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Published on: 16/09/2019
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1.
An army contingent of 104 members is to march behind an army band of 96 members in a parade. The two groups are to march in the same number of columns. What is the maximum number of columns in which they can march?
2.
The length, breadth and height of a room are 8m 50 cm, 6 m 25 cm and 4 m 75 cm respectively. Find the length of the longest rod that can measure the dimensions of the room exactly.
3.
Check whether (15)n can end with digit 0 for any \(\\ \\ n\varepsilon N\)
4.
Can two numbers have 15 as their HCF and 175 as their LCM ? Give reasons.
5.
Explain whether \(3\times 12\times 101+4\) is a prime number or a composite number
6.
Complete the following factor tree and find the composite number x :
7.
Complete the following factor tree and find the composite number x.
8.
Given that HCF (306, 1,314) = 18. Find LCM (306, 1,314).
9.
Find HCF of the numbers given below: k, u, 3k, 4k and 5k, where k is any positive integer.
10.
Find the HCF of 52 and 117 and also find the values of x and y, if it express in the form 52x + 117y.
11.
Find (HCF x LCM) for the numbers 100 and 190.
12.
Find the LCM of x and y, if xy = 180 and HCF of (x, y) = 5
13.
If n is an odd integer, then show that n2 - 1 is divisible by 8.
14.
A rational number in its decimal expansion is 327.7081. What can you say about the prime factors of q. when this number is expressed in the form \(\frac{p}{q}\)? Give reason.
15.
Write the HCF and LCM of the smallest odd composite number and the smallest odd prime number. If an odd number p divides q2, then will it divide q3 also? Explain.
1.
Let the number of columns be x.
x is the largest number, which should divide both 104 and 96
104 = 96 x 1 + 8 1
96 = 8 x 12 + 0
∴ HCF of 104 and 96 is 8
Hence, 8 columns are required.
2.
Length = 8 m 50 cm = 850 cm
breadth = 6 m 25 cm = 625 cm
height = 4 m 75 cm = 475 cm
length of the longest rod is equal to

HCF (625, 850) = 25
∵ 25 divides 475
∴ HCF(62, 850, 475)=25
3.
(15)n can end with the digit 0 only if (15)" is divisible by 2 and 5.
But prime factors of (15)n are 3n x 5n
By Fundamental theorem of Arithmetic, there is no natural number II for which (15)n ends with the digit zero.
4.
No.
15 does not divide 175.
LCM of two numbers should be exactly divisible by their HCF.
∴ Two numbers cannot have their HCF as 15 and LCM as 175
5.
\(3\times 12\times 101+4=4(3\times 3\times 101+1)\)
= 4(909+1)
= 4(910)
= a composite number
[∵ Product of more than two factors]
6.

∴ Composite number, x = 6762
7.
y = 5 x 13 = 65
and x = 3 x 195 = 585
8.
Given HCF (306, 1,314) =18
LCM (306, 1,314) = ?
Let a = 306
b = 1,314
We know that
a x b = LCM (a, b) x HCF (a, b)
⇒ 306 x 1,314 = LCM (a, b) x 18
\(\Rightarrow \quad LCM(a,b)=\frac { 306\times 1,314 }{ 18 } \)
∴ LCM (306, 1,314) = 22,338
9.
HCF of K
k.2
k.3
k.22
k.5 is k
10.
13; x = - 2, y =1
11.
19000
12.
36
13.
Let a = n2 - 1, where n = 1, 3, 5, ...
At n = 1, then a = (1)2 - 1 = 1 - 1 = 0
At n = 3, then a = (3)2 - 1 = 9 - 1 = 8
At n = 5, then a = (5)2 - 1 = 25 - 1 = 24
which is divisible by 8.
Hence, n is an odd integer.
14.
Here, 327.7081 is terminating. So, it represents a rational number.
Thus, \(327.7081=\frac { 3277081 }{ 10000 } =\frac { p }{ q } \)
Here, q = 104 = 2 x 2 x 2 x 2 x 5 x 5 x 5 x 5
= 24 x 54 = (2 x 5)4
So, the prime factors of q are 2 and 5.
15.
\(\because \) smallest odd composite number = 9
and smallest odd prime number = 3.
\(\therefore \) HCF of 9 and 3 = 3
and LCM of 9 and 3 = 9
Now, if an odd number p divides q2, then p is one of the factors of q2, i.e. q2 = pm, for some integer m. .... (i)
Now, q3 = q2 . q \(\Rightarrow \) q3 = pm . q [from Eq.(i)]
\(\Rightarrow \) q3 = p (mq)
\(\Rightarrow \) p is a factor of q3 also \(\Rightarrow \) p divides q3
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