some integer m, but not of the form $3m+2$.

Accepted Answer

Let $a$ be the positive integer and $b=3$.We know $a=bq+r$, $0 &#8804; r < b$Now, $a=3q+r$, $0 &#8804; r < 3$The possibilities of remainder is $0,1,$ or $2$.Case 1 : When $a=3q$$a^2 = (3q)^2=9q^2=3q\times 3q=3m$ where $m = 3q^2$Case 2 : When $a = 3q+1$$a^{ 2 }=(3q+1)^{ 2 }=(3q)^{ 2 }+(2&#215;3q&#215;1)+(1)^{ 2 }=3q(3q+2)+1=3m+1$ where $m= q(3q+2)$Case 3: When $a= 3q+2$$a^{ 2 }=(3q+2)^2=(3q)^{ 2 }+(2&#215;3q&#215;2)+(2)^{ 2 }=9q^{ 2 }+12q+4=9q^{ 2 }+12q+3+1=3(3q^{ 2 }+4q+1)+1=3m+1$ where $m= 3q^2+4q+1$Hence, from all the above cases, it is clear that square of any positive integer is of the form $3m$ or $3m+1$.

Question 3

The length and breadth of a rectangular field is 110 m and 30 m respectively. Calculate the length of the longest rod which can measure the length and breath of the field exactly.

Accepted Answer

Given that the length and breadth of a rectangular field is $110$ m and $30$ m respectively. The length of the longest rod that can measure both the length and breadth of the field exactly will be the highest common factor $(HCF)$ of the length and breadth.Therefore, factorise $110$ as follows:$110=2&#215;5&#215;11$Now, factorise $30$ as follows:$30=2&#215;3&#215;5$We know that $HCF$ is the highest common factor.Since the common factor between the numbers $110$ and $30$ is $2\times 5=10$.Hence, the length of the longest rod which can measure the length and breadth of the field exactly, is $10\ m$

Question 4

Show that any positive even integer is of the form $4q$ or $4q+2$, where $q$ is a whole number.

Accepted Answer

Let $a$ be any positive integer and $b=4$, then by Euclid's algorithm,$a=4q+r$, for some integer $q\ge 0$ and $r=0,1,2,3$So, $a=4q$ or $4q+1$ or $4q+2$ or $4q+3$ because $0 &#8804; r < 4$.Now, $4q$ that is $(2\times 2q)$ is an even number.Therefore, $4q+1$ is an odd number.Now, $4q+2$ that is $2(2q+1)$ which is also an even number.Therefore, $(4q+2)+1=4q+3$ is an odd number.Hence, we can say that any even integer can be written in the form of $4q$ or $4q+2$ where $q$ is a whole number.

Question 5

There are 75 rose and 45 lily flowers. These are to be made into bouquets containing both the flowers. All the bouquets should contain the same number of flowers. Find the number of bouquets with maximum number of flowers that can be formed and the number of flowers in them.

Accepted Answer

Since it is given that there are $75$ rose and $45$ lily flowers. Therefore,Factorise $75$ as follows:$75=3&#215;5&#215;5$Now, factorise $45$ as follows:$45=3&#215;3&#215;5$We know that $HCF$ is the highest common factor.Since the common factor between the numbers $75$ and $45$ is $3\times 5=15$.Hence, the number of banquets that can be formed is $15$ and total number of flowers are $5+3=8$.

Question 6

What is the greatest number which divides $392,486$ and $627$ so as to leave the same remainder in each case?

Accepted Answer

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Question 7

Express $3825$ as a product of prime factors

Accepted Answer

Factorise $3825$ as follows:$3825=3&#215;3&#215;5&#215;5&#215;17=3^{ 2 }&#215;5^{ 2 }&#215;17$Since, $3,5$ and $17$ are prime numbers as these are the numbers that are only divisible by themselves. Hence, $3825=3^{ 2 }&#215;5^{ 2 }&#215;17$

Question 8

Use Euclids division algorithm to find the HCF of:(i) $135$ and $225$(ii) $196$ and $38220$(iii) $867$ and $225$

Accepted Answer

(i) $135$ and $225$Step 1: First find which integer is larger.$225 > 135$Step 2: Then apply the Euclid's division algorithm to $225$ and $135$ to obtain $225 = 135 \times 1 + 90$Repeat the above step until you will get remainder as zero.Step 3: Now consider the divisor $135$ and the remainder $90,$ and apply the division lemma to get$135 = 90 \times 1 + 45$$90 = 2 \times 45 + 0$Since the remainder is zero, we cannot proceed further.Step 4: Hence, the divisor at the last process is $45$So, the H.C.F. of $135$ and $225$ is $45.$(ii) $196$ and $38220$ Step 1: First find which integer is larger.$38220 > 196$Step 2: Then apply the Euclid's division algorithm to 38220 and 196 to obtain$38220 = 196 \times 195 + 0$Since the remainder is zero, we cannot proceed further.Step 3: Hence, the divisor at the last process is 196.So, the H.C.F. of $196$ and $38220$ is $196.$(iii) $867$ and $225$Step 1: First find which integer is larger.$867 > 255$Step 2: Then apply the Euclid's division algorithm to $867$ and $255$ to obtain$867 = 255 \times 3 + 102$Repeat the above step until you will get remainder as zero.Step 3: Now consider the divisor $225$ and the remainder $102,$ and apply the division lemma to get$255 = 102 \times 2 + 51$$102 = 51 \times 2 = 0$Since the remainder is zero, we cannot proceed further.Step 4: Hence the divisor at the last process is 51.So, the H.C.F. of $867$ and $255$ is $51.$

Question 9

Prove that the product of three consecutive positive integers is divisible by 6.

Accepted Answer

Let the three consecutive positive integers be $n$, $n+1$ and $n+2$.Whenever a number is divided by $3$, the remainder obtained is either $0,1$ or $2$.Therefore, $n=3p$ or $3p+1$ or $3p+2$, where $p$ is some integer.If $n=3p$, then $n$ is divisible by $3$.If $n=3p+1$, then $n+2=3p+1+2=3p+3=3(p+1)$ is divisible by $3$.If $n=3p+2$, then $n+1=3p+2+1=3p+3=3(p+1)$ is divisible by $3$.So, we can say that one of the numbers among $n, n+1$ and $n+2$ is always divisible by $3$ that is:$n(n+1)(n+2)$ is divisible by $3$.Similarly, whenever a number is divided by $2$, the remainder obtained is either $0$ or $1$.Therefore, $n=2q$ or $2q+1$, where $q$ is some integer.If $n=2q$, then $n$ and $n+2=2q+2=2(q+1)$ is divisible by $2$.If $n=2q+1$, then $n+1=2q+1+1=2q+2=2(q+1)$ is divisible by $2$.So, we can say that one of the numbers among $n$, $n+1$ and $n+2$ is always divisible by $2$.Since, $n(n+1)(n+2)$ is divisible by $2$ and $3$.Hence, $n(n+1)(n+2)$ is divisible by $6$.

Question 10

Show that any positive odd integer is of the form $6q + 1$, or $6q + 3$, or $6q + 5$, where $q$ is some integer.

Accepted Answer

Using Euclid division algorithm, we know that $a = bq + r,$ $0 \leq r \leq b$ ----(1)Let $a$ be any positive integer and $b = 6$.Then, by Euclid&#8217;s algorithm, $a = 6q + r$ for some integer $q\geq 0$, and $r = 0, 1, 2, 3, 4, 5$ ,$\:or$ $0 &#8804; r < 6$.Therefore, $a = 6q \:or\: 6q + 1 \:or\: 6q + 2 \:or\: 6q + 3 \:or\: 6q + 4 \:or\: 6q + 5 $$6q+0: 6$ is divisible by $2$, so it is an even number.$6q+1: 6$ is divisible by $2$, but $1$ is not divisible by $2$ so it is an odd number.$6q+2: 6$ is divisible by $2$, and $2$ is divisible by $2$ so it is an even number.$6q+3: 6$ is divisible by $2$, but $3$ is not divisible by $2$ so it is an odd number.$6q+4: 6$ is divisible by $2$, and $4$ is divisible by $2$ so it is an even number.$6q+5: 6$ is divisible by $2$, but $5$ is not divisible by $2$ so it is an odd number.And therefore, any odd integer can be expressed in the form $6q + 1\:or\: 6q + 3\; or\: 6q + 5$

Real Numbers

Maths
8317 Views

Use Euclids division lemma to show that the cube of any positive integer is of the form $9 m, 9 m + 1$ or $9 m + 8 .$

Use Euclid's division lemma to show that the square of any positive integer is either of the form $3 m$ or $3 m + 1$ for some integer m, but not of the form $3 m + 2$ .

The length and breadth of a rectangular field is 110 m and 30 m respectively. Calculate the length of the longest rod which can measure the length and breath of the field exactly.

Show that any positive even integer is of the form $4 q$ or $4 q + 2$ , where $q$ is a whole number.

There are 75 rose and 45 lily flowers. These are to be made into bouquets containing both the flowers. All the bouquets should contain the same number of flowers. Find the number of bouquets with maximum number of flowers that can be formed and the number of flowers in them.

What is the greatest number which divides $392, 486$ and $627$ so as to leave the same remainder in each case?

Express $3825$ as a product of prime factors

Use Euclids division algorithm to find the HCF of:
(i) $135$ and $225$

(ii) $196$ and $38220$
(iii) $867$ and $225$

Prove that the product of three consecutive positive integers is divisible by 6.

Show that any positive odd integer is of the form $6 q + 1$ , or $6 q + 3$ , or $6 q + 5$ , where $q$ is some integer.

Real Numbers

Maths 8317 Views

Use Euclids division lemma to show that the cube of any positive integer is of the form 9m,9m+1 or 9m+8.

Use Euclid's division lemma to show that the square of any positive integer is either of the form 3m or 3m+1 for some integer m, but not of the form 3m+2.

The length and breadth of a rectangular field is 110 m and 30 m respectively. Calculate the length of the longest rod which can measure the length and breath of the field exactly.

Show that any positive even integer is of the form 4q or 4q+2, where q is a whole number.

There are 75 rose and 45 lily flowers. These are to be made into bouquets containing both the flowers. All the bouquets should contain the same number of flowers. Find the number of bouquets with maximum number of flowers that can be formed and the number of flowers in them.

What is the greatest number which divides 392,486 and 627 so as to leave the same remainder in each case?

Express 3825 as a product of prime factors

Use Euclids division algorithm to find the HCF of:(i) 135 and 225 (ii) 196 and 38220(iii) 867 and 225

Prove that the product of three consecutive positive integers is divisible by 6.

Show that any positive odd integer is of the form 6q+1, or 6q+3, or 6q+5, where q is some integer.

Maths
8317 Views

Use Euclids division lemma to show that the cube of any positive integer is of the form $9 m, 9 m + 1$ or $9 m + 8 .$

Use Euclid's division lemma to show that the square of any positive integer is either of the form $3 m$ or $3 m + 1$ for some integer m, but not of the form $3 m + 2$ .

Show that any positive even integer is of the form $4 q$ or $4 q + 2$ , where $q$ is a whole number.

What is the greatest number which divides $392, 486$ and $627$ so as to leave the same remainder in each case?

Express $3825$ as a product of prime factors

Use Euclids division algorithm to find the HCF of:
(i) $135$ and $225$

(ii) $196$ and $38220$
(iii) $867$ and $225$

Show that any positive odd integer is of the form $6 q + 1$ , or $6 q + 3$ , or $6 q + 5$ , where $q$ is some integer.