Smallest Digit Sum

Note that $2025000 = 2^3 * 3^4 * 5^5$

Now we can also see that if we were to not use these prime factors as the digits themselves and instead multiplied a few together, that it would increase the digit sum or mean that the product of the digits changes.

If we multiplied anything with the $5$ then we get a two digit number and so we can't do that to get a digit of the number.

So we can either do $2*3$ or $2*2$ or $2*2*2$ or $3*3$

In each case the digit we end up with is either equal to or larger than the sum of the factors used to produce the digit, and hence we will get the smallest digit sum if all the digits are just the primes factors of $2025000$ .

This gives our digit sum to be $2*3 + 3*4 + 5*5 = 43$

Since the products of the digits must equal 2025000, we first want to find the factors of 2025000. To do this, we prime factorize.

2025000 = 1000 \times 2025 = 1000 \times 45^{2} = 10^{3} \times 9^{2} \times 5^{2} = 5^{5} \times 2^{3} \times 3^{4}.

So we know the digits have to comprise of multiples of those prime factors. Now the question becomes "To find the least digit sum, should the number be composed of all prime factors as digits, or should we combine some prime factors to make a single digit(such as 2*3 = 6)?"

Looking at 2 and 3, we see that if we have them as separate digits the digit sum would be 5, but if we combine them their digit sum would be 6.

We conjecture that ab \geq a+b for all integers a, b \geq 2. To prove ab \geq a+b: (a-1)(b-1) \geq 1 for all integers a,b \geq 2 as our bounds suggest.

ab - a - b + 1 \geq 1

ab \geq a + b, and the proof is complete.

Therefore, we know that to get the lowest digit sum, we should NOT combine our prime factors.

Thus, each prime factor of 5^{5} \times 3^{4} \times 2^3 is a digit in our number with the lowest digit sum, so the digit sum is simply

5\times 5 + 3\times 4 + 2\times 3 = 43.

In terms of its prime factors, $2025000 = 2^3 \times 3^4 \times 5^5$ . Because the product of the digits is nonzero, all the digits must be nonzero. Moreover, there can't be 1s because the integer must have the smallest digit sum. There are 2 possibilities for the digits: either the digits are the factors themselves, i.e., 222333355555 arranged in no particular order, or the digits contain product between the factors, which would only allow $6=2 \times 3$ since $2 \times 5 = 10$ and $3 \times 5 = 15$ are no longer digits. Knowing that $2 + 3 = 5 < 6$ , the digit sum is increased by 1 for every pair of 2 and 3 replaced with a 6 and therefore the digit sum for the second possibility must be greater than that of the first. Hence, the smallest digit sum is that of the first possibility, which is $2 + 2 + 2 + 3 + 3 +3 + 3 + 5 + 5 + 5 + 5 + 5 = 43$ .

We have $202500 = 2^3 \cdot 3^4 \cdot 5^5.$ First consider the digits whose product gives the factors of $5.$ These digits can only can be $5$ s, since all other positive multiples of $5$ are not digits. So we must have five $5$ s in the number.

Now we must determine which numbers contribute the factors of $2$ and $3.$ The digits with only factors of $2$ and $3$ are $2, 3, 4, 6, 8, 9$ ; we use casework and our goal (have the lowest digit sum) to determine which digits to put in the number.

• If a $4$ is in the number, having two $2$ s gives the same factors and keeps the same digit sum, so we may safely ignore $4$ s.

• If a $6$ is in the number, having a $2$ and a $3$ contributes the same factors and lowers the digit sum from $6$ to $2+3=5,$ so we can ignore $6$ s.

• If a $8$ is in the number, having three $2$ s contributes the same factors and lowers the digit sum from $8$ to $2+2+2=6,$ so we can ignore $8$ s.

• If a $9$ is in the number, we add two factors of $3,$ but having two $3$ s themselves lowers the digit sum from $9$ to $3+3=6,$ so we can ignore $9$ s.

Thus, the only digits we need consider are $2$ and $3.$ We must have three $2$ s, four $3$ s, and five $5$ s, so the minimal digit sum is $3\times2+4\times3+5\times5 = \boxed{43}.$

We know that $10 = 2 \times 5$ First of all, $1000 = 2^{3} \times 5^{3} = 2 \times 2 \times 2 \times 5 \times 5 \times 5$ Next, $2025 = 81 \times 25$ $2025 = 3 \times 3 \times 3 \times 3 \times 5 \times 5$ Now, we got 3 $2$ 's, 4 $3$ 's and 5 $5$ 's. Therefore, $3 \times 2 + 4 \times 3 + 5 \times 5 = \boxed{43}$

We have to simply find the (lowest multiple) of the given number (2025000) ,so as to find the smallest digit sum whose product of digit would be equal to the given number. In this case, 2025000 = 2^3 × 3^4 × 5^5 or 2025000= 2×2×2×3×3×3×3×5×5×5×5×5

When looking for the smallest digit sum whose product equal something, the prime factorization of that number (2025000) can be really helpful because the prime factorization essentially wen multiplied gets the number, and assuming those primes are not too big, which they should not be because the target has numbers like 2, 5, and 0. After taking the prime factorization of the number you find that it is 2 2 2 3 3 3 3 5 5 5 5*5. As assumed the primes were not too large ones. The smallest digit sum results from simply adding the numbers in the prime factorization, because trying to multiply the numbers together would only increase the digits sum, thus the answer is 43 after addition.

Since every integer greater than $1$ can be written as a product of prime numbers, the set of all integers whose product of their digits equals $2025000$ is formed by the subset of all possible arrangements made using all the prime factors of $2025000$ for which the digit sum will be the same, and the subset of all integers whose some of its digits are all the prime factors of $2025000$ and the rest are $1$ 's (we can add one, two or more $1$ 's and try all possible permutations). It is straightforward that the smallest digit sum is the one derived from the integers of the first subset.

So:

$2025000 = 2^3 \times 3^4 \times 5^5$ and $DigitSum_{min}(2025000) = (2 \times 3) + (3 \times 4) + (5 \times 5) = 43$

To solve this problem you must divide the given number repeatedly by prime numbers. Since prime numbers can't be reduced any further, i.e. 9 can be reduced to 3^2, but 7 can't be reduced any further, therefore by dividing by prime numbers you will receive the smallest possible digit sum. By initially dividing 2025000 by 2, (since 2 is the smallest prime number), once you can no longer divide by 2 to produce a whole number, you move onto the next smallest prime number. You then repeat these steps moving onto the next prime number each time it is no longer divisible into a whole number until you reach 1. By doing this you will receive the following result: 2025000/2 = 1012500 1012500/2 = 506250 506250/2 = 253125 253125/3 = 84375 84375/3 = 28125 28125/3 = 9375 9375/3 = 3125 3125/5 = 625 625/5 = 125 125/5 = 25 25/5 = 5 5/5 = 1 Once you have reached one, you simply add together the number you divided by each time. So in this case: 2+2+2+3+3+3+3+5+5+5+5+5 = 43

To acquire the smallest possible digit, Take the number 2025000 and take its L.C.M starting from the smallest divisible integer which in this case is 2.After you get the smallest divisors, simple add them because they will be the smallest possible set of numbers which, if you multiply, can give you 2025000. In this case you will get 3 2's , 4 3's and 5 5's. So the Digit Number is -222333355555. When you multiply the digits it gives you 2025000 as the answer and when you add them (i.e The DIGIT SUM) you get 43 as the answer. Here we should not directly take 4/5 while dividing 2025000 the 1st time because there exists a positive integer below 4/5 which can divide 2025000 and yield the smallest possible answer. Its just a matter of simple L.C.M (LEAST COMMON MULTIPLE)

we have to find smallest possible digit sum amongst all integers whose product of their digits equals 2025000

first we have to factorise this we get 2x2x2x5x5x5x5x5x3x3x3x3

now we have to consider factors in one and two digits only because if we consider three digit factor like 100 then sum will obviously gonna go more than 100

now 3+3+3+3+5+5+5+5+5+2+2+2=43

consider case like 9+9+5+5+5+5+5+2+2+2=49

9+3+3+5+5+5+5+5+2+2+2=46

5^2=25 > 5+5 =10

2^3 = 8 >2+2+2=6

2^2=2+2 (which is same in both the cases sum will be 43 only)

therefore minimum sum will be when all factored prime numbers are in there least powers

therefore minimum sum will be 43

2025000 = 2 * 2 * 2 * 3 * 3 * 3 * 3 * 5 * 5 * 5 * 5 * 5

2 + 2 + 2 + 3 + 3 + 3 + 3 + 5 + 5 + 5 + 5 + 5 = 43

First note that the prime factorization of 2025000 is (2^3) * (3^4) * (5^5). Now, what about the integers with digit product 2025000? From the prime factorization, we can write down an example, which is 222333355555. The digit sum of this number is 43. Can we do better?

It turns out, no, we can't. Because primes are like the "atoms" of the natural numbers, they can't be broken up any further. So any number with digit product 2025000 must have all the primes listed above present.

first,we need to get the prime factorization of 2025000 2025000=(2)(2)(2)(5)(5)(5)(5)(5)(3)(3)(3)(3)

so,the smallest possible digit sum amongst all integers is 2+2+2+5+5+5+5+5+3+3+3+3=43

2025000 = 2^3 3^4 5^5

answer is 2+2+2+3+3+3+3+5+5+5+5+5+5 = 43

Factorising 2025000 into its constituent primes gives you: 2025000 = $2^3$ $\times$ $3^4$ $\times$ $5^5$ .

Each prime factor is a digit of the smallest integer with product of digits equaling 2025000.

By adding all the factors together, you get: 3 ­ $\times$ 2 + 4 $\times$ 3 + 5 $\times$ 5 = 43.

First factorise 2025000 Which comes out to be

$2^{3}*3^{4}*5^{5}$

smallest digit sum is obtained by writing each of its prime factors 1 time so we have the sum as

$2*3+3*4+5*5$

Prime factorization of $2025000 = 2^3 \times 3^4 \times 5^5$ . Adding up all the digit gives us $( 3 \times 2 ) + ( 4 \times 3 ) + ( 5 \times 5 ) = \boxed{43}$

Let the set of integers which have digit product $2025000$ and minimum digit sum be $N$ .

Note that the prime factorisation of $2025000$ is $2^{3} \cdot 3^{4} \cdot 5^{5}$ . Thus an integer with digits 3 2's, 4 3's and 5 5's in any permutation has digit product $2025000$ and digit sum $2 \cdot 3 + 3 \cdot 4 + 5 \cdot 5 = 43$ .

We claim the set of such integers to be $N$ .

Assume there is an integer with digit product $2025000$ and smaller digit sum, $P$ . We construct it by altering the number 222333355555 (an element of $N$ ).

To construct $P$ , we must replace any 2 of the prime digits 2,3,or 5 with their product digit. However, replacing 2 and 3 with 6 merely increases the digit sum by 1 while 10 and 15, product of pairs (2,5) and (3,5) respectively cannot be done.

Therefore there is no method to construct any such $P$ . We conclude the claim to be true and the minimum digit sum to be 43.

2025000=5^5 2^3 3^4,AS THIS ARE ALL PRIME FACTORS OF 2025000.THEREFORE,THERE CANNOT BE FURTHER ANY FACTORIZATION.SO, THESE ARE SMALLEST NUMBERS WHOSE PRODUCT IS 2025000.THEREFORE,ADDITION=5 5+2 3+3*4=43

2025000 = 2 x 2 x 2 x 3 x 3 x 3 x 3 x 5 x 5 x 5 x 5 x 5

Sum of digit = 3(2) + 4(3) + 5(5) = 43

Here, it can be observed that $prime factorisation$ of

$2025000 = 3^4.5^5.2^3$

$\Rightarrow$ smallest digit sum = sum of prime factors multiplied by their respective exponents. $= 3.4 + 5.5 + 2.3 = \boxed {43}$ .

We just need to prime factorize 2025000

We have (2^3)(3^4)(5^5)

Thus , in order to have the product of their digits be 2025000

It must have 3 2 s, 4 3 s , and 5 5`s.

So their digit sum is 3(2) + 4(3) + 5(5) = 43

get the prime factors of 2025000..

$2^{3}$ + $5^{5}$ + $3^{4}$

then get their sum

$\boxed{43}$

we can factorize 2025000=10^3 3^4 5^2 =5^5 3^4 2^3 hence we got the digit sum is 43

2025000=5^5times3^4times2^3

Thus one of the numbers whose product of digits is 2025000 could be 555553333222 , thus sum of the digits = 5times5+3times4+2times3=43