Question

Find the smallest number which when divided by 8, 9, 10, 15, 20 gives a remainder of 5 every time.

Answer 1

✰ AnSwer :

• The smallest number which when divided by 8, 9, 10, 15, 20 gives a remainder of 5 every time is 365.

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☯ GiVen :

• Numbers are 8 , 9 , 10 , 15 , 20

☯ To Find :

• We need to find the smallest number when divided by 8 , 9 , 10 , 15 , 20 leaves remainder as 5.

☯ SoluTion :

• Let us find the LCM of these numbers.

★ By prime factorization method,

• 8 = 2 × 2 × 2
• 9 = 3 × 3
• 10 = 2 × 5
• 15 = 3 × 5
• 20 = 2 × 2 × 5

★ LCM = 2 × 5 × 2 × 3 × 2 × 3 = 360

So,

➠Required number = LCM + 5

➠Required number = 360 + 5

➠Required number = 365

Therefore, 365 is the smallest number when divided by 8 , 9 , 10 , 15 , 20 gives remainder as 5.

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Answer 2

$\sf{\underline{\boxed{\green{\large{\bold{ Given}}}}}}$

• Numbers are 8 , 9 , 10 , 15 , 20
• reminder left everytime = 5

$\sf{\underline{\boxed{\green{\large{\bold{ To\: find}}}}}}$

• smallest number that divided by 8 , 9 , 10 , 15 , 20 leaving remainder as 5.

$\sf{\underline{\boxed{\green{\large{\bold{ solution}}}}}}$

• LCM of the given terms

. | 8 , 9 , 10 , 15 , 20

2 | 4 , 9 , 5 , 15 , 10

2 | 2 , 9 , 5 , 15 , 5

2 | 1 , 9 ,5 ,15 , 5

3 | 1 , 3 , 5 , 5 , 5

3 | 1 , 1 , 5 , 5 , 5

5 | 1 , 1 , 1 , 1 , 1

LCM = 2 x 2 x 2 x 3 x 3 x 5

LCM = 8 x 9 x 5

LCM = 72 x 5

LCM = 360

So,

⟹Required number = LCM + 5

⟹Required number = 360 + 5

⟹Required number = 365

$\sf{\underline{\boxed{\green{\large{\bold{ Answer}}}}}}$

Therefore, 365 is the smallest number which is divisible by 8 , 9 , 10 , 15 , 20 leaving remainder as 5.

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