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Question 26 : How many pairs (a,b) of positive integers are there such that a ≤ b and ab = 4^{2017}?

- 2019
- 2018
- 2020
- 2017

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4^{2017} = 2^{2×2017}

Number of factors of 2^{2×2017} = 2×2017 + 1= 4035

Number of pairs of a,b such that a×b = 2^{2×2017} is equal to half the Number of factors of 2^{2×2017}

Number of pairs of a,b such that a×b = 2^{2×2017}

= \\frac{4035}{2})

You observe that \\frac{4035}{2}) is not an integer, because 2^{2×2017} is a perfect square and one of the pairs of (a,b) is (2017 , 2017).

So, the number of such pairs is the highest integer roundoff of \\frac{4035}{2}) = 2018.

Among all these 2018 pairs, one of the integer is less than or equal to the other. (Equal in the case of (2017,2017))

We assume that a is the least one of the two...

Hence there are 2018 pairs that satify this condition.

**Alternate Method:**

a × b = 4^{2017}

a × b = 2^{2×2017}

a and b are of the form 2^{x} and 2^{y} respectfully...

Since a ≤ b;

2^{x} ≤ 2^{y}

x ≤ y

Also, a × b = 2^{2×2017}

2^{x} + 2^{y} = 2^{2×2017}

x + y = 2×2017

We have 2 conditions to deal with...

x ≤ y and x + y = 2×2017

Let's start with x = y:

Here, x = y = 2017

From here on, we decrement x and increment y to maintain the conditions x ≤ y and x + y = 2×2017

We can keep doing this until x = 0, because if x is negative, a which is 2^{x} will not remain an integer.

x | y |
---|---|

2017 | 2017 |

2016 | 2018 |

2015 | 2019 |

⁝ | ⁝ |

⁝ | ⁝ |

⁝ | ⁝ |

2 | 4033 |

1 | 4034 |

0 | 4035 |

Hence x can range from 2017 to 0; and thereby x can take 2018 values.

Therefore, there can be 2018 pairs of (a,b) that sitisfy a ≤ b and ab = 4^{2017}

The question is **"How many pairs (a,b) of positive integers are there such that a ≤ b and ab = 4 ^{2017}?" **

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