When I started seeing all such numbers, my expression was almost like this and the case would be similar for most of the mortal newbies.

Don’t worry, it is just a very common initial crisis most of the CAT aspirants face (even for people from math background) and with very minimal effort, you can also put a cape and come up with some fancy names for yourself.

In this article we will learn some simple and interesting concepts to solve 10 such "Monster" number waala questions. If you know how to solve all the given questions without taking more than a minute per question, you need not spend time on this page and can explore other interesting articles we have in MBAtious. For others, let’s have some fun!

- What is the remainder when 88888….. 300 digits is divided by 111.
- What is the remainder when 123123123123… 600 digits divided by 1001.
- What is the remainder when 5555…. 900 digits is divided by 7.
- What is the remainder when 123123… 300 digits is divided by 999
- What is the remainder when 1111… 111 digits is divided by 625
- What is the remainder when 3^1 + 33^2 + 333^3 + 3333^4 + … 33 terms is divided by 6
- What is the remainder when 123456789^2 – 987654321^2 divided by 8
- What is the remainder when 123456…. 9899100 is divided by 16
- Which is greater, 49^51 or 51^49
- What is the number of digits in 3^40

**Soltuions:**

- Concept 1: Any digit written 3 times will be divisible by 3 and 37

Given question, 111 = 37 * 3. As the given number is divisible by both 3 and 37, remainder is 0.

- Concept 2: Any number of the form abcabc is divisible by 7, 11 and 13.

Given question, 1001 = 7 * 11 * 13 and the given number is divisible by 7, 11 and 13. So the remainder when divided by 1001 is 0

- Concept 3: Any digit repeated P times is divisible by P, where P is a prime > 5

here, P = 7 and hence 555555 (repeated 6 times) is completely divisible by 7

900 is divisible by 6, so 555555… repeated 900 times is completely divisible by 7. Remainder is 0

- Concept 4: A number is divisible by 99.. n digits if the sum of the digits taken n at a time from right is divisible by 99.. n digits.

Here, sum of digits taken 3 at a time for 123123… 300 digits = (123) * 100 = 12300

12300/99 gives a remainder of 312

- Concept 5: A number is divisible by 5
^{n }if the last n digits are divisible by 5^{n }

so to check for the divisibility with 625 ( 5^4) we need to check the last 4 digits, 1111

1111/625 will give a remainder of 486

- Concept 6: Any power of a number formed by the digit 3 repeated any times will give a remainder of 3 when divided by 6

[ ( 33… n terms ) ^ any power ] / 6 gives a remainder of 3

so 3/6, 33^2/6, 333^3/6.. will give a remainder of 3

so we are looking at 3 + 3 + 3 … 33 terms / 6 = 99/6 gives a remainder of 3

- Concept 7: (Any odd number)^2 / 8 gives a remainder of 1

here both terms are odd and hence will give a remainder of 1 each with 8

so net remainder is 1 – 1 = 0

- Concept 8: A number is divisible by 2
^{n }if the last n digits are divisible by 2^{n }

Here as we need to check with the divisibility by 16 (= 2^4) we are worried only about last 4 digits, which are 9100

9100 will give a remainder of 12 when divided by 16, so 12 is our answer

- Concept 9: To compare a^b and b^a, closer the base (a or b) is to e, higher the value will be.

here 49 is closer to e (= 2.71) so 49^51 > 51^49

- Concept 10: To find the number of digits of an expression, take the integer part of the log of the number and add one.

Here, log (3^40) = 40 log 3 = 40 x 0.477 = 19.xx

Integral part = 19, number of digits = 19 + 1 = 20

Learn the basic log values like log2, log3 etc.

Please let me know in case of errors or better methods :)

Happy learning!

]]>When I started seeing all such numbers, my expression was almost like this and the case would be similar for most of the mortal newbies.

Don’t worry, it is just a very common initial crisis most of the CAT aspirants face (even for people from math background) and with very minimal effort, you can also put a cape and come up with some fancy names for yourself.

In this article we will learn some simple and interesting concepts to solve 10 such "Monster" number waala questions. If you know how to solve all the given questions without taking more than a minute per question, you need not spend time on this page and can explore other interesting articles we have in MBAtious. For others, let’s have some fun!

- What is the remainder when 88888….. 300 digits is divided by 111.
- What is the remainder when 123123123123… 600 digits divided by 1001.
- What is the remainder when 5555…. 900 digits is divided by 7.
- What is the remainder when 123123… 300 digits is divided by 999
- What is the remainder when 1111… 111 digits is divided by 625
- What is the remainder when 3^1 + 33^2 + 333^3 + 3333^4 + … 33 terms is divided by 6
- What is the remainder when 123456789^2 – 987654321^2 divided by 8
- What is the remainder when 123456…. 9899100 is divided by 16
- Which is greater, 49^51 or 51^49
- What is the number of digits in 3^40

**Soltuions:**

- Concept 1: Any digit written 3 times will be divisible by 3 and 37

Given question, 111 = 37 * 3. As the given number is divisible by both 3 and 37, remainder is 0.

- Concept 2: Any number of the form abcabc is divisible by 7, 11 and 13.

Given question, 1001 = 7 * 11 * 13 and the given number is divisible by 7, 11 and 13. So the remainder when divided by 1001 is 0

- Concept 3: Any digit repeated P times is divisible by P, where P is a prime > 5

here, P = 7 and hence 555555 (repeated 6 times) is completely divisible by 7

900 is divisible by 6, so 555555… repeated 900 times is completely divisible by 7. Remainder is 0

- Concept 4: A number is divisible by 99.. n digits if the sum of the digits taken n at a time from right is divisible by 99.. n digits.

Here, sum of digits taken 3 at a time for 123123… 300 digits = (123) * 100 = 12300

12300/99 gives a remainder of 312

- Concept 5: A number is divisible by 5
^{n }if the last n digits are divisible by 5^{n }

so to check for the divisibility with 625 ( 5^4) we need to check the last 4 digits, 1111

1111/625 will give a remainder of 486

- Concept 6: Any power of a number formed by the digit 3 repeated any times will give a remainder of 3 when divided by 6

[ ( 33… n terms ) ^ any power ] / 6 gives a remainder of 3

so 3/6, 33^2/6, 333^3/6.. will give a remainder of 3

so we are looking at 3 + 3 + 3 … 33 terms / 6 = 99/6 gives a remainder of 3

- Concept 7: (Any odd number)^2 / 8 gives a remainder of 1

here both terms are odd and hence will give a remainder of 1 each with 8

so net remainder is 1 – 1 = 0

- Concept 8: A number is divisible by 2
^{n }if the last n digits are divisible by 2^{n }

Here as we need to check with the divisibility by 16 (= 2^4) we are worried only about last 4 digits, which are 9100

9100 will give a remainder of 12 when divided by 16, so 12 is our answer

- Concept 9: To compare a^b and b^a, closer the base (a or b) is to e, higher the value will be.

here 49 is closer to e (= 2.71) so 49^51 > 51^49

- Concept 10: To find the number of digits of an expression, take the integer part of the log of the number and add one.

Here, log (3^40) = 40 log 3 = 40 x 0.477 = 19.xx

Integral part = 19, number of digits = 19 + 1 = 20

Learn the basic log values like log2, log3 etc.

Please let me know in case of errors or better methods :)

Happy learning!

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