Now, HCF (56, 96)= 8
Applying Euclid’s division algorithm on 8 and 404, we get
Now, HCF (404, 8) = 4
Hence, H.C.F. of 56, 96 and 404 is 4. Ans.
Problems Based on Fundamental theorem of Arithmetic
Since,
1296 = 2 × 2 × 2 × 2 × 3 × 3 × 3 × 3
= 24 × 34
and 2520 = 2 × 2 × 2 × 3 × 3 × 5 × 7
= 23 × 32 × 5 × 7
∴ LCM = Product of each prime factor with highest powers
= 24 × 34 × 5 × 7 = 45360
i.e., LCM (1296, 2520)= 45360
H.C.F. = Product of common prime factors with lowest powers
= 23 × 32 = 8 × 9 = 72
i.e., H.C.F. (1296, 2520) = 72.
Using prime factorization method, Find the LCM of
(i) 12, 15, 20, 27 (ii) 21, 28, 36, 45.
Since, 12 = 2 × 2 × 3 = 22 × 3
15 = 3 × 5
20 = 2 × 2 × 5 = 22 × 5
and 27 = 3 × 3 × 3 = 33
∴ LCM = Product of each prime factor with highest power
= 22 × 33 × 5 = 540
i.e., LCM (12, 15, 20, 27) = 540.
(ii) 21, 28, 36, 45
Since, 21 = 3 × 7
28 = 2 × 2 × 7 = 22 × 7>
36 = 2 × 2 × 3 × 3 = 22 × 32
and 45 = 5 × 3 × 3 = 5 × 32
∴ LCM = Product of each prime factor with highest power
= 32 × 7 × 22 × 5
= 1260
i.e. LCM (21, 28, 36, 45)= 1260.
In order to arrange the books as required, we have to find the largest number that divides 105, 140 and 175 exactly.
Clearly, required number is the HCF of 105, 140 and 175.
Case I :
I. Finding HCF of 105 and 140 by applying Euclid’s division lemma, we get
II. Since, the remainder 35 ≠ 0, we apply division lemma to get
Since, the remainder at this stage is zero, so the divisor i.e., 35 at this stage is the HCF of 105 and 140.
Case II :
I. Finding the HCF of 35 and 175 by applying Euclid’s division lemma, we get
II. Since the remainder at this stage is zero, so the divisor i.e., 35 at this stage is the HCF of 35 and 175.
Thus, HCF of 105, 140) and 175 is 35.
Now,
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