What is the tens digit of $11!+12!$?

Question

In the following problem

What is the tens digit of $11!+12!$?

I have tried too many times and wasted much time too to find the easiest way by logically. Please give some idea to find out it and also to find out hundreds digit quickly. I have these kind of problems too :

Hundred digit of $123456789^2 $?

Answer 1

HINT : try using modular arithmetic ;

For the first one you need to find ;

$11!+12!\mod{100}$

$\implies 11!(13)\equiv0\mod{100}$

so the last 2 digits of $11!+12! $ is $00$

For the second one;

find $(123456789)^2\mod1000$

$(123456000+789)^2\mod1000$

$(123456000)^2+(789)^2+2\cdot123456000\cdot789\equiv (789)^2\mod1000$

$\quad$

we can write $(789)^2 = (700+89)(700+89) = 490000+124600+7921$

$(789)^2\equiv(490000+124600+7921)\mod1000$

$(789)^2\equiv 0+ 600+921 \equiv 1521\equiv521\mod1000$

So last 3 digits are 521.

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Hope I helped