Calculation & Estimation Techniques

1) Find the last digit of the product of 24 x 36 x 38 x 52 x 63 x 57 x 94

2

4

6

8

To find the last digit of the product, we only need to consider the last digits of each number being multiplied. The last digit of 24 is 4, 36 is 6, 38 is 8, 52 is 2, 63 is 3, 57 is 7, and 94 is 4. The product of these last digits is 4 x 6 x 8 x 2 x 3 x 7 x 4, which equals 8064. The last digit of 8064 is 6. Therefore, the last digit of the product is 6.

Explanation

To find the last digit of the product, we only need to consider the last digits of each number being multiplied. The last digit of 24 is 4, 36 is 6, 38 is 8, 52 is 2, 63 is 3, 57 is 7, and 94 is 4. The product of these last digits is 4 x 6 x 8 x 2 x 3 x 7 x 4, which equals 8064. The last digit of 8064 is 6. Therefore, the last digit of the product is 6.

2) A four digit perfectly square number has its first two digits same and also the last two digits same. Find the number

8833

6655

7744

9922

The number 7744 is a four-digit perfectly square number because its first two digits (77) are the same and its last two digits (44) are also the same.

Explanation

The number 7744 is a four-digit perfectly square number because its first two digits (77) are the same and its last two digits (44) are also the same.

3) Find the last 2 digits of 47 ^ 89

66

67

68

69

To find the last two digits of a number, we only need to consider the remainder when dividing the number by 100. In this case, we need to find the remainder of 47^89 divided by 100. By using the concept of modular arithmetic, we can simplify the calculation. The last two digits of 47^89 will be the same as the last two digits of (47^20)^4 multiplied by 47^9. Simplifying further, we find that (47^20)^4 ends in 00, so we only need to find the last two digits of 47^9. By calculating the powers of 47 modulo 100, we find that 47^9 ends in 67. Therefore, the last two digits of 47^89 is 67.

Explanation

To find the last two digits of a number, we only need to consider the remainder when dividing the number by 100. In this case, we need to find the remainder of 47^89 divided by 100. By using the concept of modular arithmetic, we can simplify the calculation. The last two digits of 47^89 will be the same as the last two digits of (47^20)^4 multiplied by 47^9. Simplifying further, we find that (47^20)^4 ends in 00, so we only need to find the last two digits of 47^9. By calculating the powers of 47 modulo 100, we find that 47^9 ends in 67. Therefore, the last two digits of 47^89 is 67.

4) Find the last digit of the following : 1⁹⁶¹ + 2⁹⁶¹ + 3⁹⁶¹ +……+ 9⁹⁶¹

3

4

5

6

The last digit of any number is determined by its remainder when divided by 10. In this case, we can observe that the last digit of each number in the given series is 1. When we add these numbers, the sum will have a last digit of 1 as well. Therefore, the last digit of the given expression is 1.

Explanation

The last digit of any number is determined by its remainder when divided by 10. In this case, we can observe that the last digit of each number in the given series is 1. When we add these numbers, the sum will have a last digit of 1 as well. Therefore, the last digit of the given expression is 1.

5) The number of trailing zeros in 6! X 8! X 10! Is

2

3

4

5

The number of trailing zeros in a factorial is determined by the number of factors of 10 in the factorial. Since 10 = 2 x 5, we need to count the number of factors of 2 and 5 in the product of 6!, 8!, and 10!. Since there are more factors of 2 than 5, we only need to count the number of factors of 5. In 6!, there is 1 factor of 5. In 8!, there are 1 factor of 5. In 10!, there are 2 factors of 5. Therefore, the total number of factors of 5 is 1 + 1 + 2 = 4, which means there are 4 trailing zeros in the product.

Explanation

The number of trailing zeros in a factorial is determined by the number of factors of 10 in the factorial. Since 10 = 2 x 5, we need to count the number of factors of 2 and 5 in the product of 6!, 8!, and 10!. Since there are more factors of 2 than 5, we only need to count the number of factors of 5. In 6!, there is 1 factor of 5. In 8!, there are 1 factor of 5. In 10!, there are 2 factors of 5. Therefore, the total number of factors of 5 is 1 + 1 + 2 = 4, which means there are 4 trailing zeros in the product.

6) How many zeros will be at the end of 124 !

26

27

28

29

To determine the number of zeros at the end of 124!, we need to find the number of times 10 can be factored out of it. Since 10 is equivalent to 2 x 5, we need to count the number of pairs of 2 and 5 in the prime factorization of 124!. Since the number of 2's is always greater than the number of 5's, we only need to count the number of 5's. To do this, we can divide 124 by 5, which gives us 24. However, there are also multiples of 5 that have additional factors of 5, such as 25 and 125. So, we divide 124 by 25, which gives us 4. Finally, we divide 124 by 125, which gives us 0. Adding up these results (24 + 4 + 0), we get 28. Therefore, there will be 28 zeros at the end of 124!.

Explanation

To determine the number of zeros at the end of 124!, we need to find the number of times 10 can be factored out of it. Since 10 is equivalent to 2 x 5, we need to count the number of pairs of 2 and 5 in the prime factorization of 124!. Since the number of 2's is always greater than the number of 5's, we only need to count the number of 5's. To do this, we can divide 124 by 5, which gives us 24. However, there are also multiples of 5 that have additional factors of 5, such as 25 and 125. So, we divide 124 by 25, which gives us 4. Finally, we divide 124 by 125, which gives us 0. Adding up these results (24 + 4 + 0), we get 28. Therefore, there will be 28 zeros at the end of 124!.

7) N= (5 + 55 +555…..5555555555) If N is divided by 8 find the remainder

2

3

4

5

not-available-via-ai

Explanation

not-available-via-ai

8) Find the unit’s digit of 13 ^ 137 + 17 ^ 731 + 18 ^ 138 + 12 ^ 432

5

6

7

8

To find the units digit of each number, we need to observe the pattern. The units digit of 13 raised to any power alternates between 3 and 9. The units digit of 17 raised to any power is always 7. The units digit of 18 raised to any power alternates between 8 and 4. The units digit of 12 raised to any power is always 2. Therefore, the units digit of the sum of these numbers will be the sum of the units digits of each number, which is 3 + 7 + 8 + 2 = 20. The units digit of 20 is 0, so the correct answer is 0.

Explanation

To find the units digit of each number, we need to observe the pattern. The units digit of 13 raised to any power alternates between 3 and 9. The units digit of 17 raised to any power is always 7. The units digit of 18 raised to any power alternates between 8 and 4. The units digit of 12 raised to any power is always 2. Therefore, the units digit of the sum of these numbers will be the sum of the units digits of each number, which is 3 + 7 + 8 + 2 = 20. The units digit of 20 is 0, so the correct answer is 0.

9) Find unit digit of the expression 123 ^123

3

7

1

None

The unit digit of any number raised to the power of 3 follows a pattern. For the digit 3, the pattern is 3, 9, 7, 1, and it repeats. Since 123 is divisible by 4, the unit digit of 123^123 will be the same as the unit digit of 123^4, which is 7.

Explanation

The unit digit of any number raised to the power of 3 follows a pattern. For the digit 3, the pattern is 3, 9, 7, 1, and it repeats. Since 123 is divisible by 4, the unit digit of 123^123 will be the same as the unit digit of 123^4, which is 7.

10) Find the difference in the number of trailing zeros between 624! And 625!

2

3

4

5

The number of trailing zeros in a factorial is determined by the number of factors of 5. Since 625 is divisible by 5, but 624 is not, the difference in the number of trailing zeros between 624! and 625! is equal to the number of factors of 5 in 625. Since 625 is equal to 5^4, there are 4 factors of 5 in 625, resulting in a difference of 4.

Explanation

The number of trailing zeros in a factorial is determined by the number of factors of 5. Since 625 is divisible by 5, but 624 is not, the difference in the number of trailing zeros between 624! and 625! is equal to the number of factors of 5 in 625. Since 625 is equal to 5^4, there are 4 factors of 5 in 625, resulting in a difference of 4.

11) Find the unit’s digit of 2018 ^ 98

4

8

2

6

The unit's digit of any power of 2 follows a pattern: 2, 4, 8, 6. Since 2018 is an even number, its unit's digit is 8. Therefore, the unit's digit of 2018^98 will be the same as the unit's digit of 8^98, which is 4.

Explanation

The unit's digit of any power of 2 follows a pattern: 2, 4, 8, 6. Since 2018 is an even number, its unit's digit is 8. Therefore, the unit's digit of 2018^98 will be the same as the unit's digit of 8^98, which is 4.

12) Given that 100! = 56^P x Q , where Q is not a multiple of 7 find P

12

14

16

18

The equation given states that 100! is equal to 56P multiplied by Q. We are told that Q is not a multiple of 7. To find P, we need to divide both sides of the equation by 56 and then divide by Q. This will leave us with P on one side of the equation. Since we are not given the value of Q, we cannot solve for P directly. Therefore, the answer cannot be determined from the information provided.

Explanation

The equation given states that 100! is equal to 56P multiplied by Q. We are told that Q is not a multiple of 7. To find P, we need to divide both sides of the equation by 56 and then divide by Q. This will leave us with P on one side of the equation. Since we are not given the value of Q, we cannot solve for P directly. Therefore, the answer cannot be determined from the information provided.

13) Find the number of zeros in 6! ^6!

120

720

6! X 6!

Cannot be determined

The number of zeros in a factorial is determined by the number of factors of 10. Since 10 can be written as 2 x 5, we need to count the number of 2s and 5s in the prime factorization of the factorial. In 6!, there is only one 5. However, in 6! x 6!, there are two 5s. Therefore, the number of zeros in 6! x 6! is 2, resulting in an answer of 720.

Explanation

The number of zeros in a factorial is determined by the number of factors of 10. Since 10 can be written as 2 x 5, we need to count the number of 2s and 5s in the prime factorization of the factorial. In 6!, there is only one 5. However, in 6! x 6!, there are two 5s. Therefore, the number of zeros in 6! x 6! is 2, resulting in an answer of 720.

14) If the number of trailing zeros of n ! is 35 then the value of n is

140

142

144

All the Three

If the number of trailing zeros in n! is 35, it means that n! has a factor of 10^35. In order for a number to have a factor of 10^35, it must have at least 35 factors of 5 and 35 factors of 2. Since every even number has at least one factor of 2, we only need to consider the factors of 5. The number of factors of 5 in n! can be found using the formula n/5 + n/25 + n/125 + ... In this case, the number of factors of 5 is greater than or equal to 35, so n can be any of the given options: 140, 142, or 144.

Explanation

If the number of trailing zeros in n! is 35, it means that n! has a factor of 10^35. In order for a number to have a factor of 10^35, it must have at least 35 factors of 5 and 35 factors of 2. Since every even number has at least one factor of 2, we only need to consider the factors of 5. The number of factors of 5 in n! can be found using the formula n/5 + n/25 + n/125 + ... In this case, the number of factors of 5 is greater than or equal to 35, so n can be any of the given options: 140, 142, or 144.

15) If there are 30 zeros at the end of n! , find n ?

89

99

100

N DOES NOT EXIST

NONE OF THE OPTIONS

not-available-via-ai

Explanation

not-available-via-ai

16) Which of the following cannot be the number of zeros at the end of n!

28

29

31

32

The number of zeros at the end of n! is determined by the number of factors of 5 in n!. Every multiple of 5 contributes at least one factor of 5. Since 29 is a prime number and not a multiple of 5, it cannot contribute any factors of 5 to n!. Therefore, 29 cannot be the number of zeros at the end of n!.

Explanation

The number of zeros at the end of n! is determined by the number of factors of 5 in n!. Every multiple of 5 contributes at least one factor of 5. Since 29 is a prime number and not a multiple of 5, it cannot contribute any factors of 5 to n!. Therefore, 29 cannot be the number of zeros at the end of n!.

17) A is the product of 1st 100 multiples of 2 and B is the product of the 1st 10 multiples of 5. How many zeros will be in the end of A/B

10

11

12

13

To find the number of zeros at the end of A/B, we need to find the number of factors of 10 in the product A and B. Since 10 can be factored into 2 and 5, we need to find the number of factors of 2 and 5 in A and B separately. A is the product of the first 100 multiples of 2, so it will have 100 factors of 2. B is the product of the first 10 multiples of 5, so it will have 10 factors of 5. Since there are more factors of 2 than 5, we can only have as many zeros at the end of A/B as there are factors of 5. Therefore, there will be 10 zeros at the end of A/B.

Explanation

To find the number of zeros at the end of A/B, we need to find the number of factors of 10 in the product A and B. Since 10 can be factored into 2 and 5, we need to find the number of factors of 2 and 5 in A and B separately. A is the product of the first 100 multiples of 2, so it will have 100 factors of 2. B is the product of the first 10 multiples of 5, so it will have 10 factors of 5. Since there are more factors of 2 than 5, we can only have as many zeros at the end of A/B as there are factors of 5. Therefore, there will be 10 zeros at the end of A/B.

18) Find the remainder (2222⁵⁵⁵⁵ + 5555²²²²) ÷ 7

0

1

3

7

The expression (22225555 + 55552222) divided by 7 can be simplified by adding the two numbers together, resulting in 77777777. When this sum is divided by 7, the remainder is 0. Therefore, the correct answer is 0.

Explanation

The expression (22225555 + 55552222) divided by 7 can be simplified by adding the two numbers together, resulting in 77777777. When this sum is divided by 7, the remainder is 0. Therefore, the correct answer is 0.

19) Find the highest power of 12 in 100!

37

47

48

8

To find the highest power of 12 in 100!, we need to determine the number of times 12 can be divided into 100!. Since 12 can be expressed as 2^2 * 3, we need to count the number of times 2^2 and 3 appear as factors in the prime factorization of 100!. The highest power of 2 that divides 100! is 97, and the highest power of 3 that divides 100! is 48. Since 12 has 2^2 and 3 as factors, we can only have a maximum power of 12^48 in 100!. Therefore, the correct answer is 48.

Explanation

To find the highest power of 12 in 100!, we need to determine the number of times 12 can be divided into 100!. Since 12 can be expressed as 2^2 * 3, we need to count the number of times 2^2 and 3 appear as factors in the prime factorization of 100!. The highest power of 2 that divides 100! is 97, and the highest power of 3 that divides 100! is 48. Since 12 has 2^2 and 3 as factors, we can only have a maximum power of 12^48 in 100!. Therefore, the correct answer is 48.

20) Find the highest power of 12 in 50!

4

8

22

None

not-available-via-ai

Explanation

not-available-via-ai

21) Find the last two digit of the expression 67¹³

87

67

47

27

The last two digits of a number can be found by taking the remainder when the number is divided by 100. In this case, we need to find the remainder when 6713 is divided by 100. By performing the division, we find that the remainder is 13. Therefore, the last two digits of the expression 6713 is 13.

Explanation

The last two digits of a number can be found by taking the remainder when the number is divided by 100. In this case, we need to find the remainder when 6713 is divided by 100. By performing the division, we find that the remainder is 13. Therefore, the last two digits of the expression 6713 is 13.