Find the number of zeroes at the end of 15!

Find the number of zeroes at the end of 15!

Solution:

Maximum power of 5 in 15

= 3

This way:

[ 15/ 5 ]  …….. [till the quotient is  greater than  zero]

= 3

Answer: 3

Find the number of zeroes at the end of 68!

Find the number of zeroes at the end of 68!

Solution:

Maximum power of 5 in 68!

= 13 + 2

= 15

This way:

[ 68/ 5 ]  +  [68/5x5]  …….. [till the quotient is  greater than  zero]

= 13 + 2

= 15

Answer: 15

Find the number of zeroes at the end of 350!.

Find the number of zeroes at the end of 350!

Solution:

Maximum power of 5 in 350!

= 70 + 14 + 2

= 86

This way:

[ 350 / 5 ]  +  [350/5x5]  +  [350 / 5x5x5] + …….. [till the quotient is  greater than  zero]

= 70 + 14 + 2

= 86

Answer: 86

Find the sum of divisors of 544 which are perfect squares.

The remainder when 10^10 + 10^100 + ......10^10000000000 is divided by 7 is

Find the last two digits of 15 x 37 x 63 x 51 x 97 x 17.

Find the last two digits of 15 x 37 x 63 x 51 x 97 x 17.

Solution:

15×37×63×51×97×17

= (10+5)*(40-3)*(60+3)*(50+1)*(100-3)*(10+7)

Multiply 5 x 3x 3 x 1 x 3 x 7

Find the remainder when (10^3 + 9^3)^752 is divided by 12^3?

Find the remainder when 7^99 is divided by 2400.

Find the remainder when 7^99 is divided by 2400.

Solution:

Using pattern recognition(cyclicity method).

1) 7^1  mod 2400 = 7
2)  7^2  mod 2400 = 49
3)  7^3 mod 2400 = 343
4)  7^4  mod 2400 = 1
After this the same pattern will be keep on repeating.
So, 

Cyclicty = 4
Power = 99

Power mod Cyclicity = 99 mod 4 = 3

Third value in the above cycle is 343.

Answer: 343

What are the last two digits of 7^2008?

What are the last two digits of 7^2008?

Solution:
7^1 = 07
7^2 = 49  
7^3 = 343   -> 49 times 7 is 343 => last two digits => 43
7^4 = 2401  -> 43 times 7 is 2401 => last  two digits => 01
7^5 = 16807
7^6 = 117649
7^7 = 823543
7^8 = 5764801 -> last two digits again 01

01 -> repeats after every 4 times

Now, 2008 mod 4 = 0

Hence, the last two digits of 7^2008 = 01

When 2^256 is divided by 17 the remainder would be

Which of the following can be a number divisible by 24?

The units digit of the expression 127^813 x 553^3703 x 4532^828 is:

Find the units digit of the expression 11^1 +12^2 + 13^3 +14^4 +15^5 +16^6

The rightmost non-zero digit of the number 30^2720 is:

Find the number of zeroes at the end of 48!.

Find the number of zeroes at the end of 48!.

Solution:

Maximum power of 5 in 48!

= 9 + 1 

= 10

This way:

[ 48 / 5 ]  +  [48/5x5]  +  [48  / 5x5x5] + …….. [till the quotient is  greater than  zero]

= 9 + 1 + 0

= 10

Answer: 10

Find the maximum power of 18 in 200!.

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What is the total number of divisors of 1728?

Find the number of divisors of 1728


Method:
Find  the prime factorization of the number n.
Take all the exponents in the factorization.
Add 1 to each
Then multiply these "exponents + 1"s together.

Solution:

Prime Factorization of 1728:

2 | 1728 
2 | 864

2 | 432

2 | 216  

2 | 108

2 | 54

3 | 27

3 | 9

3 | 3

   | 1

2^6, 3^3
Exponents: 6 & 3

Add 1 to each exponent (6+1) (3+1)
Multiple “exponents+1” together = (6+1) x (3+1) =7 x 4 = 28

Total number of divisors of 1728 = 28

What is the total number of divisors of 600?

Numbers: Find the number of divisors of 600

Method:
Find  the prime factorization of the number n.
Take all the exponents in the factorization.
Add 1 to each
Then multiply these "exponents + 1"s together.

Example:
Find the number of divisors of 600

Solution:
Prime Factorization of 600:

2 | 600 
2 | 300
2 | 150 
3 | 75
5 | 25
5 | 5
   | 1 
 
2^3, 3^1 & 5^2

Exponents: 3, 1 & 2
Add 1 to each exponent = (3+1) (1+1) (2+1)
Multiple “exponents+1” together = (3+1) x (1+1) x (2+1) = 4 x 2 x 3 = 24

Total number of divisors of 600 = 24

Find the sum of divisors of 544

Find the sum of divisors of 544.

Method:
Find the the prime factorization of the number
Calculate the sum of its divisors
Take each different prime factor
Add together all its powers up to the one that appears in the prime factorization
And then multiply all these sums together!

Solution:

Prime Factorization of 544:

2  | 544  

2  | 272

2  | 136

2  | 68

2  | 34
17 | 17

     | 1

2^5  x   17^1
Exponents: 5 & 1

X	2^0	2^1	2^2	2^3	2^4	2^5	Row Sum
17^0	1	2	4	8	16	32	63
17^1	17	34	68	136	272	544	1071
Col  Sum	18	36	72	144	288	576	1134

Sum of divisors of 544 = 1134