Dhaka Secondary 2009/6
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Please don't post problems (by starting a topic) in the "Secondary: Solved" forum. This forum is only for showcasing the problems for the convenience of the users. You can post the problems in the main Divisional Math Olympiad forum. Later we shall move that topic with proper formatting, and post in the resource section.
Determine the unit’s digit (one’s digit) of the sum of the expression: \[ (1!)^3 + (2!)^3 + (3!)^3 + \cdots + (13!)^3 + (14!)^3 + (15!)^3 \]
- leonardo shawon
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Re: Dhaka Secondary 2009/6
${15! (15!+1) \div 2}^2$
Ibtehaz Shawon
BRAC University.
long way to go .....
BRAC University.
long way to go .....
Re: Dhaka Secondary 2009/6
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
remember that every integer from $5!$.......... is divisible by 10.so just calculate the first four.
বড় ভালবাসি তোমায়,মা
- leonardo shawon
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Re: Dhaka Secondary 2009/6
sorry Moon bhaia, i forgot to correct it....
Ibtehaz Shawon
BRAC University.
long way to go .....
BRAC University.
long way to go .....
Re: Dhaka Secondary 2009/6
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 $a<10$)
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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Nur Muhammad Shafiullah | Mahi
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- afif mansib ch
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Re: Dhaka Secondary 2009/6
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?
\[(1!)^3+(2!)^3+(3!)^3+(4!)^3=14049\equiv 9(mod10)\]
so the last digit should be 9.
am i correct?
Re: Dhaka Secondary 2009/6
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 $9$
$(1!)^3\equiv 1(mod 10)$
$(2!)^3= 2^3\equiv 8(mod 10)$
$(3!)^3=6^3\equiv 6(mod 10)$( 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 $9$