**Questions on Fundamental Theorem of Arithmetic :**

Here we are going to see, some practice questions based on fundamental theorem of arithmetic.

**Question 1 :**

For what values of natural number n, 4^{n} can end with the digit 6?

**Solution :**

4^{n}

if n = 1, then 4^{1} = 4

if n = 2, then 4^{2} = 16

if n = 3, then 4^{3} = 64

if n = 4, then 4^{4} = 256

if n = 5, then 4^{5} = 1024

if n = 6, then 4^{6} = 4096

When n is even, 4^{n} ends with 6.

**Question 2 :**

If m, n are natural numbers, for what values of m, does 2^{n} x 5^{m} ends in 5?

**Solution :**

For any value of n, 2^{n} will become even.

For any value m, 5^{m} ends with 5. The product of even number and a number ends with the digit 5, we get the answer ends with 0.

We should not apply the value 0 for n and m, because n and m are natural numbers.

**Question 3 :**

Find the HCF of 252525 and 363636.

**Solution :**

252525 = 5^{2}⋅ 3 ⋅ 7 ⋅ 13 ⋅ 37

363636 = 2^{2}⋅ 3^{3 }⋅ 7 ⋅ 13 ⋅ 37

H.C.F = 3 ⋅ 7 ⋅ 13 ⋅ 37 = 10101

**Question 4 :**

If 13824 = 2^{a} ×3^{b} then find a and b.

**Solution :**

To find the values of a and b, we have decompose 13824 as the product of prime factors.

13824 = 2^{9} x 3^{3}

Hence the value of a is 9 and b is 3.

**Question 5 :**

**Solution :**

113400 = 2^{3} x 3^{4 }x 5^{2} x 7^{1}

The values of p_{1}, p_{2}, p_{3} and p_{4} are 2, 3, 5 and 7 respectively.

The values of x_{1}, x_{2}, x_{3} and x_{4} are 3, 4, 2 and 1 respectively.

**Question 6 :**

Find the LCM and HCF of 408 and 170 by applying the fundamental theorem of arithmetic.

**Solution :**

408 = 2^{3} x 3 x 17

170 = 2 x 5 x 17

Common factors are 2 and 17

H.C.F = 34

L.C.M = 2^{3} x 3 x 5 x 17

L.C.M = 2040

**Question 7 :**

Find the greatest number consisting of 6 digits which is exactly divisible by 24,15,36?

**Solution :**

We will find out the LCM of 24,15 and 36 is 360.

The greatest 6-digit number is 999999.

Now, We will divide this number by LCM of 24,15 and 36, we will get,

999999/360 will get remainder 279.

Now 999999 – 279 = 999720.

Now we will check it out for the numbers we will get,

999720/24 = 41655

999720/15 = 66648

999720/36 = 27770

999720 is the greatest number 6-digit number divisible by 24,15 and 36.

**Question 8 :**

What is the smallest number that when divided by three numbers such as 35, 56 and 91 leaves remainder 7 in each case?

**Solution :**

**The smallest number which when divided by 35, 56 and 91 = LCM of 35, 56 and 91**

**35 = 5 x 7**

**56 = 2 x 2 x 2 x 7**

**91 = 7 x 13LCM = 7 x 5 x 2 x 2 x 2 x 13 = 3640**

**The smallest number that when divided by 35, 56, 91 leaves a remainder 7 in each case = 3640 + 7 = 3647.**

**Question 9 :**

Find the least number that is divisible by the first ten natural numbers.

**Solution :**

first 10 natural numbers:

1, 2, 3, 4, 5, 6, 7, 8, 9, 10

L.C.M of these natural numbers:

Factors = 2 × 2 × 2 × 3 × 3 × 5 × 7

= 2520

The smallest number is 2520.

After having gone through the stuff given above, we hope that the students would have understood, "Questions on Fundamental Theorem of Arithmetic".

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