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 13
LCM = 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.
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