Math, asked by chaudharymehak2800, 6 hours ago

The smallest number which when divided by 17, 23 and 29 leaves a remainder 11 in each case is: (a) 493 (b) 11350 (c) 11339 (d) 667​

Answers

Answered by mathdude500
22

\large\underline{\sf{Solution-}}

Since, we have to find the smallest number which when divided by 17, 23 and 29 leaves a remainder 11 in each case.

So, let assume that the smallest number be 'x' which when divided by 17, 23 and 29 leaves a remainder 11 in each case.

So, it means

➢ x when divided by 17, leaves the remainder 11.

➢ x when divided by 23, leaves the remainder 11.

➢ x when divided by 29, leaves the remainder 11.

\bf\implies \:x - 11 \: is \: divisible \: by \: 17

\bf\implies \:x - 11 \: is \: divisible \: by \: 23

\bf\implies \:x - 11 \: is \: divisible \: by \: 29

Thus,

\bf\implies \:x - 11 \: is \: divisible \: by \: 17, \: 23, \: 29

\bf\implies \:x - 11 \:  = LCM( \: 17, \: 23, \: 29)

\bf\implies \:x - 11 \:  = 17 \times 23 \times 29

\bf\implies \:x - 11 \:  = 11339

\bf\implies \:x \:  = 11339 + 11

\bf\implies \:x \:  = 11350

Hence,

The smallest number is 11350 which when divided by 17, 23 and 29 leaves a remainder 11 in each case.

So, option (b) is correct.

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

MORE TO KNOW

1. HCF (a, b) × LCM(a, b) = a × b

2. LCM is always divisible by HCF as well as by numbers.

3. HCF always divides LCM as well as numbers.

4. HCF of two consecutive positive integers is always 1.

5. HCF of two consecutive even positive integers is 2.

6. HCF of two consecutive odd positive integers is 1.

7. LCM of prime numbers is equals to their product. It means if a and b are prime numbers, then LCM(a, b) = a × b.

Answered by RvChaudharY50
9

To Find :- The smallest number which when divided by 17, 23 and 29 leaves a remainder 11 in each case is : (a) 493 (b) 11350 (c) 11339 (d) 667

Concept used :- The least number which when divided by x, y and z leaves the same remainder ‘r’ each case is :-

  • (LCM of x, y and z) + r .

Solution :-

from above told concept,

→ Required number = (LCM of 17, 23 and 29) + 11

So, finding prime factors of 17, 23 and 29 we get,

→ 17 = 1 × 17

→ 23 = 1 × 23

→ 29 = 1 × 29

then,

→ LCM = 17 × 23 × 29 = 11339

therefore,

→ Required number = (LCM of 17, 23 and 29) + 11

→ Required number = 11339 + 11

→ Required number = 11350 (b) (Ans.)

Hence, The smallest number which when divided by 17, 23 and 29 leaves a remainder 11 in each case is equal to 11350 .

Learn more :-

Find the least number which when divide by 626, 618 and 676 leaves a remainder 3 in each

case.

https://brainly.in/question/23444967

find the smallest perfect square number that is divisible by 6 ,7 ,8 and 27

https://brainly.in/question/23226397

Similar questions