MHB How to Solve for 10000a + 100b + c Given Specific Equations?

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The discussion focuses on solving the equations 109a + 991b + 101c = 44556 and 1099a + 901b + 1110c = 59800 for positive integers a, b, and c, each less than or equal to 100. Participants explore the possibility of using modular arithmetic to simplify the problem. One contributor expresses gratitude for the insights shared, indicating that the approach clarified their understanding. The conversation highlights the collaborative nature of problem-solving in mathematics. Ultimately, the goal is to evaluate the expression 10000a + 100b + c based on the solutions found.
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If $a, b, c$ are positive integer such that $a, b, c \le 100$ satisfy

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

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

Evaluate $10000a+100b+c$.
 
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Re: Evaluate 10000a+100b+c

anemone said:
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
$$10000a+100b+c = 34316$$
 
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:
Re: Evaluate 10000a+100b+c

anemone said:
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
 
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