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Evaluate 10000a + 100b + c

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anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,685
If $a, b, c$ are positive integer such that $a, b, c \le 100$ satisfy

$109a+991b+101c=44556$

$1099a+901b+1110c=59800$

Evaluate $10000a+100b+c$.
 

Klaas van Aarsen

MHB Seeker
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Mar 5, 2012
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Re: Evaluate 10000a+100b+c

If $a, b, c$ are positive integer such that $a, b, c \le 100$ satisfy

$109a+991b+101c=44556$

$1099a+901b+1110c=59800$

Evaluate $10000a+100b+c$.
Mod 10:
\begin{array}{}
-a+b+c &\equiv& 6 \pmod{10} \\
-a+b+0 &\equiv& 0 \pmod{10} \\
\therefore c &\equiv& 6 \pmod{10}
\end{array}

Mod 100:
\begin{array}{lcr}
9a-9b+c &\equiv& 56 \pmod{100} \\
-a+b+10c &\equiv& 0 \pmod{100} \\
\therefore a-b &\equiv& 60 \pmod{100} \\
\therefore c &\equiv& 16 \pmod{100} \\
\therefore c = 16 &\wedge& (a = b+60 &\vee& a = b - 40)
\end{array}

Substituting in the original equations yields as only solution:
$$a = 3, \quad b = 43, \quad c = 16$$

Therefore
$$10000a+100b+c = 34316$$
 
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anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,685
Re: Evaluate 10000a+100b+c

Thanks for participating, I like Serena! I was wondering at first if this is a problem that could be approached using modular arithmetic and you proved it to me and hence thanks to your solution!

My method:

Multiply the first given equation by 11 we get

$1199a+10901b+1111c=490116$

Now subtracting the above equation from the second given equation, i.e.

$1099a+901b+1110c=59800$, we have

$100a+10000b+c=430316$

But observe that $100a+10000b+c=10000b+100a+c=430316=430000+300+16=43(10000)+3(100)+16(1)$, we can say that

$b=43, a=3, c=16$ and hence we obtain $10000a+100b+c=10000(3)+100(43)+16=34316$.
 
Last edited:

kaliprasad

Well-known member
Mar 31, 2013
1,309
Re: Evaluate 10000a+100b+c

Thanks for participating, I like Serena! I was wondering at first if this is a problem that could be approached using modular arithmetic and you proved it to me and hence thanks to your solution!

My method:

Multiply the first given equation by 11 we get

$1199a+10901b+1111c=490116$

Now subtracting the above equation from the second given equation, i.e.

$1099a+901b+1110c=59800$, we have

$10a+10000b+c=430316$

But observe that $10a+10000b+c=10000b+10a+c=430316=430000+300+16=43(10000)+3(100)+16(1)$, we can say that

$b=43, a=3, c=16$ and hence we obtain $10000a+100b+c=10000(3)+100(43)+16=34316$.
for once I was stumped.
neat ans