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Kids maths homework -

ABCDE +  
 BCDE +  
  CDE  
______  
74915

We're told A = 7. I don't even know how to go about working this out. Help?!

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  • $\begingroup$ Go from back to front. As a start, note that $E$ is $5$ and $D$ is $0$. $\endgroup$ – André Nicolas Sep 10 '15 at 4:29
  • $\begingroup$ To be honest, when I read the title, I understood it as "I've been stuck on this homework problem for 10 years". $\endgroup$ – Trogdor Sep 10 '15 at 5:01
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As noted by Andre Nicholas in the comment: start from the back.
Note that $E+E+E$ gives a value which ends with 5. Only the number 5 has such a property.
Now we know that $E+E+E=15$. Since $DE+DE+DE$ gives a value which ends with a 15 at the end. That means D is 0.
$DE+DE+DE$ is 15, which doesn't carry any value to the next number, so, $C+C+C=9$ and $C=3$. Similarly, $C+C+C=9$ doesn't carry any value to the next number. Hence, $B+B=4$ and $B=2$. Therefore $ABCDE=72305$.

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I'll set the problem up. I trust you can follow and apply the logic to the rest of the problem. From the rightmost column, we have that $E+E+E = 5$ (maybe after carrying) - equivalently $E+E+E \equiv 5 \pmod {10}$. The only way for this to happen is if $E = 5$.

Then $E+E+E = 15$ and so we have to carry, giving that $1+D+D+D = 1$ or equivalently $D+D+D = 0$ (again, maybe after carrying). What value(s) of $D$ give you that $3D$ is a multiple of $10$? (Go case by case, starting at $0$ to see what the values of $D$ have to be if you're not sure.) Can you take it from here?

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Instead from the back, you can also start from the front!

Since $A$ is given to be $7$, the condition is equivalent to $$2000 \times B + 3 \times \underline{CDE} = 4915$$ Now looking at everything modulo $3$, we get $$2 \times B \equiv 4915 \equiv 1 \pmod 3$$

Since $B \le \frac{4915}{2000} = 2.4575$, $B$ can only be $2$. This leaves us $3 \times \underline{CDE} = 915 \implies \underline{CDE} = 305$.

Combine all this, we have $\underline{ABCDE} = 72305$.

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