The least number which when divided by 4 5 and 6 leaves remainder 1.2 and 3 respectively is

Solution:

The required number when divides 285 and 1249, should leave remainder 9 and 7 respectively,

285 – 9 = 276 and 1249 -7 = 1242 can divide them exactly.

The required number is equivalent to the H.C.F. of 276 and 1242.

Applying Euclid’s division lemma on 276 and 1242 , we get

1242 = 276 x 4 + 138

276 = 138 x 2 + 0. 

Since, the remainder is now 0,

Therefore, the H.C.F.(req. number) = 138

Question 12. Find the largest number which exactly divides 280 and 1245 leaving remainders 4 and 3, respectively.

Solution:

The required number when divides 280 and 1245, should leave remainder 4 and 3 respectively,

280 – 4 = 276 and 1245 – 3 = 1242 has to be exactly divisible by the number.

The required number is equivalent to the H.C.F. of 276 and 1242.

Applying Euclid’s division lemma on 276 and 1242 , we get

1242 = 276 x 4 + 138

276 = 138 x 2 + 0 (the remainder becomes 0 here)

Since, the remainder is now 0,

Therefore, the H.C.F.(req. number) = 138

Question 13. What is the largest number which that divides 626, 3127 and 15628 and leaves remainders of 1, 2 and 3 respectively?

Solution:

The required number when divides 626, 3127 and 15628, should leave remainder 1,2 and 3 respectively,

626 – 1 = 625, 3127 – 2 = 3125 and 15628 – 3 = 15625 has to be exactly divisible by the number.

The required number is equivalent to the H.C.F of 625, 3125 and 15625.

Considering 625 and 3125 first , we apply Euclid’s division lemma

3125 = 625 x 5 + 0

∴ H.C.F (625, 3125) = 625

Applying Euclid’s division lemma on 15625 and 625 , we get

15625 = 625 x 25 + 0

Now, 

∴ H.C.F. (625, 3125, 15625) = 625

Question 14. Find the greatest number that will divide 445,572 and 699 leaving remainders 4, 5 and 6 respectively.

Solution:

The required number when divides 445,572 and 699, should leave remainder 4, 5 and 6 respectively,

445 – 4 = 441, 572 – 5 = 567 and 699 – 6 = 693 has to be exactly divisible by the number.

The required number is equivalent to the H.C.F of 441, 567 and 693.

Applying Euclid’s division lemma on 441 and 567 , we get

567 = 441 x 1 + 126

441 = 126 x 3 + 63

126 = 63 x 2 + 0.

∴ H.C.F (441 and 567) = 63

Applying Euclid’s division lemma on 63 and 693 , we get

693 = 63 x 11 + 0.

Now,

∴ H.C.F. (441, 567 and 693) = 63

Question 15. Find the greatest number which divides 2011 and 2623 leaving remainders 9 and 5 respectively.

Solution:

The required number when divides 2011 and 2623 should leave remainder  9 and 5  respectively,

2011 – 9 = 2002 and 2623 – 5 = 2618 has to be exactly divisible by the number.

The required number is equivalent to the H.C.F. of 2002 and 2618

Applying Euclid’s division lemma, we get,

2618 = 2002 x 1 + 616

2002 = 616 x 3 + 154

616 = 154 x 4 + 0. 

The remainder becomes 0, 

Therefore, the H.C.F. (2002, 2618) = 154

Solution:

The required number when divides 1251, 9377 and 15628 should leave remainder 1, 2 and 3 respectively,

1251 – 1 = 1250, 9377 – 2 = 9375 and 15628 – 3 = 15625 has to be exactly divisible by the number.

The required number is equivalent to the H.C.F of 1250, 9375 and 15625.

Applying Euclid’s division lemma on  1250, 9375 , we get,

9375 = 1250 x 7 + 625

1250 = 625 x 2 + 0

∴ H.C.F (1250, 9375) = 625

Applying Euclid’s division lemma on  625 and 15625 , we get,

15625 = 625 x 25 + 0

Since, the remainder is now 0,

∴ H.C.F. (1250, 9375, 15625) = 625

So, the required number is 625.

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?

Solution:

Number of chocolates of 1st brand in a pack = 24

Number of chocolates of 2nd brand in a pack = 15.

The least number of both brands of chocolates is equivalent to their LCM.

L.C.M. of 24 and 15 = 2 x 2 x 2 x 3 x 5 = 120

Now,

The number of packets of 1st brand to be bought = 120 / 24 = 5

The number of packets of 2nd brand to be bought = 120 / 15 = 8

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?

Solution:

Size of bathroom = 10 ft. by 8 ft.

Converting from ft. to inch.

= (10 x 12) inch by (8 x 12) inch 

= 120 inch by 96 inch

The largest size of tile required is equal the HCF of the numbers 120 and 96.

Applying Euclid’s division lemma on 120 and 96 ,we get,

120 = 96 x 1 + 24

96 = 24 x 4 + 0

⇒ HCF = 24

Thus, the largest size of tile which required is 24 inches.

Also,

Number of tiles required = (area of bathroom) / (area of a tile)

= (120 x 96) / (24×24)

= 5 x 4

We obtain,

= 20 tiles

Therefore, 20 tiles each of size 24 inch by 24inch are required to be cut.

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?

Solution:

We have,

Number of pastries = 15

Number of biscuit packets = 12

The required number of boxes containing an equal number of both pastries and biscuits will be equal to the HCF of the numbers 15 and 12.

Applying Euclid’s division lemma on 15 and 12, we get

15 = 12 x 1 + 3

12 = 3 x 4 = 0

Therefore, the required boxes = 3

Now,

∴ Each box will contain 15/3 = 5 pastries and 12/3 = 4 biscuit packs.

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?

Solution:

We have,

Number of goats = 105

Number of donkeys = 140

Number of cows = 175

The largest number of animals in one trip is equivalent to the HCF (105, 140 and 175).

Applying Euclid’s division lemma on 105 and 140, we get

140 = 105 x 1 + 35

105 = 35 x 3 + 0

Therefore, the HCF (105 and 140) = 35

Applying Euclid’s division lemma on 35 and 175, we get

175 = 35 x 5 +0

Hence, the HCF (105, 140, 175) = 35.

Therefore, 35 animals went on each trip.

Solution:

We have,

Length of the room = 8m 25 cm = 825 cm (in cm)

Breadth of the room = 6m 75cm = 675 cm

Height of the room = 4m 50cm = 450 cm

The longest rod which can measure the room will be exactly equivalent to the HCF of given measurements 825, 675 and 450.

Applying Euclid’s division lemma on 675 and 450, we get

675 = 450 x 1 + 225

450 = 225 x 2 + 0

Therefore, the HCF (675, 450) = 225

Applying Euclid’s division lemma on 225 and 825, we get

825 = 225 x 3 + 150

225 = 150 x 1+75

Applying Euclid’s division lemma on 150 and 75, we get

150 = 75 x 2 + 0 

Thus, HCF (225, 825) = 75.

Also,

 HCF of 825, 675 and 450 is 75.

Therefore, the length of the longest required rod is 75 cm.

Question 22. Express the HCF of 468 and 222 as 468x + 222y where x, y are integers in two different ways.

Solution:

We have the integers, 468 and 222, where 468 > 222

Applying Euclid’s division lemma on 468 and 222, we get

468 = 222 x 2 + 24……… (1)

Applying Euclid’s division lemma on 222 and remainder 24, we get

222 = 24 x 9 + 6………… (2)

Applying Euclid’s division lemma on 24 and remainder 6, we get

24 = 6 x 4 + 0……………. (3)

The remainder now is 0.

Therefore, the H.C.F. of 468 and 222

We can express the HCF as a linear combination of 468 and 222, by

6 = 222 – 24 x 9 [from (2)]

= 222 – (468 – 222 x 2) x 9 [from (1)]

= 222 – 468 x 9 + 222 x 18

6 = 222 x 19 – 468 x 9 = 468(-9) + 222(19)

∴ 6 = 468x + 222y, where x = -9 and y = 19.


Answer

The least number which when divided by 4 5 and 6 leaves remainder 1.2 and 3 respectively is
Verified

Hint: 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.