$2$ s in Divisor Product of $100$

Consider the multiples of two in the prime factors of each divisor. These will determine the number of 2s in the product. We can ignore 1, 5, and 25 as they are not multiples of two.

Thus there are 9 in total.

As we know that from number theory, the product of positive divisors of $n∈N =n^{\frac{τ(n)}{2}}$ where $τ(n)=$ number of positive divisors of $n∈N$ .

So, the product of positive divisors of $100$ is $= 100^{\frac{τ(100)}{2}}$
$= 100^{\frac{9}{2}}$
$= (2^2 × 5^2)^{\frac{9}{2}}$ $= 2^9 × 5^9$ Which clearly imply the product is divisible by $2^9$ so that $m=9$

Edit : adding more details about τ(n)= number of divisors of $n∈N$ .

To find τ(n) First express $n$ in the prime factorization form! Say,

$n= p^a × q^b × r^c ×...$ where $p,q,r,....$ are prime numbers! then,

$τ(n)= (a+1)(b+1)(c+1)....$

So that, $τ(100) = τ(2^2 × 5^2)=(2+1)(2+1)=9$

Generalization : $\begin{aligned} N & = p_1^{e_1}\cdot p_2^{e_2}\cdots p_n^{e_n} & \text{prime decomposition of a number}\ N \\ D & = \prod_{\stackrel{d\geq 1}{d|N}} d & \text{product of its divisors} \\ D & = p_1^{m_1}\cdot p_2^{m_2}\cdots p_n^{m_n} & \text{its prime content} \\ m_i & = \tfrac12e_i \cdot \prod_{j=1}^n (e_j+1) & \text{maximum power of}\ p_i.\end{aligned}$

In this case, $100 = N = 2^2\cdot 5^2$ ; so $m = \tfrac12\cdot 2\cdot (2+1)\cdot (2+1) = \boxed{9}.$

The positive divisors of 100 are 1, 2, 4, 5, 10, 20, 25, 50, and 100.

Using prime factorization:

$2=2^1$

$4=2^2$

$5=5^1$

$10=2^1\cdot5^1$

$20=2^2\cdot5^1$

$25=5^2$

$50=2^1\cdot5^2$

$100=2^2\cdot5^2$

Because we are multiplying these factors together, we can add identical numbers' powers. Thus, our product is equal to $1 \cdot 2^9 \cdot 5^9$

And we can easily see that the largest $2^m$ number that this product is divisible by is $2^9$

$m=9$

Note that there are 9 total divisors $d$ of 100 . Total divisors $1 \leq d \leq 100$ are $1, 2, 4, 5, 10 , 20, 25, 50, 100$ . $\begin{aligned}& \therefore P_{100} = 1\times 2\times 4 \times 5 \times 10 \times 20 \times 25 \times 50 \times 100 =2^9 \times k \end{aligned}$ Hence required highest integer value for $m =9$

The divisors of 100 is 1,2,4,5,10,20,25,50,100. And their product is 200,000,000 or 2x10^8 which is 2x(2x5)^8 = 2^9x5^8. And the largest power of two that is a divisor of the product of the divisors of 100 is 9

Positive divisors of 100 are:

1, 2, 4, 5, 10, 20, 25, 50, 100.

So by factorizing each of them we get:

1= 2^0

2= 2^1

4= 2^2

5= 5 × 2^0

10= 5 × 2^1

20= 5 × 2^2

25= 5^2 × 2^0

50= 5^2 × 2^1

100= 5^2 × 2^2.

We need to take their product that will consist of adding all the powers of 5 and 2. Since, 2^m divides the product so the maximum value for m can be the power of 2 obtained on factorizing the product. As we know on multiplication the power of common bases get adds up , therefore, largest value for m = 9 .

100 = 5^2x2^2 or 5x5x2x2, so only the combinations of 5 and 2 from exp 0 to exp 2 can be divisors. Then I thought that I could put all possible combinations in a table, so:

BASE 3:

And I count all the twos (the first column) 0+0+0+1+1+1+2+2+2=9 So 2^9 is the maximum 2^m

m = 9

The general expresion to 10^n would be (n+1)^2 • n/2

As $100=2^2\cdot5^2$ , every factor of 100 will be of the form $2^{k_{1}}\cdot5^{k_{2}}$ , where $0 \le k_{1},k_{2} \le 2$ . The three factors for which $k_{1}=0$ will contribute no 2's, the three factors for which $k_{1}=1$ will each contribute one 2 (for a combined contribution of $2^3)$ , and the three factors for which $k_{1}=2$ will each contribute two 2's (for a combined contribution of $2^6)$ . Thus the total contribution of all factors will be $2^9$ , so that will be the highest power of 2 dividing the product in question.

List all the factors: 1, 2, 4, 5, 10, 20, 25, 50, 100

Regroup the possible combinations: 1x100 = 100 | 2x50 = 100 | 4x25 = 100 | 5x20 = 100 | 10 = 10

By counting the number of "0"s, we know that the product of all factors is 10^9

Break down the product into a combination of 2 times 5: "10^9 = 2^9 x 5^9"

This shows that the maximum value for m in 2^m is 9

By filling in the blanks, we would have the product of all the positive divisors of 100 listed as: 1 x 2 x 4 x 5 x 10 x 20 x 25 x 50 x 100. By using prime factorization of each of these we can rewrite this product as 1 x 2 x $2^2$ x 5 x $2*5$ x $2^2 * 5$ x $5^2$ x $2*5^2$ x $2^2 * 5^2$ . Then we see that $2^9$ would be a factor of this product.

$100 = 2^2×5^2$ , thus, 100 has (2+1)(2+1) = 9 divisors. From these, only 1, 5 and 25 are not multiples of 2 ( $5^0, 5^1, 5^2$ respectively). This means there are 6 multiples of two in the product. But then, we've got to count one 2 more for each multiple of four, which are 4, 20 and 100 ( $2^2, 2^2×5, 2^2×5^2$ respectively) so there are three 2s more to count. It is easy to see there are no multiples of other powers of two in this product, so it is divisible by $2^{6+3} = 2^9$ , and we can see that 9 is maximum $m$ . Hence, the answer is $\boxed 9$ .

Here's one I haven't read here (maybe I was just too lazy). We can combine the Divisors in pairs: 1x100, 2x50, 4x25, 5x20 and 10x10. As all these Products are 100 we see that in each pair the factor 2 occurs twice however we only have one 10 so we subtract the one two we counted double. 5x2-1=9

First, we consider all the divisors that are of the form $2^1\cdot k$ where $k\nmid 2$ . There are three values of $k$ that are divisors of $100$ : $k=1,5,25$ . Thus, three divisors have $2^1$ as a factor. Similarly, with the same values of $k$ , there are three divisors of the form $2^2\cdot k$ . Adding together the exponents of all these divisors, we get $m=1\cdot 3+2\cdot 3=9.$

(1 * 2 * 4 * 5 ... * 100)/2^m = ((25^3) * (5^3))/2^(m-9) Therefore m=9

$100$ can be written as $2^2\times5^2$ . The powers of two can be generated with two cases: $2\times5^n$ or $2^2\times5^n$ . The first case has $3$ powers of 2 because there are $3$ ways to multiply a power of five by 2, those ways being $5^0,5^1,$ and $5^2$ . The second case has $2\times3=6$ powers of 2 because there are three ways to multiply a power of five by 2, then doubled because we are multiplying by a power of $2$ . Adding these cases up gets us a solution of $3+6=$ 9 .

Positive divisors of 100: 1, 2, 4, 5, 10, 20, 25, 50, 100

Product of the positive divisors of 100 = $1 \cdot 2 \cdot 4 \cdot 5 \cdot 10 \cdot 20 \cdot 25 \cdot 50 \cdot 100 = 10^9$

We know that, $1000 = 10^3 = 8 \cdot 125 = 2^3 \cdot 5^3$

Therefore, $10^9 = (10^3)^3 = (2^3 \cdot 5^3)^3 = 2^9 \cdot 5^9$

Hence, m = 9 .

Matlab solution: sum(mod(prod(divisors(100)),2.^[1:100])==0)

Another programmatic solution (JavaScript):

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const product = new Array(100) // Empty array for numbers 1..100
    .fill(0) // Fill it so it can be mapped
    .map((e, i) => i + 1) // [1..100]
    .filter(e => 100 % e == 0) // [1,2,4,5,..,100]
    .reduce((acc, x) => acc * x, 1); // Calculate product

/** Infinite sequence of integers. */
function* integers()
{
    let i = 1;
    while (true)
        yield i++;
}

const mValues = [];
for (let m of integers())
{
    const value = Math.pow(2, m);
    // Stop if power exceeds product.
    if (value > product)
        break;

    // Add m to list of potential solutions if the power divides the product.
    if (product % value == 0)
        mValues.push(m);
}
// The last value in the list is the solution.
console.log(mValues[mValues.length - 1]);

Mathematica

Here are the prime factors of the final product

FactorInteger[Times @@ Divisors@100]

{{2, 9}, {5, 9}} which means *2^9 * 5^9 *