Exercise 1.2
1. Express each number as a product of its prime
factors:
(i) 140
= 22 x
5 x 7
(ii) 156 = 22
x 3 x 13
(iii)
3825 = 32
x 52 x 17
(iv)
5005 = 5
x 7 x
11 x 13
(v) 7429 =
17 x 19 x 23
2. Find the LCM and HCF of the following pairs of
integers. Verify that
LCM x HCF = product of the two numbers.
LCM x HCF = product of the two numbers.
(i) 26 and
91
Solution
: Prime factors are
26 =
2 x 13
91 =
7 x 13
HCF (26, 91) = 13
LCM (26, 91)
= 2 x
7 x 13 =
182
Verification:
LCM (26,
91) x
HCF (26, 91) = 26
x 91
182 x
13 = 26
x 91
2366 = 2366
(ii) 510 and
92
Solution
: Prime factors are
510 =
2 x 3 x 5
x 17
92 = 2 x 2 x 23 = 22 x 23
HCF (510, 92) = 2
LCM (510, 92)
= 22 x 3
x 5 x 17 x 23 = 23460
Verification:
LCM (510,
92) x
HCF (510, 92) = 510 X 92
23460 x 2 = 510 X 92
46920 = 46920
(iii) 336 and 54
Solution
: Prime factors are
336 = 24 x 3 x 7
54 = 2 x 33
HCF (336, 54) = 2 x
3 = 6
LCM (336, 54)
= 24 x 33 x 7 = 3024
Verification:
LCM (336,
54) x
HCF (336, 54) = 336
x 54
3024 x 6 = 336 x 54
18144 = 18144
3. Find the
HCF and LCM of the following integers by applying the prime factorization
method.
(i) 12, 15
and 21
Solution
: Prime factors are,
12 = 22 x 3
15 =
3 x 5
21 =
3 x 7
HCF (12, 15, 21)
= 3
LCM (12, 15, 21)
= 22 x
3 x 5
x 7 = 420
(ii) 17, 23
and 29
Solution
: Prime factors are,
17 = 1
x 17
23 = 1 x 23
29 = 1 x 29
HCF (17, 23, 29)
= 1
LCM (17, 23, 29)
= 17 x 23 x 29 = 11339
(iii) 8, 9
and 25
Solution
: Prime factors are,
8 = 23
9 = 32
25 = 52
HCF (8, 9, 25)
= 1
LCM (8, 9, 25)
= 23 x 32 x 52
= 1800
4. Given that HCF (306, 657) = 9, find the LCM ( 306, 657)
Solution
:
LCM (
306, 657) = 306 x 657 = 306 x 657
= 34 x 657 = 22338
HCF(306, 657) 9
5. Check whether 6n can end with the
digit zero for any natural number n.
Solution : If the number 6n,
for any n, were to end with the digit zero, then it would be divisible by 5
i.e., the prime factorization of 6n would contain the prime 5. This
is not possible because,
6n = (2 x 3) n
So, the primes in the factorization of 6n
is 2 and 3.
So, the uniqueness of the fundamental theorem of
arithmetic guarantees that there are no other primes in the factorization of 6n.
So, there is no natural number n for which 6n ends with the digit
zero.
6. Explain
why 7 x
11 x 13 +
13 and 7
x 6 x 5 x
4 x 3
x 2 x
1 + 5 are composite numbers
Solution :
Consider 7 x 11
x 13 + 13
= 13
( 7 x 11 + 1 )
= 13
x 78 =
13 x 13
x 2 x 3
The prime
factors are 2, 3 and 13. As it contains
these prime factors, it cannot be prime. Hence it is a composite number.
Also consider,
7 x 6
x 5 x 4 x
3 x 2
x 1 + 5
= 5 (7 x
6 x 4
x 3 x
2 x 1 + 1) = 5
x 1009
The prime
factors are 5 and 1009. As it contains these prime factors, it cannot be prime.
Hence it is a composite number.
7. There is a
circular path around a sports field. Sonia takes 18 minutes to drive one round
of the field while Ravi takes 12 minutes for the same. Suppose they both start
at the same point at the same time, and go in the same direction. After how
many minutes will they meet again?
Solution
: Given that Sonia takes 18 minutes to drive
one round of the field while Ravi takes 12 minutes for the same. Also, both are
moving in the same direction.
i.e., Ravi has taken less time when compared to
Sonia to drive one round of the field. We need to find the LCM of the time
taken by Sonia and Ravi to find the time at which they meet again,
Prime factors of 12 = 22 x 3
and that of 18
= 2 x 32
\
LCM ( 12, 18 ) = 22 x 32 = 4
x 9 = 36 minutes
\
Both Ravi and Sonia will meet each other after 36 minutes.
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