Number System is an alltime favourite with the competitive exam setters as well as the students. The topic is known for intriguing conceptual problems that test the best brains.
Algebra is a branch of mathematics that substitutes letters for numbers. An algebraic equation depicts a scale, what is done on one side of the scale with a number is also done to either side of the scale.
Let's have a look at some solved examples:
Example 1: Find the highest power of 7 in 100!.
Highest power of 7 in 100! = [100/7] + [100/49] = 16
Example 2: What is the number of trailing zeroes in 23!?
[23/5] = 4. It is less than 5, so we stop here. The answer is 4.
Example 3: What would be the sum of all the numbers which can be formed by using the digits 1, 3, 5, 7 all at a time and which have no digits repeated?
The sum of the numbers formed by taking all the given n digits is (sum of all the n digits) * (n  1)! * (111…..n times).
Here n = 4, and Sum of 4 digits = 16
The sum of all the numbers which can be formed by using the digits 1, 3, 5, 7= (16) * (4 – 1)! * ( 1111) = 16 * 3! * 1111
Example 4: Find the number of factors of 98 and also find the sum and product of all factors.
First, write the number 98 into prime factorization.
i.e. 98 = 2 × 49 = 2 × 7 × 7 = 2^{1} x 7^{2 }
Here A = 2, B = 7, p = 1, q = 2
Number of factors for 98 = (p + 1)(q +1) = 2 × 3 = 6
Sum of all factors of 98= 3 × 57 = 171
Product of all factors of number 98 = (98)^{6/2} = 941192
Example 5: Find the number of odd, even, perfect square, perfect cube factors of 4500.
4500 = 45 × 100 = 9 × 5 × 10 × 10 = 3 × 3 × 5 × 5 × 2 × 5 × 2 = 2^{2} × 3^{2} × 5^{3}
Here, consider A = 2 , B = 3 , C = 5 , p= 2 , q = 2 and r = 3
Here identifying that odd number are 3 and 5.
Numbers of odd factors of number 4500 = (q + 1) (r + 1) = 3 × 4 = 12
∴ Total number of factors = (p + 1)(q +1)(r +1) = 3 × 3 × 4 =36
Numbers of even factors of number = (Total number of factors – Numbers of odd factors) = 36 – 12 = 24
Number of perfect square factors of number 4500 = 2 x 2 x 2 = 8
Number of perfect cube factors of number 4500 = 1 x 1 x 2 = 2
Example 6: In how many ways you can express 36 as the product of two of its factors?
Prime factorization of 36 i.e. we write 36 = 2^{2} × 3^{2}
Number of factors of 36 will be (2+1)(2+1) = 9
(i.e. factors are 1, 2,3, 4, 6, 9, 12, 18, 36).
Since we are asked the total number of ways hence we include the square root of 36 i.e. 6 as well.
Thus the number of ways you can express 36 as the product of two of its factors is (9+1)/2 = 5.
How many positive integral solutions are possible for the equation X^{2 } Y^{2} = N?
When we are asked to calculate how many positive integral solutions are possible for the equation X^{2}  Y^{2} = N, there can be 4 cases.
Case 1: N is an odd number and not a perfect square
Case 2: N is an odd number and a perfect square
Case 3: N is an even number and not a perfect square
Case 4: N is an even number and a perfect square
Example 7: How many positive integral solutions are possible for the equation X^{2 }– Y^{2 }= 135?
The total number of factors of 135 = 1, 3, 5, 9, 15, 27, 45 and 135 is 8.
So, total number of positive integral solutions = 8/2 = 4.
Example 8: How many positive integral solutions are possible for the equation X^{2}– Y^{2 }= 121?
The total number of factors of 121 = 1, 11 and 121 is 3.
So, the total number of positive integral solutions = (31)/2 = 1
Example 9: How many positive integral solutions are possible for the equation X^{2}– Y^{2 }= 160?
Total number of factors of 40 = 1, 2, 4, 5, 8, 10, 20 and 40 is 8 (as N = 160 and N/4 = 40)
So, total number of positive integral solutions = 8/2 = 4
Example 10: How many positive integral solutions are possible for the equation X^{2}– Y^{2}= 256?
The total number of factors of 64 = 1, 2, 4, 8, 16, 32 and 64 is 7.
So, (71)/2 = 3 positive integral solutions.
EduRev Tip: If a given number is of the form 4k + 2, it cannot be expressed as difference of two squares.
But how does this help us in finding the last two digits of any square?
(50n ± x)^{2} = 2500n^{2} ± 100nx + x^{2}
The last two digits of each of 2500n^{2} and 100nx will be 00. Thus the last two digits of the RHS, and hence of the LHS, will be the last two digits of x^{2}.
Example 11: What are the last two digits of 268^{2}?
268 = 50 × 5 + 18.
Thus, the last two digits of 268^{2} will be same as the last two digits of 18^{2} i.e. 24.
Example 12: What are the last two digits of 278^{2}?
While you can consider 278 = 50 × 5 + 28, it will consider
278 = 50 × 6 – 22, to be x in the range 0 to 25.
Thus, the last two digits of 278^{2} will be the same as the last two digits of 22^{2} i.e. 84.
Q.1. Is 7248 is divisible (i) by 4, (ii) by 2, and (iii) by 8?
Answer:
(i) The number 7248 has 48 on its extreme right side which is exactly divisible by 4. When we divide 48 by 4 we get 12.
Therefore, 7248 is divisible by 4.
(ii) The number 7248 has 8 on its unit place which is an even number so, 7248 is divisible by 2.
(iii) 7248 is divisible by 8 as 7248 has 248 at its hundred place, tens place and unit place which is exactly divisible by 8.
Q.2. A number is divisible by 4 and 12. Is it necessary that it will be divisible by 48? Give another example in support of your answer.
Answer: 48 = 4 × 12 but 4 and 12 are not coprime.
Therefore, it is not necessary that the number will be divisible by 48.
Let us consider the number 72 for an example
72 ÷ 4 = 18, so 72 is divisible by 4.
72 ÷ 12 = 6, so 72 is divisible by 12.
But 72 is not divisible by 48.
Q.3. Without actual division, find if 235932 is divisible (i) by 4 and (ii) 8.
Answer:
(i) The number formed by the last two digits on the extreme right side of 235932 is 32
32 ÷ 4 = 8, i.e. 32 is divisible by 4.
Therefore, 235932 is divisible by 4.
(ii) The number formed by the last three digits on the extreme right side of 235932 is 932
But 932 is not divisible by 8.
Therefore, 235932 is not divisible by 8.
Example 1: Find the HCF of 96, 36, and 18.
96 = 2 × 3 × 2 × 2 × 2 × 2
36 = 2 × 3 × 2 × 3
18 = 2 × 3 × 3
Therefore, the HCF of 96, 36 and 18 is the product of the highest number of common factors in the given numbers i.e., 2 × 3 = 6.
In other words, 6 is the largest possible integer, which can divide 96, 36 and 18 without leaving any remainder.
Example 2: Find the HCF of 42 and 70.
42 = 3 × 2 × 7
70 = 5 × 2 × 7
Hence, HCF is 2 x 7 = 14.
Example 3: Find the HCF of numbers 144, 630, and 756.
144 = 2^{4} × 3^{2}
630 =2 × 3^{2 }× 5 × 7
756 = 2^{2} x 3^{3} x 7
Hence, HCF of 144, 630, 756 = 2 × 3^{2} = 18.
Example 4: Find the LCM of 96, 36, and 18.
96 = 2 × 2 × 2 × 2 × 2 × 3 = 2^{5 }× 3^{1}
36 = 2 x 2 x 3 x 3 = 2^{2} x 3^{2}
18 = 2 x 3 x 3 = 2^{1} x 3^{2}
Therefore, LCM of 96, 36 and 18 is the product of the highest powers of all the prime factors, i.e. 2^{5} x 3^{2 }= 32 x 9 = 288.
That is, 288 is the smallest integer which is divisible by 96, 36 and 18 without leaving any remainder.
Example 5: Find the LCM of 42 and 70.
42 = 3 × 2 × 7
70 = 5 × 2 × 7
Hence, LCM is 2 × 3 × 5 × 7 = 210.
Apart from the method of prime factorization, there is another method of finding the LCM of given numbers, and the method is known as the long division method. This method is quite helpful in getting LCM quickly if there are three or more than three numbers.
EduRev Tip:
HCF of two prime numbers is always 1.
HCF of coprime numbers is always 1.
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