matyhs
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REAL NUMBERS
Exercise 1.1
Q.1. Use Euclid’s division algorithm to find the HCF of:
(i) 135 and 225
(ii) 196 and 38220
(iii) 867 and 255
Solution. (i) In 135 and 225, 225 is larger integer.
Using Euclid’s division algorithm,
[Where 135 is divisor, 90 is remainder]
225 = 135 × 1 + 90
Since, remainder 90 ≠ 0 , by applying Eudid’s division algorithm to 135 and 90
∴
135 = 90 × 1 + 45
Again since, remainder 45 ≠ 0 , by applying Eudid’s division algorithm to 90 and 45
∴
90 = 45 × 2 + 0
Now, the remainder is zero so, our procedure stops.
Hence, HCF of 135 and 225 is 45.
(ii) In 196 and 38220, 38220 is larger integer.
Using Euclid’s division algorithm,
38220 = 196 × 195 + 0
⇒ remainder = 0
The remainder is zero so our procedure stops.
Hence, HCF of 196 and 38220 is 196.
(iii) In 867 and 255, 867 is larger integer.
Using Euclid’s division algorithm
867 = 255 × 3 + 102 [Where 255 is divisor, 102 is remainder]
Since, remainder 102 ≠ 0 , by applying Eudid’s division algorithm to 255 and 102
∴ 255 = 102 × 2 + 51
Since, remainder 51 ≠ 0 , by applying Eudid’s division algorithm 102 and 51
So,
102 = 51 × 2 + 0 remainder = 0
The remainder is zero so our procedure stops.
Hence, HCF of 867 and 255 is 51.
Q.2. Show that any positive odd integer is of the form 6q + 1 or 6q + 3 or 6q + 5, where q is some integer.
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Solution. Let a be any positive odd integer. We apply Euclid’s division algorithm with a and b = 6 .
Since 0 ≤ r < 6, the positive remainders are 0, 1, 2, 3, 4 and 5.
i.e., a can be 6q or 6q + 1, or 6q + 2 , or 6q + 3 , or 6q + 4 , or 6q + 5 where q is the quotient. But, a is odd so a cannot be equal to 6q, 6q + 2, 6q + 4 which are even numbers. (divisible by 2)
⇒ Any odd integer is of the form 6q + 1 or 6q + 3 or 6q + 5.
Q.3. An army contingent of
CBSEPracticalSkills.com
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Edulabz International
REAL NUMBERS
Exercise 1.1
Q.1. Use Euclid’s division algorithm to find the HCF of:
(i) 135 and 225
(ii) 196 and 38220
(iii) 867 and 255
Solution. (i) In 135 and 225, 225 is larger integer.
Using Euclid’s division algorithm,
[Where 135 is divisor, 90 is remainder]
225 = 135 × 1 + 90
Since, remainder 90 ≠ 0 , by applying Eudid’s division algorithm to 135 and 90
∴
135 = 90 × 1 + 45
Again since, remainder 45 ≠ 0 , by applying Eudid’s division algorithm to 90 and 45
∴
90 = 45 × 2 + 0
Now, the remainder is zero so, our procedure stops.
Hence, HCF of 135 and 225 is 45.
(ii) In 196 and 38220, 38220 is larger integer.
Using Euclid’s division algorithm,
38220 = 196 × 195 + 0
⇒ remainder = 0
The remainder is zero so our procedure stops.
Hence, HCF of 196 and 38220 is 196.
(iii) In 867 and 255, 867 is larger integer.
Using Euclid’s division algorithm
867 = 255 × 3 + 102 [Where 255 is divisor, 102 is remainder]
Since, remainder 102 ≠ 0 , by applying Eudid’s division algorithm to 255 and 102
∴ 255 = 102 × 2 + 51
Since, remainder 51 ≠ 0 , by applying Eudid’s division algorithm 102 and 51
So,
102 = 51 × 2 + 0 remainder = 0
The remainder is zero so our procedure stops.
Hence, HCF of 867 and 255 is 51.
Q.2. Show that any positive odd integer is of the form 6q + 1 or 6q + 3 or 6q + 5, where q is some integer.
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Solution. Let a be any positive odd integer. We apply Euclid’s division algorithm with a and b = 6 .
Since 0 ≤ r < 6, the positive remainders are 0, 1, 2, 3, 4 and 5.
i.e., a can be 6q or 6q + 1, or 6q + 2 , or 6q + 3 , or 6q + 4 , or 6q + 5 where q is the quotient. But, a is odd so a cannot be equal to 6q, 6q + 2, 6q + 4 which are even numbers. (divisible by 2)
⇒ Any odd integer is of the form 6q + 1 or 6q + 3 or 6q + 5.
Q.3. An army contingent of