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## Number System: Divisibility Test-1

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Question 1 |

If n is a natural number what is the remainder when (33*45*67*89)

^{n}is divided by 2?0 | |

1 | |

2 | |

none of these |

Question 1 Explanation:

33, 45, 67, 89 are all odd numbers and we know that the product of any number of odd numbers is odd.

By extension, the product of any odd number raised to any power will be odd again.

Now when an odd number is divided by 2,

the remainder is always 1.

By extension, the product of any odd number raised to any power will be odd again.

Now when an odd number is divided by 2,

the remainder is always 1.

Question 2 |

If n is a natural number what is the remainder when (23*35*57*79)

^{5}is divided by 10?0 | |

1 | |

3 | |

5 |

Question 2 Explanation:

The remainder of any number divided by 10 is its units digit.

Since the product 23*35*57*79 has one 5 in one of the numbers,

it will end in 5 (there is no even number in the given product and any odd number

when multiplied by 5, the product ends in 5).

Since this product is raised to power 5, the final number has to

end is five going by the above logic (5*5*5*5*5 will always end with a 5).

Thus the unit digit will be 5 and that will also be the remainder.

Since the product 23*35*57*79 has one 5 in one of the numbers,

it will end in 5 (there is no even number in the given product and any odd number

when multiplied by 5, the product ends in 5).

Since this product is raised to power 5, the final number has to

end is five going by the above logic (5*5*5*5*5 will always end with a 5).

Thus the unit digit will be 5 and that will also be the remainder.

Question 3 |

If n is a natural number what is the remainder when (23*35*57*79)

^{4}is divided by 100?25 | |

75 | |

3 | |

5 |

Question 3 Explanation:

In this question, the remainder is nothing else but the last 2 digits.

For the GRE exam, you can use the on-screen calculator to calculate

the product and figure out the value. Else, you can solve the question logically.

The product of the terms is 3624915 and the question effectively becomes the remainder when (3624915)

Now we basically need to figure out 15

15

Thus, the last digits in this case would be 25 and this would be remainder with 100 as well.

For the GRE exam, you can use the on-screen calculator to calculate

the product and figure out the value. Else, you can solve the question logically.

The product of the terms is 3624915 and the question effectively becomes the remainder when (3624915)

^{4}is divided by 100.Now we basically need to figure out 15

^{4}to figure out the answer.15

^{4}is equal to 50625.Thus, the last digits in this case would be 25 and this would be remainder with 100 as well.

Question 4 |

What is the least value must be given to A so that the number 78A3945 is divisible by 11?

0 | |

1 | |

3 | |

5 |

Question 4 Explanation:

We have number 78A3945

(5 + 9 + A + 7) – (4 +3 + 8 ) = (21 + A) – 15

(6 + A ) so the smallest value of A will be 5

(5 + 9 + A + 7) – (4 +3 + 8 ) = (21 + A) – 15

(6 + A ) so the smallest value of A will be 5

Question 5 |

The number 6n

^{2}+ 6n for all natural numbers n, is always divisible by 6 only | |

6 and 12 | |

12 only | |

18 only |

Question 5 Explanation:

6n

This number is clearly divisible by 6 and 12,

Let’s check If n = 1

The value of 6n(n + 1) = 12 divisible by 6 and 12

If n = 2

The value of 6n(n + 1) = 36 also divisible by 6 and 12

^{2}+ 6n = 6n(n + 1)This number is clearly divisible by 6 and 12,

Let’s check If n = 1

The value of 6n(n + 1) = 12 divisible by 6 and 12

If n = 2

The value of 6n(n + 1) = 36 also divisible by 6 and 12

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