It is 98910.
A 2 digit number divided by a four digit number, such as 2345, will leave the whole 2-digit number as a remainder. It cannot leave a remainder of 1.
This is an extremely poor question. There are 28 pairs of numbers and, in less than 10 minutes, I could find reasons why 17 pair were different from the other numbers in the set. A bit more time and I am sure I could do the rest. So here they are: 53 and 72: leave a remainder of 15 when divided by 19. 53 and 77: leave a remainder of 5 when divided by 12. 53 and 82: leave a remainder of 24 when divided by 29. 53 and 87: leave a remainder of 2 when divided by 17. 53 and 95: leave a remainder of 11 when divided by 21. 53 and 97: leave a remainder of 9 when divided by 11. 67 and 95: leave a remainder of 11 when divided by 28. 67 and 87: leave a remainder of 7 when divided by 17. 67 and 97: leave a remainder of 7 when divided by 30. 72 and 82: leave a remainder of 0 when divided by 2 (are even). 72 and 87: leave a remainder of 0 when divided by 3 (multiples of 3). 72 and 95: leave a remainder of 3 when divided by 23. 72 and 97: leave a remainder of 22 when divided by 25. 77 and 95: leave a remainder of 5 when divided by 9. 77 and 97: leave a remainder of 17 when divided by 20. 82 and 95: leave a remainder of 4 when divided by 13. 87 and 95: leave a remainder of 7 when divided by 8.
Well, isn't that a happy little math problem! When a number is divided by another number, we get a quotient and a remainder. In this case, we are looking for natural numbers that leave a remainder of 5 when divided into 47. So, we start by listing some numbers that fit this description: 5, 10, 15, 20, 25, 30, 35, 40, and 45. That's 9 different numbers in total that will leave a remainder of 5 when divided into 47.
10
To find the sum of all 3-digit numbers that leave a remainder of 5 when divided by 7, we first identify the smallest and largest 3-digit numbers in this category. The smallest is 105 (since (105 \mod 7 = 5)) and the largest is 995 (since (995 \mod 7 = 5)). These numbers form an arithmetic sequence with a common difference of 7. The number of terms can be found using the formula for the nth term of an arithmetic sequence, and the sum can then be calculated using the sum formula (S_n = \frac{n}{2} (a + l)), where (a) is the first term, (l) is the last term, and (n) is the number of terms. The sum of these numbers is 71,500.
The infinitely many numbers that leave a remainder when divided by 9. The infinitely many numbers that leave a remainder when divided by 9. The infinitely many numbers that leave a remainder when divided by 9. The infinitely many numbers that leave a remainder when divided by 9.
A 2 digit number divided by a four digit number, such as 2345, will leave the whole 2-digit number as a remainder. It cannot leave a remainder of 1.
All numbers that leave a remainder when divided by 35.
This is an extremely poor question. There are 28 pairs of numbers and, in less than 10 minutes, I could find reasons why 17 pair were different from the other numbers in the set. A bit more time and I am sure I could do the rest. So here they are: 53 and 72: leave a remainder of 15 when divided by 19. 53 and 77: leave a remainder of 5 when divided by 12. 53 and 82: leave a remainder of 24 when divided by 29. 53 and 87: leave a remainder of 2 when divided by 17. 53 and 95: leave a remainder of 11 when divided by 21. 53 and 97: leave a remainder of 9 when divided by 11. 67 and 95: leave a remainder of 11 when divided by 28. 67 and 87: leave a remainder of 7 when divided by 17. 67 and 97: leave a remainder of 7 when divided by 30. 72 and 82: leave a remainder of 0 when divided by 2 (are even). 72 and 87: leave a remainder of 0 when divided by 3 (multiples of 3). 72 and 95: leave a remainder of 3 when divided by 23. 72 and 97: leave a remainder of 22 when divided by 25. 77 and 95: leave a remainder of 5 when divided by 9. 77 and 97: leave a remainder of 17 when divided by 20. 82 and 95: leave a remainder of 4 when divided by 13. 87 and 95: leave a remainder of 7 when divided by 8.
Eight.
They are the numbers that when divided by 2 leave a remainder of 1
0.1522
There are 11 such numbers.
Well, isn't that a happy little math problem! When a number is divided by another number, we get a quotient and a remainder. In this case, we are looking for natural numbers that leave a remainder of 5 when divided into 47. So, we start by listing some numbers that fit this description: 5, 10, 15, 20, 25, 30, 35, 40, and 45. That's 9 different numbers in total that will leave a remainder of 5 when divided into 47.
0.2857
Bn = 37n+1, where n is a positive integer from 1 to infinity, therefore there exists an infinite number of numbers that can be divided into 37 and leave 1 as a remainder
10