West Bengal Pre RMO 2015

Number Theory Level 3

What is the trailing number of zeros at the end of the integer

${ 10! }^{ 2 } + { 11! }^{ 2 } + { 12! }^{ 2 } \cdots+ { 99! }^{ 2 }$ ?

The answer is 4.

2 solutions

Md Zuhair
Oct 23, 2016

The number of trailing zeroes at the end of $(10!)^2 + (11!)^2 + ..... + (99!)^2$ is to be finded out.

No of zeroes at the end of $10! = [10/5] = 2$ or $(10!)^2 = 4$ zeroes. Now, $(11!)^2$ similarly has $4$ zeroes at the end uptill $(15!)^2$ which has $9$ zeroes. Hence Now $(10!)^2$ also have a non zero part, which when added to the others will leave with only ${4}$ zeroes at the end.

Hence Ans is $\boxed{4}$ .

This is a good start. However, your explanation has a common misconception.

The sum of several numbers that end with 4 zeros, could end with more than 4 zeroes. For example, $10000 + 990000 = 1000000$ which has 6 zeros.

Do you see how to fix this gap in your solution?

Keep writing more solutions and you will get the hang of this!

Calvin Lin Staff - 4 years, 7 months ago

how will i fix this gap for this problem?

Md Zuhair - 4 years, 7 months ago

Well, think about the gap.

What have you actually shown?
What else do we need to show?

Calvin Lin Staff - 4 years, 7 months ago

Geeta .
Jun 20, 2018

My solution is approximately same as MD Zuhair's but to complete the work from where he left ,we need to prove that

1 + (11! /10!)×(11! / 10!) + ...... + (14! / 10!)×(14! / 10!) is not divisible by 10 which can be shown using modular arithmetic.

West Bengal Pre RMO 2015

The answer is 4.

2 solutions

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