``` I give you 20 point but give correct answer``` 2. Find the LCM and HCF of the following pairs of integers and verify that LCMx HCF = product of the two numbers.
(1) 26 and 91
(ii) 510 and 92
(ii) 336 and 54
3. Find the LCM and HCF of the following integers by applying the prime factorisation method.
(1) 12, 15 and 21
(ii) 17, 23 and 29
(iii) 8,9 and 25
Answers
Answer:
2)1) 26 - 2 × 13. LCM -- 2 × 7 × 13 = 182
91 - 7 × 13. HCF --- 13
* LCM × HCF = 1st no. × 2nd no.
182 × 13 = 26 × 91
2366. = 2366
2 ) 510 -- 2×3×5×17. LCM --- 2^2×3×5×17×23 =
23460
92 --- 2^2 × 23. HCF --- 2
* LCM × HCF = 1st no. × 2nd no.
23460 × 2 = 510 × 92
46920. = 46920
3) 336 --- 2^4 × 3 × 7 LCM --- 2^4 × 3^3 × 7 =
3024
54 ------ 2× 3^3. HCF ---- 2 × 3 = 6
* LCM × HCF = 1st no. × 2nd no.
3024 × 6 = 336 × 54
18144. = 18144
3) 1) 12 --- 2^2 × 3. LCM --- 2^2 × 3 × 5 × 7 =
15----- 3 × 5 420
21 ----- 3 × 7. HCF --- 3
2) 17 - 17 × 1. LCM --- 17 × 23 × 29 =
23 - 23 × 1. 11339
29 - 29 × 1. HCF ---- 1
3) 8 - 2^3. LCM --- 2^3 × 3^2 × 5^2 =
9 - 3^2. 1800
25 - 5^2. HCF ---- 1
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-- :
Using prime factorisation method:
(i) 12, 15 and 21
Factor of 12=2×2×3
Factor of 15=3×5
Factor of 21=3×7
HCF (12,15,21)=3
LCM (12,15,21)=2×2×3×5×7=420
(ii) 17, 23 and 29
Factor of 17=1×17
Factor of 23=1×23
Factor of 29=1×29
HCF (17,23,29)=1
LCM (17,23,29)=1×17×23×29=11,339
(iii) 8, 9 and 25
Factor of 8=2×2×2×1
Factor of 9=3×3×1
Factor of 25=5×5×1
HCF (8,9,25)=1
LCM (8,9,25)=2×2×2×3×3×5×5=1,800