Dhaka Secondary 2009/6

Problem for Secondary Group from Divisional Mathematical Olympiad will be solved here.
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BdMO
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Dhaka Secondary 2009/6

Unread post by BdMO » Fri Jan 21, 2011 6:17 pm

Determine the unit’s digit (one’s digit) of the sum of the expression: \[ (1!)^3 + (2!)^3 + (3!)^3 + \cdots + (13!)^3 + (14!)^3 + (15!)^3 \]

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leonardo shawon
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Re: Dhaka Secondary 2009/6

Unread post by leonardo shawon » Sat Jan 29, 2011 11:59 pm

${15! (15!+1) \div 2}^2$
Ibtehaz Shawon
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Moon
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Re: Dhaka Secondary 2009/6

Unread post by Moon » Tue Feb 01, 2011 11:46 pm

Your solution is not correct; look, it asks for the last digit. It's actually a modular arithmetic exercise...:)
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Tahmid Hasan
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Re: Dhaka Secondary 2009/6

Unread post by Tahmid Hasan » Tue Feb 01, 2011 11:48 pm

remember that every integer from $5!$.......... is divisible by 10.so just calculate the first four. ;)
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leonardo shawon
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Re: Dhaka Secondary 2009/6

Unread post by leonardo shawon » Wed Feb 02, 2011 1:51 pm

sorry Moon bhaia, i forgot to correct it.... :)
Ibtehaz Shawon
BRAC University.

long way to go .....

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*Mahi*
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Re: Dhaka Secondary 2009/6

Unread post by *Mahi* » Fri Feb 04, 2011 12:55 pm

It's like this-
As for every integer $n>5,10\mid n!$ so the last digit (suppose $a$) will be-
$a\equiv 1!+2!+3!+4!\equiv 1+2+6+4(mod 10)$
$a\equiv 3(mod 10)$
Or,$a=3$(as $a<10$)
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afif mansib ch
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Re: Dhaka Secondary 2009/6

Unread post by afif mansib ch » Thu Aug 18, 2011 1:36 am

shouldn't the solution be like this?
\[(1!)^3+(2!)^3+(3!)^3+(4!)^3=14049\equiv 9(mod10)\]
so the last digit should be 9.
am i correct? :mrgreen: :oops: :shock: :D :mrgreen:

Shifat
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Re: Dhaka Secondary 2009/6

Unread post by Shifat » Thu Aug 25, 2011 4:01 pm

Yes the answer is correct, but I could not do such multiplications in the hall, so here is another solution frm me

$(1!)^3\equiv 1(mod 10)$
$(2!)^3= 2^3\equiv 8(mod 10)$
$(3!)^3=6^3\equiv 6(mod 10)$( 6 last digit amon number er jekono power last digit 6)
$(4!)^3=24^3\equiv 4(mod 10)$( 4 last digit hole odd power e last digit hoy 4, even power e last digit hoy 6)
after these $(n!)^3\equiv 0(mod 10)$(last digit 0)
so we find the last digit = $19\equiv 9(mod10)$
the answer is $9$

:idea: :mrgreen:

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