Solution: Since, Both Sonia and Ravi move a similar way and simultaneously, the technique to make when they will meet again at the beginning stage is LCM of 18 and 12. Consequently, LCM(18,12) =...
Clarify why 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 are composite numbers.
Solution: By the meaning of composite number, we know, assuming a number is composite, it implies it has factors other than 1 and itself. In this way, for the given articulation; 7 × 11 × 13 + 13...
Check whether 6n can end with the digit 0 for any regular number n.
Solution: If the number 6n closures with the digit zero (0), then, at that point it ought to be distinct by 5, as we probably are aware any number with unit place as 0 or 5 is detachable by 5. Prime...
Considering that HCF (306, 657) = 9, discover LCM (306, 657).
Solutiont: As we realize that, HCF×LCM=Product of the two given numbers Accordingly, 9 × LCM = 306 × 657 LCM = (306×657)/9 = 22338 Consequently, LCM(306,657) = 22338
Discover the LCM and HCF of the accompanying numbers by applying the great factorisation technique.
(I) 12, 15 and 21 (ii) 17, 23 and 29 (iii) 8, 9 and 25 Solution: (I) 12, 15 and 21 Composing the result of prime components for every one of the three numbers, we get, 12=2×2×3 15=5×3 21=7×3...
Discover the LCM and HCF of the accompanying sets of numbers and check that LCM × HCF = result of the two numbers.
(I) 26 and 91 (ii) 510 and 92 (iii) 336 and 54 Solution: (I) 26 and 91 Communicating 26 and 91 as result of its excellent elements, we get, 26 = 2 × 13 × 1 91 = 7 × 13 × 1 In this manner, LCM (26,...
Express each number as a result of its superb elements:
(I) 140 (ii) 156 (iii) 3825 (iv) 5005 (v) 7429 Solution (I) 140 By Taking the LCM of 140, we will get the result of its great factor. Subsequently, 140 = 2 × 2 × 5 × 7 × 1 = 22×5×7 (ii) 156 By...