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Find 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 lessUse Euclid’s division algorithm to find the HCF of 867 and 255
867 can be written in multiples of 255 as below. 867 = 255 * 3 + 102 255 = 102 * 2 + 51 102 = 51 * 2 + 0 As you can see the remainder has become zero now and the highest common factor is 51.
867 can be written in multiples of 255 as below.
867 = 255 * 3 + 102
255 = 102 * 2 + 51
102 = 51 * 2 + 0
As you can see the remainder has become zero now and the highest common factor is 51.
See lessUse Euclid’s division algorithm to find the HCF of 135 and 225
225 = 135*1+90 135=90*1+45 90=45*2+0 Therefore HCF is 45
225 = 135*1+90
135=90*1+45
90=45*2+0
Therefore HCF is 45
See lessUse Euclid’s division algorithm to find the HCF of 196 and 38220
According to Euclid's theorem x = y*q + r. 38220 = 196×195 + 0 So the HCF is 196.
According to Euclid’s theorem x = y*q + r.
38220 = 196×195 + 0
So the HCF is 196.
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