- Question 12. Find the largest number which exactly divides 280 and 1245 leaving remainders 4 and 3, respectively.
- Question 13. What is the largest number which that divides 626, 3127 and 15628 and leaves remainders of 1, 2 and 3 respectively?
- Question 14. Find the greatest number that will divide 445,572 and 699 leaving remainders 4, 5 and 6 respectively.
- Question 15. Find the greatest number which divides 2011 and 2623 leaving remainders 9 and 5 respectively.
- Question 17. Two brands of chocolates are available in packs of 24 and 15 respectively. If I need to buy an equal number of chocolates of both kinds, what is the least number of boxes of each kind I would need to buy?
- Question 18. A mason has to fit a bathroom with square marble tiles of the largest possible size. The size of the bathroom is 10ft. by 8ft. What would be the size in inches of the tile required that has to be cut and how many such tiles are required?
- Question 19. 15 pastries and 12 biscuit packets have been donated for a school fete. These are to be packed in several smaller identical boxes with the same number of pastries and biscuit packets in each. How many biscuit packets and how many pastries will each box contain?
- Question 20. 105 goats, 140 donkeys and 175 cows have to be taken across a river. There is only one boat which will have to make many trips in order to do so. The lazy boatman has his own conditions for transporting them. He insists that he will take the same number of animals in every trip and they have to be of the same kind. He will naturally like to take the largest possible number each time. Can you tell how many animals went on each trip?
- Question 22. Express the HCF of 468 and 222 as 468x + 222y where x, y are integers in two different ways.
## Question 12. Find the largest number which exactly divides 280 and 1245 leaving remainders 4 and 3, respectively.
## Question 13. What is the largest number which that divides 626, 3127 and 15628 and leaves remainders of 1, 2 and 3 respectively?
## Question 14. Find the greatest number that will divide 445,572 and 699 leaving remainders 4, 5 and 6 respectively.
## Question 15. Find the greatest number which divides 2011 and 2623 leaving remainders 9 and 5 respectively.
## Question 17. Two brands of chocolates are available in packs of 24 and 15 respectively. If I need to buy an equal number of chocolates of both kinds, what is the least number of boxes of each kind I would need to buy?
## Question 18. A mason has to fit a bathroom with square marble tiles of the largest possible size. The size of the bathroom is 10ft. by 8ft. What would be the size in inches of the tile required that has to be cut and how many such tiles are required?
## Question 19. 15 pastries and 12 biscuit packets have been donated for a school fete. These are to be packed in several smaller identical boxes with the same number of pastries and biscuit packets in each. How many biscuit packets and how many pastries will each box contain?
## Question 20. 105 goats, 140 donkeys and 175 cows have to be taken across a river. There is only one boat which will have to make many trips in order to do so. The lazy boatman has his own conditions for transporting them. He insists that he will take the same number of animals in every trip and they have to be of the same kind. He will naturally like to take the largest possible number each time. Can you tell how many animals went on each trip?
## Question 22. Express the HCF of 468 and 222 as 468x + 222y where x, y are integers in two different ways.
Answer VerifiedHint: If we find the least common multiple of the numbers, we get the least possible number with remainder 0. So, now we find (Least common multiple + 1) as this leaves the remainder 1. Complete step-by-step answer: Now we will find Least Common Multiple of the following numbers: 2, 3, 4, 5 and 6.Firstly, we do prime factorization of the following numbers 2, 3, 4, 5 and 6Case 1: Prime factorization of 2As 2 is itself a prime number we write 2 = 2Case 2: Prime factorization of 3As 3 is also itself a prime number we write 3 = 3Case 3: Prime factorization of 4As 4 can be written in terms of prime number 24 = 2.2$4={{2}^{2}}$ Case 4: Prime factorization of 5As 5 is itself a prime number we write 5 = 5Case 5: Prime factorization of 6As 6 can be written in terms of 2 and 3 we write:6 = 2.3Now we will write all numbers with their prime factorized form, together:2 = 23 = 34 = 2 x 25 = 56 = 2 x 3So, we will remove the terms which are repeated. Now we will write the repeated terms: One of the 2 is repeated thrice (in 2, 4, 6). The number 3 is repeated twice (in 3, 6). By removing the repeated terms and writing them only once we get:Least Common Multiple = 2. 3. 2. 5By simplifying we get:Least common multiple = 6. 10Least common multiple = 60As 60 is divisible by the numbers 2, 3, 4, 5, 6 we get:a 0 remainder when 60 is divided by numbers 2, 3, 4, 5, 6.So, if we add 1 to the Least Common Multiple, we get:Least Common Multiple + 1 = 6161 when divided by 2, 3, 4, 5, 6 gives remainder 1.but the second question was the number must be divisible by 7.So, now we need to multiply an integer to least common multiple and then add 1, to make both conditions satisfy.\[Required\text{ }number=\left[ \left( Least\text{ }Common\text{ }Multiple \right).n \right]+1\] Case 1: n = 2By substituting n value in above equation, we get:Required number = (60 x 2) + 1 = 121121 is not divisible by 7.Case 2: n = 3By substituting n value in above equation, we get:Required number = (60 x 3) + 1 = 181181 is not divisible by 7.Case 3: n = 4By substituting the value of n in above equation, we get:Required number = (60 x 4) + 1 = 241241 is not divisible by 7.Case 4: n = 5By substituting the value of n in above equation, we get:Required number = (60 x 5) + 1 = 301301 is divisible by 7.Therefore 301 is the least number which leaves remainder as 1 when divided by 2, 3, 4, 5, 6 and leaves no remainder when divided by 7.Option (b) is correct.Note: While taking Least Common Multiple be careful to remove only repeated numbers.Getting an idea to add 1 to the Least common multiple is crucial. So, be careful while applying it. |