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.
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.
171 cannot go into ANY two digit number - its smallest positive multiple is 171. Therefore the remainder must be the original 2 digit number. However, the question states that the remainder is 6. That is a contradiction. Consequently, there cannot be any solution to the question as stated.
103
111
-3
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.
171 cannot go into ANY two digit number - its smallest positive multiple is 171. Therefore the remainder must be the original 2 digit number. However, the question states that the remainder is 6. That is a contradiction. Consequently, there cannot be any solution to the question as stated.
8. For a remainder of 2 when divided by 4, they must be 10, 14, 18, ..., 98 For a remainder of 1 when divided by 3, they must be 10, 13, 16, ..., 97 The overlap is 10, 22, 34, 46, 58, 70, 82, 94 - 8 numbers
17
103
The simplest answer is 10124 divided by 125.
Every 2 digit number, when divided by the number one less than it, will result in a remainder of 1.
100
111
12.5