Lost your password? Please enter your email address. You will receive a link and will create a new password via email.
Please briefly explain why you feel this question should be reported.
Please briefly explain why you feel this answer should be reported.
Please briefly explain why you feel this user should be reported.
Show that any positive odd integer is of the form 6q+1, or 6q+3, or 6q+5, where q is some integer.
Sol:- Let us start with taking a, where a is a positive odd integer. We apply the division algorithm with a and b = 6. Since 0 ≤ r < 6, the possible remainders are 0, 1, 2 , 3, 4 and 5. That is, a can be 6q, or 6q + 1, or 6q + 2, or 6q + 3 or 6q + 4 or 6q + 5, where q is the quotient. However, siRead more
Sol:-
Let us start with taking a, where a is a positive odd integer. We apply the division algorithm with a and b = 6.
Since 0 ≤ r < 6, the possible remainders are 0, 1, 2 , 3, 4 and 5.
That is, a can be 6q, or 6q + 1, or 6q + 2, or 6q + 3 or 6q + 4 or 6q + 5, where q is the quotient.
However, since a is odd, a cannot be 6q or 6q + 2 or 6q + 4 (since they are divisible by 2). Therefore, any odd integer is of the form 6q + 1 or 6q + 3 or 6q + 5.
See lessA sweetseller has 420 kaju barfis and 130 badam barfis. She wants to stack them in such a way that each stack has the same number, and they take up the least area of the tray. What is the maximum number of barfis that can be placed in each stack for this purpose?
Sol:- This can be done by trial and error. But to do it systematically, we find HCF (420, 130). Then this number will give the maximum number of barfis in each stack and the number of stacks will then be the least. The area of the tray that is used up will be the least. Now, let us use Euclid’s algoRead more
Sol:-
This can be done by trial and error. But to do it systematically, we find HCF (420, 130). Then this number will give the maximum number of barfis in each stack and the number of stacks will then be the least. The area of the tray that is used up will be the least.
Now, let us use Euclid’s algorithm to find their HCF. We have :
420= 130×3+30
130= 30×4+10
30= 10×3+0
So, the HCF of 420 and 130 is 10.
See lessTherefore, the sweetseller can make stacks of 10 for both kinds of barfi.
Show that any positive odd integer is of the form 4q + 1 or 4q + 3, where q is some integer.
Sol:- Let us start with taking a, where a is a positive odd integer. We apply the division algorithm with a and b = 4. Since 0 ≤ r < 4, the possible remainders are 0, 1, 2 and 3. That is, a can be 4q, or 4q + 1, or 4q + 2, or 4q + 3, where q is the quotient. However, since a is odd, a cannot be 4Read more
Sol:-
Let us start with taking a, where a is a positive odd integer. We apply the division algorithm with a and b = 4.
Since 0 ≤ r < 4, the possible remainders are 0, 1, 2 and 3.
That is, a can be 4q, or 4q + 1, or 4q + 2, or 4q + 3, where q is the quotient.
However, since a is odd, a cannot be 4q or 4q + 2 (since they are both divisible by 2). Therefore, any odd integer is of the form 4q + 1 or 4q + 3.
See lessFind the LCM and HCF of 6 and 20 by the prime factorisation method.
Sol:-
Sol:-
See lessFind the HCF of 96 and 404 by the prime factorisation method. Hence, find their LCM.
Sol:- The prime factorisation of 96 and 404 gives : 96 = $2^5$ × 3, 404 = $2^2$ × 101 Therefore, the HCF of these two integers is $2^2$ = 4. Also,
Sol:-
The prime factorisation of 96 and 404 gives :
96 = × 3,
404 = × 101
Therefore, the HCF of these two integers is = 4.
Also,
See lessFind the HCF and LCM of 6, 72 and 120, using the prime factorisation method.
Sol:- We can write 6 , 72 , 120 as below 6 = 2 × 3 72 = $2^3$× $3^2$ (72 = 2×2×2×3×3) 120 = $2^3$×3×5 (120 = 2×2×2×3×5) Here, $2^1$ and $3^1$ are the smallest powers of the common factors 2 and 3 respectively. So, HCF(6,72,120)= $2^1$ × $3^1$ = 2×3 = 6 $2^3$, $3^2$ and $5^1$ are the greatest powersRead more
Sol:-
We can write 6 , 72 , 120 as below
6 = 2 × 3
72 = × (72 = 2×2×2×3×3)
120 = ×3×5 (120 = 2×2×2×3×5)
Here, and are the smallest powers of the common factors 2 and 3 respectively.
So, HCF(6,72,120)= × = 2×3 = 6
, and are the greatest powers of the prime factors 2, 3 and 5 respectively involved in the three numbers.
So, LCM(6,72,120)= × × =360
See lessShow that every positive even integer is of the form 2q, and that every positive odd integer is of the form 2q + 1, where q is some integer.
Sol:- Let a be any positive integer and b = 2. Then, by Euclid’s algorithm, a = 2q + r for some integer q ≥ 0, and r = 0 or r = 1, because 0 ≤ r < 2. So, a = 2q or 2q + 1. If a is of the form 2q, then a is an even integer. Also, a positive integer can be either even or odd. Therefore, any positivRead more
Sol:-
Let a be any positive integer and b = 2. Then, by Euclid’s algorithm,
a = 2q + r
for some integer q ≥ 0, and r = 0 or r = 1, because 0 ≤ r < 2. So, a = 2q or 2q + 1.
If a is of the form 2q, then a is an even integer. Also, a positive integer can be either even or odd. Therefore, any positive odd integer is of the form 2q + 1.
See lessUse Euclid’s algorithm to find the HCF of 4052 and 12576.
Sol:- Since 12576 > 4052 so, we will divide 12576 by 4052 and follow below steps till we will get 0 as a remainder. 12576 = 4052×3 + 420 4052 = 420×9 + 272 420 = 272×1 + 148 272 = 148×1 + 124 148 = 124×1 + 24 124 = 24×5 + 4 24 = 4×6 +0 The remainder has now become zero, so our procedure stops herRead more
Sol:- Since 12576 > 4052 so, we will divide 12576 by 4052 and follow below steps till we will get 0 as a remainder.
12576 = 4052×3 + 420
4052 = 420×9 + 272
420 = 272×1 + 148
272 = 148×1 + 124
148 = 124×1 + 24
124 = 24×5 + 4
24 = 4×6 +0
The remainder has now become zero, so our procedure stops here. Since the divisor at this stage is 4, the HCF of 12576 and 4052 is 4.
See less