MATH COUNTS QUESTION #3

Algebra Level 2

If $20! = 2^{a} \times b$ for some integer $b$ , what is the greatest possible value of $a$ ?

The answer is 18.

1 solution

Jordan Cahn
Oct 30, 2018

The factor $2$ appears once in every multiple of $2$ less than (or equal to) 20, an additional time in every multiple of $4$ , yet another time in every multiple of $8$ , etc. Thus $a=\left\lfloor \frac{20}{2} \right\rfloor + \left\lfloor \frac{20}{4} \right\rfloor + \left\lfloor \frac{20}{8} \right\rfloor + \left\lfloor \frac{20}{16} \right\rfloor + \left\lfloor \frac{20}{32} \right\rfloor + \cdots = 10 + 5 + 2 + 1 + 0 + \cdots = \boxed{18}$

I don't get how?

James Bacon - 2 years, 6 months ago

$20! = 20\times 19 \times 18 \times 17 \times \cdots \times 3 \times 2 \times 1$ . We want to count how many times $2$ appears as a prime factor of this number. We know every even number has a factor of two, so we count each of those once. Every multiple of four has two factors of two, so we count those twice. Every multiple of 8 has three multiples of two, so we count those three times, etc.

There are 10 multiples of 2, five multiples of 4, two multiples of 8, and one multiple of 16 in our starting product. So there are 18 total 2s.

Jordan Cahn - 2 years, 6 months ago

MATH COUNTS QUESTION #3

The answer is 18.

1 solution

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