Find square out of the sum of factorials!

Number Theory Level 2

If A A is a positive integer, how many values of n n satisfy 1 ! + 2 ! + ⋯ + n ! = A 2 1! + 2! + \cdots + n! = A^2 ?


The answer is 2.

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Anurag Pandey by Anurag Pandey , 23, India

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2 solutions

Anurag Pandey
Sep 18, 2016

The question is basically asking the number of values n for which 1 ! + 2 ! + ⋯ + n ! 1! + 2! + \cdots + n! is a perfect square.

Let's observe :

1 ! = 1 1! = 1 ( which is a perfect square )

1 ! + 2 ! = 3 1! + 2! = 3

1 ! + 2 ! + 3 ! = 9 1! + 2! + 3! = 9 (which is a perfect square)

1 ! + 2 ! + 3 ! + 4 ! = 33 1! + 2! + 3! + 4! = 33 ( observe that the last digit is 3)

1 ! + 2 ! + 3 ! + 4 ! + 5 ! = 120 + 33 = 153 1! + 2! + 3! + 4! + 5! = 120 + 33 = 153 .

Now observe that now for every next term the unit digit would be 3 and we know that a number with unit digit 3 can't be a perfect square. And thus our answer is 2 \boxed {2}

7 Helpful 0 Interesting 0 Brilliant 0 Confused
Rohan Jasani
Sep 18, 2016

Perfect squares are 1( mod3) so through this we deduce that A is equal to 1 thus n is equal to 0 & 1

0 Helpful 0 Interesting 0 Brilliant 0 Confused

Actually in the question we have started from 1 and so the answer would be n =1 and 3.

Anurag Pandey - 4 years, 9 months ago

This solution is wrong. For one, not all perfect squares are 1 modulo 3. They may be divisible by 3.

Satvik Golechha - 3 years ago

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