Impossible. Anything divided by 1 is itself.
The smallest 4 digit number divisible by 50 is 1000, so to have a remainder of 17, it is 1000 + 17 = 1017.
If a 3-digit number is estimated to have a quotient of 50 when divided by a 1-digit number, we can express it as ( n \div d \approx 50 ), where ( n ) is the 3-digit number and ( d ) is the 1-digit divisor. This implies that ( n ) is approximately ( 50d ). For example, if ( d = 2 ), then ( n ) could be around 100 (since ( 50 \times 2 = 100 )), and if ( d = 3 ), then ( n ) could be around 150 (since ( 50 \times 3 = 150 )). Thus, two possible pairs are (100, 2) and (150, 3).
55 52+52 = 50
1/80
A multiple of 50 is any number in the 50 times table. So the 3 digit multiples of 50 are: 100, 150, 200 ... 850, 900, 950.
10017
1017
4000
The smallest 4 digit number divisible by 50 is 1000, so to have a remainder of 17, it is 1000 + 17 = 1017.
The number 50 is composite. Any two-digit (or more-digit) number that ends in zero can be divided evenly by 10 (and 5 and 2, for that matter). That means no two- or more-digit number ending in 0 can be prime.
To find the product of a 2-digit number and a 4-digit number that is approximately 500,000, we need to consider the magnitude of the numbers involved. Since a 2-digit number ranges from 10 to 99 and a 4-digit number ranges from 1000 to 9999, their product will be in the range of 10,000 to 99,000,000. To get a product around 500,000, we can estimate that the 2-digit number is around 50 and the 4-digit number is around 10,000. Therefore, the product of a 50 and 10,000 is 500,000.
55 52+52 = 50
100
1/80
A multiple of 50 is any number in the 50 times table. So the 3 digit multiples of 50 are: 100, 150, 200 ... 850, 900, 950.
50
5.5556