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- #1

- Feb 14, 2012

- 3,685

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

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

Evaluate $10000a+100b+c$.

- Thread starter anemone
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- Thread starter
- Admin
- #1

- Feb 14, 2012

- 3,685

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

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

Evaluate $10000a+100b+c$.

- Admin
- #2

- Mar 5, 2012

- 8,780

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

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

Evaluate $10000a+100b+c$.

\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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- #3

- Feb 14, 2012

- 3,685

Thanks for participating,

My method:

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

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:

- Mar 31, 2013

- 1,309

for once I was stumped.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 $10000a+100b+c=10000(3)+100(43)+16=34316$.

neat ans