Question

83. Find the least number which when divided separately by 15, 20, 36, and 48 leaves 3 as remainder in each case.​

Answer 1

Step-by-step explanation:

Find LCM of 15, 20, 36, and 48

$\begin{array}{r|l} 2 & 15,20,36,48 \\\cline{1-2} 3 & 15,10,18,24 \\\cline{1-2} 5 & 5,10,6,8 \\\cline{1-2} 2 & 1,2,6,8\\\cline{1-2} 3 & 1,1,3,4\\\cline{1-2}2&1,1,1,4 \\\cline{1-2}2&1,1,1,2 \\\cline{1-2} & 1,1,1,1 \end{array}$

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LCM = 2 × 3 × 5 × 2 × 3 × 2 × 2

$\longrightarrow$ 6 × 10 × 6 × 2

$\longrightarrow$ 60 × 12

$\longrightarrow$ 720

Least common multiple of 15, 20, 36, and 48 is 720.

Add 3 to their LCM

$\longrightarrow$ 720 + 3

$\longrightarrow$ 723

Therefore, 723 is the least number which when divided separately by 15, 20, 36, and 48 leaves 3 as remainder in each case.

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Verification:

• 723 ÷ 15

$\longrightarrow$ Quotient = 48

$\longrightarrow$ Remainder = 3

• 723 ÷ 20

$\longrightarrow$ Quotient = 36

$\longrightarrow$ Remainder = 3

• 732 ÷ 36

$\longrightarrow$ Quotient = 20

$\longrightarrow$ Remainder = 3

• 732 ÷ 48

$\longrightarrow$ Quotient = 15

$\longrightarrow$ Remainder = 3

Answer 2

Given :-

Numbers = 15,20,36 and 48 leaves 3 as remainder

To Find :-

Least number

Solution :-

At first, we need to find the LCM of 15,20,36 and 48. Then, we have to add 3 in it

(LCM of 15,20,36 and 48) + 3

By prime factorization

15 = 3 × 5

20 = 2 × 2 × 5

36 = 2 × 2 × 3 × 3

48 = 2 × 2 × 2 × 2 × 3

LCM = 2 × 2 × 2 × 2 × 3 × 3 × 5

LCM = 720

Number = (LCM of 15,20,36 and 48) + 3

Number = 720 + 3

Number = 723