Chisenbop!

Algebra Level 3

Chisenbop is a way of counting from 0-99 on your fingers.

You can represent any number from 0-99 by having each finger (including thumbs) represent a number as follows:

and you put up the fingers you need to add up to the number.

For example, for the number 52 you would raise your thumb on your left hand (50) and two fingers on your right hand (2).

When counting from 0-99, how many of those numbers require exactly 5 fingers to be raised?

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The answer is 18.

1 solution

Jordan Cahn
Oct 31, 2018

We count by range:

0-9 has one number requiring five fingers: 9
10-19 has two: 14 and 18
20-29 has two: 23 and 27
30-39 has two: 32 and 36
40-49 has two: 41 and 45
50-59 has two: 54 and 58
60-69 has two: 63 and 67
70-79 has two: 72 and 76
80-89 has two: 81 and 85
Finally, 90-99 has one: 90 itself

Thus there are $1 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 1 = \boxed{18}$ numbers with five fingers raised.

Is there any general method to do this, without any counting?

Parth Sankhe - 2 years, 7 months ago

In general a digit $d$ will require $F(d) = \left\lfloor \frac{d}{5} \right\rfloor + (d\bmod 5)$ fingers on the appropriate hand. Thus, given $a$ with $1\leq a < 9$ , there will be precisely two $b$ for which $\overline{ab}$ will require raising five fingers: $b=5-F(a)$ and $10-F(a)-1$ .

We need to consider $a=0$ and $a=9$ separately, since in each case one of those two $b$ fails:

If $a=0$ then the first equation yields $\overline{ab}=0$
If $a=9$ then the second equation yields $\overline{ab} = 94$ , which actually requires nine fingers.

Jordan Cahn - 2 years, 7 months ago

Thank you!

Parth Sankhe - 2 years, 7 months ago

Chisenbop!

The answer is 18.

1 solution

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