5. How many five-digit natural numbers are there in which there are no zeros and the sum of all digits is greater than their product?

The use of a pocket calculator or any manuals is not allowed.

## Ministry of Science and Education of the Republic of Croatia Agency for Education and Education Croatian Mathematical Society

## County Competition in Mathematics February 26, 2024. 7th grade - elementary school

Each task is worth 10 points. In addition to the final result, the procedure is also graded. To earn all points, it is necessary to find all solutions and confirm that there are no others, write down the procedure, and justify your conclusions.
5. How many five-digit natural numbers are there in which there is no zero and the sum of all digits is greater than their product?

Solution. It is known that none of the digits of the desired number is zero. The product of the digits will be smaller if there is a one among them. We will prove that the desired numbers have at least three ones.

First, we will prove that there are no desired numbers in which no digit is one.

In the case that no digit of the number is one, the smallest product of the digits would be achieved for the number 22222, but then $32=2 \cdot 2 \cdot 2 \cdot 2 \cdot 2>2+2+2+2+2=10$, so the product of the digits is greater than their sum.

The sum of the digits of a five-digit number can be at most $9+9+9+9+9=45$. If we increase any digit of the number 22222 by at least 1, the product of the digits will be at least $2 \cdot 2 \cdot 2 \cdot 2 \cdot 3=48$, which is more than the maximum possible sum of the digits.

Therefore, at least one digit of the desired five-digit number must be one.

2 POINTS

Second, we will show that there are no desired numbers that have exactly one one. The smallest product in this case is achieved with four twos and one one. In this case, the product is 16, and the sum is 9, so this possibility is ruled out. By increasing any digit by 1, the sum increases by 1, but the product increases by more than 1, so there are no such possibilities.

1 POINT

Third, we will show that there are no desired numbers that have exactly two ones. The smallest product in this case is achieved with three twos and two ones. In this case, both the product and the sum are equal to 8, so this possibility is ruled out. By increasing any digit by 1, the sum increases by 1, but the product increases by more than 1, so there are no such possibilities.

1 POINT

Consider the cases where at least three digits of the desired number are ones.

| Number of ones | Desired number | Sum greater than product? | Number of cases |
| :---: | :---: | :---: | :---: |
| 5 | 11111 | YES $(5>1)$ | 1  |
| 4 | 21111,12111,11211,11121,11112 | YES $(6>2)$ | 5 |
|  | 31111,13111,11311,11131,11113 | YES $(7>3)$ | 5 |
|  | 41111, ... | YES $(8>4)$ | 5 |
|  | 51111, | YES $(9>5)$ | 5 |
|  | 61111, ... | YES $(10>6)$ | 5 |
|  | 71111, ... | YES $(11>7)$ | 5 |
|  | 81111, ... | YES $(12>8)$ | 5 |
|  | 91111, | YES $(13>9)$ | 5 |
|  | Total with four ones: |  | $8 \cdot 5=40$ |
| 3 | 22111,21211,21121,21112,12211,  | YES $(7>4)$ | $10^{*}$ |
|  | 12121,12112,11221,11212,11122 |  |  |
|  | 32111,31211,31121,31112,13211,  | YES $(8>6)$ | $20^{**}$ |
|  | 13121,13112,11321,11312,11132,  |  |  |
|  | 23111,21311,21131,21113,12311,  |  |  |
|  | 12131,12113,11231,11213,11123 |  |  |
|  | 42111,41211,41121,41112,14211,  | YES $(8>6)$ | $20^{***}$ |
|  | 14121,14112,11421,11412,11142,  |  |  |
|  | 24111,21411,21141,21114,12411,  |  |  |
|  | 12141,12114,11241,11214,11124 |  |  |
|  | Total with three ones: |  | 50 |

In total, there are $1+40+10+20+20=91$ such five-digit numbers.

1 POINT

* The student does not have to list all possibilities but can determine their number as follows: one two can be placed in any of the 5 positions, the second two in any of the remaining 4 (fill the remaining three positions with ones). However, this counts the possibilities twice because the two twos are identical digits. Therefore, the number of possibilities must be divided by 2, i.e., we have $\frac{5 \cdot 4}{2}=10$ possibilities.

** The student does not have to list all possibilities but can determine their number as follows: a two can be placed in any of the 5 positions, and a three in any of the remaining 4 (fill the remaining three positions with ones). Thus, we have $5 \cdot 4=20$ possibilities.

*** The student can refer to the previous case.

TOTAL 10 POINTS

Note. All desired numbers can be determined in different ways, for example, by considering the product of the digits. Possible products are from 1 to 9. Determining all numbers earns 5 POINTS, and 1 POINT is earned for: the idea of considering the product along with the number 11111, determining all numbers for which the product of the digits is 2, 3, 5, 7, or 9, determining all numbers with a product of the digits 4, determining all numbers with a product of the digits 6, determining all numbers with a product of the digits 8. The answer to how many desired numbers there are in total earns 1 POINT (and this point is awarded if the answer is given without justification). Proving that there is no desired number with two threes and three ones earns 1 POINT, and proving that the product cannot be greater than 9 earns 3 POINTS.

## COUNTY COMPETITION IN MATHEMATICS