Question

What will be greatest 5-digit number which is exactly divisible by 7, 10, 15, 21 and 28?

Answer 1

║⊕ANSWER⊕║

The First step is to find the LCM of the given numbers

LCM of 7, 10, 15, 21 and 28  

7 is a  prime  number

10 = 2*5  

15 = 3*5

21 = 3*7

28 = 2²*7

LCM = 2² × 3 × 5 × 7 = 420

So the number should be divisible by 420

Largest five digit number = 99999

99999 ÷ 420 = $238 \frac{39}{420}$

So remainder is 39.

The largest number divisible = 99999 - 39 = 99960

Answer 2

Answer:

The largest number divisible is 99960.

Step-by-step explanation:

Solution :

• Step 1 :
• Find the LCM of 7, 10, 15, 21 and 28.

★ 7 :

7 is the prime number.

★ 10 :

$\begin{array}{r|l} 2 & 10\\ \cline{1-2} 5 & 5 \\\cline{1-2} & 1 \end{array}$

★ 15 :

$\begin{array}{r|l} 3 & 15 \\\cline{1-2} 5 & 5 \\\cline{1-2} & 1 \end{array}$

★ 21 :

$\begin{array}{r|l} 3 & 21 \\ \cline{1-2} 7 & 7 \\\cline{1-2} & 1 \end{array}$

★ 28 :

$\begin{array}{r|l} 2 & 28 \\ \cline{1-2} 2 &14 \\\cline{1-2} 7 & 7\\\cline{1-2} & 1 \end{array}$

• Step 2 :
• Write their factors together.

7 is prime  number.

10 = 2 × 5  

15 = 3 × 5

21 = 3 × 7

28 = 2 × 2 × 7

LCM = 2² × 3 × 5 × 7 = 420.

∴ The number should be divisible by 420.

• Step 3 :
• Divide the LCM with the Largest five digit number.

As we know,

$\bigstar\;\boxed{\textsf{Largest five digit number = 99999}}}$

So,

$\sf \\ \\\implies \dfrac{99999}{420}\\ \\ \\\implies 238 \dfrac{39}{420}.$

• Step 4 :
• Minus the Remainder from the Largest five digit number.

So, The remainder is 39.

⇒ 99999 - 39.

⇒ 99960.

∴ The largest number divisible = 99960