Math Parlor Trick

Question

A magician asks a person in the audience to think of a number $\overline {abc}$. He then asks them to sum up $\overline{acb}, \overline{bac}, \overline{bca}, \overline{cab}, \overline{cba}$ and reveal the result. Suppose it is $3194$. What was the original number?

The obvious approach was modular arithmetic.

$(100a + 10c + b) + (100b + 10a + c) + (100b + 10c + a) + (100c + 10a + b) + (100c + 10b + a) = 3194$

$122a + 212b + 221c = 3194$

Since $122, 212, 221 \equiv 5 (mod\space9)$ and $3194 \equiv 8 (mod\space9)$

$5(a + b + c) \equiv 8 (mod\space9)$

So, $a + b + c = 7$ or $16$ or $26$

Hit and trial produces the result $358$. Any other, more elegant method?

score 4 · Accepted Answer · answered Jun 29 '13 at 05:20

4

The sum of all six combinations is $222(a+b+c)$

So, $3194+100a+10b+c=222(a+b+c)$

As $3194/222>14$

If $a+b+c=15, 100a+10b+c=222(15)-3194=136$ $\implies a+b+c=1+3+6=10\ne 15$

If $a+b+c=16, 100a+10b+c=222(16)-3194=358$ $\implies a+b+c=3+5+8=16$ as needed

answered Jun 29 '13 at 05:20

lab bhattacharjee

274,582

Which combinations are you referring to? – Gerard Jun 29 '13 at 05:23
1

@AnanayAgarwal, the last one $a=3,b=5,c=8$ – lab bhattacharjee Jun 29 '13 at 05:24

score 3 · Answer 2 · answered Jun 29 '13 at 05:34

Let $S$ be the sum,

$$S \text{ mod} 10=A$$

$$S \text{ mod} 100=B$$

$$A=2b+2a+c$$ $$\frac{B-A}{10}=(2c+2a+b)$$ $$\frac{S-B}{100}=(a+2b+2c)$$

$$\text{Now just solve the system of equations for $a$ $b$ and $c$}$$ $$\text{ The original number will be a+10b+100c}$$ $$\text{ Now memorize this formula and do the addition in your head}$$

Math Parlor Trick

2 Answers2