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Get and Sign to REPORT ADDITIONAL INFORMATION CPD 11 383 Rev 513 20182022
Lesion o aver a p rdida o da o ha sido notificado al Departamento de Polic a de Chicago puede obtenerse despu s de 14 d as de trabajo desde la fecha en que el incidente fue reportado. Para obtener una copia del informe mande un cheque o giro pagadero al DEPARTMENT OF REVENUE  CITY OF CHICAGO por el importe de. Escrito arriba sea suficiente a prop sito del seguro sin embargo podr n haber Instancias en las que una copia del reporte informe del caso sea deseado. Una copia del reporte del caso que...
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Mathematical Puzzles: What is () + () + () = 30 using 1,3,5,7,9,11,13,15?
My question had been merged with another one and as a result, I have added the previous answer to the present one. Hopefully this provides a clearer explanation. Just using the numbers given there, it's not possible, because odd + odd = even, even + odd = odd. 30 is an even number, the answer of 3 odd numbers must be odd, it's a contradiction. If what people say is true, then the question is wrongly phrased its any number of operations within those three brackets must lead to 30. Then it becomes a lot easier. Such as 15 + 7 + (7 + 1). That would give 30. But it assumes something that the question does not state explicitly and cannot be done that way. I still stick to my first point, it can't be done within the realm of math and just using three numbers, if not, then the latter is a way to solve it.EDIT: This question has come up many times, Any odd number can be expressed as the following, Let [math]n, m, p[/math] be an odd number, [math] n = 1 (mod[/math] [math]2), m = 1 (mod[/math] [math]2), p = 1 (mod[/math] [math]2)[/math][math]n+m+p = 1 + 1 + 1 (mod[/math] [math]2)[/math]Let's call [math]n+m+p[/math] as [math]x[/math][math]=> x = 3 (mod[/math] [math]2)[/math]Numbers in modulo n can be added, I'll write a small proof for it below, [math]a = b (mod[/math] [math]n), c = d (mod[/math] [math]n)[/math][math]a+c = b+d (mod[/math] [math]n)[/math]We can rewrite [math]b[/math] and [math]d[/math] in the following way, [math]n  (b  a) => ba = n*p[/math] (for some integer p) [math]b = a + np[/math][math]b = a + np, d = c + nq[/math][math]b + d = a + np + c + nq[/math][math]b+d = a + c + n(p + q)[/math]Now we have shown that our result is true, moving forward, [math]3 = 1 (mod[/math] [math]2)[/math][math]x = 1 (mod[/math] [math]2)[/math]Therefore the sum of three odd numbers can never be even. It will always be congruent to 1 in mod 2.(This was what I wrote for a merged answer).Modular arithmetic  Link on modular arithmetic, the basic operations. Modular multiplicative inverse  The multiplicative inverse in modular operations.Congruence relationFermat's little theorem Modular exponentiation  As title suggests.Good luck! 
Can you add 5 odd numbers to get 30?
It is 7,9 + 9,1 + 1 + 3 + 9 = 30Wish you can find the 7,9 and 9,1 in the list of1,3,5, 7,9 ,11,13,151,3,5,7, 9,1 1,13,15 
Adding 5 odd numbers, the number starting 1 to 20, the result should be 50. How we will solve this?
So what exactly is an odd number? Well before we define an odd number, let's think about what an even number is. A number is even if it is divisible by 2. Or in other words, a number n is even iff n = 2k for some integer k. So now that we have a definition of what it means to be even, we can easily define what an odd number is. Any odd number is always one more than an even number, so a number n is odd iff n = 2k + 1 for some integer k.So you are asking what five odd numbers can we add together that will have a sum of 50? Well let's first add any two odd numbers together. Call them 2k + 1 and 2l + 1 because k and l may be different. 2k + 1 + 2l + 1 = 2k + 2l + 2. Which equals 2(k + l + 1). But k + l + 1 is just some integer so we can rewrite 2(k + l + 1) as just 2k. So the sum of two odd numbers must be even. What about the sum of three odd numbers?Well we now know that the sum of two odds must be even so the sum of three odds is the same as asking the sum of an even and an odd.2k + 2l + 1 = 2(k+l) + 1. But k + l is just some integer so we can rewrite that as 2k + 1. So the sum of an even and an odd must be odd. This means that the sum of three odds must be odd.So if the sum of any three odds is odd, the sum of four odds is the same as saying what is the sum of an odd plus an odd. And we already saw that this must be even. So if the sum of four odds must be even, then the sum of five odds is the same as saying what's the sum of an even and an odd. We saw that this must be odd.So it does not matter which 5 odd numbers you select. When we add any 5 odd numbers their sum must be odd so therefore it is impossible to find 5 odd numbers who have a sum of 50 because 50 = 2(25) is even. 
How do I get the 30 as addition of any five odd numbers? If numbers are 1,3,5,7,9,11,13,15 you can repeat the numbers.
There is no mention if you can use sign or not and thus the only possible way to get this is as follows: (15  9) + (13  7) + (7  1) + (9  1) + (13  9). If you solve inside the brackets, you will get the following equation 6 + 6 + 6 + 8 + 4. Adding all these numbers will give you 30if the question would beQ4: ( ) + ( ) + ( ) + ( ) + ( ) = 30This is what you have for the equation. The following are the numbers that you can use to fill in the brackets: 1, 3, 5, 7, 9, 11, 13 and 15. You can repeat the numbers if required. The resulting sum should be 30. 
How do I get to 50 as addition of any five odd numbers? If numbers are 1, 3, 5, 7, 9, 11, 13, 15, 17, 19 you can repeat the numbers.
Whether it’s impossible or trivial depends on what’s meant by “you can repeat the numbers.” If you read this as allowing only 5 of those odd numbers to be added, whether or not any is repeated as part of the five, then it’s not possible since the sum of any odd number of odd numbers will always be odd. However, if you read this as picking 5 of those odd numbers to sum, and repeating any one of them at least once, then one answer is 19+17+9+3+1+1=50. 
How do banks calculate education loans in India? I have a loan of 4 lakh rupees from a government bank in India at 13.75% interest. When I calculated the EMI for 5 years it came out to be around 11,000; but the bank says it is 12,500 for 7 years.
In education loans the repayment does not start from the very next month. Also not all the loan amount is disbursed in one go.Suppose an engineering student has applied and has been sanctioned an education loan. She has already passed first year and the loan is for the remaining three years of study. If the repayment period is 7 years the total loan tenure would be 11 years that is 4 years moratorium period plus 7 years repayment period.The moratorium for installments would be for three and a half years or four years, from the date of first disbursement, as the case may be. If the student gets job soon after the completion of course the moratorium is six months after completion of course. If it takes longer than six months to get the job the moratorium is one year after completion of course. During moratorium it is a holiday for repayment of installments. Borrower student or her family are not supposed to repay installments of principal loan amount during this period.Also they can choose from the two options available for repayment of interest during moratorium. Firstly, they can pay the interest, calculated on simple rate during moratorium or Secondly choose not to pay any amount and pay all the interest portion also after getting job along with principal. In case the second option is chosen the interest would be added to the principal to fix the monthly installment amount for repayment for the total amount (Principal plus interest during moratorium). Usually the education loans, less than 10 lakhs, are repaid in seven years that is 84 monthly installments starting from the next month just AFTER the moratorium period.
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