n-25r=17 I think. I did this a while ago.
The smallest 4 digit number divisible by 50 is 1000, so to have a remainder of 17, it is 1000 + 17 = 1017.
n + 20 - 17 + 13 - n = 20 - 17 + 13 = 3 + 13 = 16
To write the remainder as a whole number, simply express it as the leftover amount after division. For instance, if you divide 17 by 5, the quotient is 3 and the remainder is 2. You can write this as 17 = 5 × 3 + 2, where 2 is the whole number remainder. Ensure that the remainder is always less than the divisor.
Since 17 is a factor of 119, the remiander will be the same modulo 17. Thus, the remainder, on division by 17 would be 19 except that 19 contains one more 17 - leaving a final remainder of 2.
A remainder is the fraction left over as a result of dividing an uneven number. 52 ÷ 3 = 17 with a remainder of 1/3.
"F2013" is not exactly a number. Nor is "F100".
The smallest 4 digit number divisible by 50 is 1000, so to have a remainder of 17, it is 1000 + 17 = 1017.
n + 20 - 17 + 13 - n = 20 - 17 + 13 = 3 + 13 = 16
When dividing any integer by 17, the remainder can range from 0 to 16, as the remainder must be less than the divisor. Therefore, the greatest possible remainder when the divisor is 17 is 16. This means that when any integer is divided by 17, the remainder will be less than 17 but could be as high as 16.
To write the remainder as a whole number, simply express it as the leftover amount after division. For instance, if you divide 17 by 5, the quotient is 3 and the remainder is 2. You can write this as 17 = 5 × 3 + 2, where 2 is the whole number remainder. Ensure that the remainder is always less than the divisor.
Since 17 is a factor of 119, the remiander will be the same modulo 17. Thus, the remainder, on division by 17 would be 19 except that 19 contains one more 17 - leaving a final remainder of 2.
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A remainder is the fraction left over as a result of dividing an uneven number. 52 ÷ 3 = 17 with a remainder of 1/3.
If x ≡ 39 mod 357 then: x = 357k + 39 for some integer k. Now 357 = 21×17, and 39 = 2×17+5 → x = 21×17×k + 2×17 + 5 → x = 17(21k + 2) + 5 → x = 17m + 5 where m = 21k + 2 (is an integer) → x ≡ 5 mod 17 → the remainder when the number is divided by 17 is 5.
17
An even number can be divided by 2 evenly. An odd number will have a remainder of 1 when divided by 2. 17 is an odd number.
85. A number when divided by 238, leaves a remainder 79. What will be the remainder when the number is divided bv 17 ? (1) 8 (2) 9 (3) 10 @) 11