Write 2 addition problems with the same number of addends, one with even numbers and one with odd numbers. Explain what is common to both answers.
121183
addends are the numbers you add to get the sum or answer in an addition problem\nEXAMPLE: 5 + 3 = 8\n5 and 3 are the addends while 8 is the sum or answer
132220
Let’s call a positive integer ambiguous if there are two different partitions of equal\nsize of it into summands whose products are the same. For example, 14 is ambiguous because\n14 = 3+3+8 = 2+6+6 and 3 · 3 · 8 = 2 · 6 · 6 = 72. A number is called doubly ambiguous if\nthere are two pairs of partitions each of which has the same product, and these two products\nare different. For example, 13 is doubly ambiguous since 13 = 1 + 6 + 6 = 2 + 2 + 9 and\n1 · 6 · 6 = 2 · 2 · 9 = 36 and at the same time 13 = 1 + 1 + 3 + 4 + 4 = 1 + 2 + 2 + 2 + 6 and\n1 · 1 · 3 · 4 · 4 = 1 · 2 · 2 · 2 · 6 = 48. Now the first mathematicians response to the second mathematicianss\ncomment is equivalent to ‘the bus number is ambiguous’. Note that if n is ambiguous, then so\nis n + 1 and if n is doubly ambiguous, then n + 1 is also doubly ambiguous. An examination\nof all partitions of 11 shows that 11 is not ambiguous. Thus every integer n at least 12 is\nambiguous, and every integer bigger than or equal to 13 is doubly ambiguous. The second\nmathematician’s comment that he knew the age of the first mathematician implies that the bus number is not\ndoubly ambiguous. Thus the bus number must be 12.
29145
I don't think you have stated the puzzle correctly.\n\nIf each letter is supposed to represent a single digit, (which is the usual way these kinds of problems are stated) and the correct statement of the arithmetic is\n XYZXYZ+ABCD-ABC=EFGHA\nthen there is no solution.\n\nXYZXYZ is a 6 digit number, since ABCD has to be larger than ABC, ABCD-ABC is a positive number.\nWhen you add that to XYZXYZ, the result will have at least 6 digits, but EFGHA only has 5 so the problem is inconsistant.\n\nThere is another problem: The puzzle as you have stated it, uses 11 symbols to represent the 10 digits. If you allow more than one letter to represent a single digit then you can say let all the letter represetn the same value. Then you can get: A=B=C=D=E=F=G=Z=Y=Z=0 and the problem is trivial.
112850
Working backwards from your conclusion, but following your logic, you have not joined 2 ones with 2 ones, but rather joined 2 tens with 2 ones.  However you have used the same symbol to represent both the 2 tens and the 2 ones, which is the flaw in your logic.\n\nIn mathematics means adding, not merely joining together, and the column placement of the inputs and the results cannot be disregarded.
104047
Each answer is 4 more than the sum of the previous two terms.  The 49 is a misprint and should be 59.\n\n(5+11) + 4 = 20\n(11+20) + 4 = 35\n(20+35) + 4 = 59\n\nso,\n\n(35+59) + 4 = 98\n(59+98) + 4 = 161
494270
If the difference is the same as one of the decimals in the problem, then a - b = b. So a = 2b. Your task is to find a way to write two decimals in a way so that one decimal is double the other.\n\nWith your teacher's hint, you only have to work with 0, 1, 3, and 5. Some trial and error later, you get 30 is twice 15. So the solution is\n\n4.30-2.15=2.15
612704
here is one\n024331\nassuming that the 2&3 numbers are the only evens\nassuming only one one\nassuming  2&3 numbers are not same numbers\nassuming only one set of repeating numbers\nalso 024551, 024771, 024991, 026331, 026551, 026771, 026991, 028331, 028551, 028771, 028991, 042331, 042551, 042771, 042991, 046331, 046551, 046771, 046991, and so on.
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It's when you make mistake in addition. But I won't say the sum can't be put correctly but It can be made correct by changing the sign You could change the sign from + to - . Now can I have my ten points ? Please ???!
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Try this:\n\nvoid addIt(int* begin, int* end, int x)\n{\nint* p;\nfor (p=begin; p!=end; ++p) {\n*p = *p + x;\n}
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It's called a coincidence, nothing else.  \n\n\nCheck this out..\n\nY is the 25 letter of the alphabet, A id the 1st, H is the 8th, and O is the 15th (two O's of course)\n\n25+1+8+15+15=64\n\n6 x 4 = 24\n\n2+4=6\n\n6 repeated three times (three being the number of math problems there were) is 666! OMG Yahoo is evil! \n\nNumbers are sooo mutable you can make them equal almost anything you wanted
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There is no simple trick for the addition of two numbers. You will have to calcuate their actual value and then add them. So for you example you would have to go to 4 + 8 = 12.
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Hi again Ken,\n\nDid you see my other answer to this similar question?  Did it help any?\n\nIf n is the first of 4 even integers, then the others are:\n\nn+2\nn+4\nn+6\n\nAnd their sum is:\n\nn + n+2 + n+4 + n+6 = 4n + 12\n\nNow, if their sum is 340, then you have the following:\n\n4n + 12 = 340\n4n = 340-12 = 328\nn = 328 ÷ 4 = 82\n\nSo your 4 even integers are:\n\nn = 82\nn+2 = 84\nn+4 = 86\nn+6 = 88\n\nPS - I think the reason you are not getting the answer you want is that the question is worded incorrectly.  Your question states "n is the first 4 even integers" when what I think you mean to say is that "n is the first OF 4 even integers"\n\n~ ♥ ~
22518
By definition, even numbers are all numbers that can be written as 2 times n where n is a whole number.\n\nOdd numbers are those that can be written as 2 times n + 1.\n\n0 = 2 times 0, so it is even.
516684
So let's say the problem looks like this\n\n   11\nx 12\n--------\n        <-----the answer from 11*2\n       <------answer from 11*1\n\nfirst hide the 1 from the 12 and work out the problem as 11*2\nthen work on multiplying the 1 with 11 but make sure you start putting your answer down below your first answer and the answer starts in the column that is below the 1 from the 12.\n\nfinally when the two answers are found, you can add them and get your final answer. SOOO here's what it looks like\n\n      11\n   x  12\n--------------\n           22    (this is from 11x2)\n +  11       (this is from 11 x1)\n------------\n     132 = final answer!
499852
Add the digits of the base 5 number. If the sum is even, the number is even. If the sum is odd, the number is odd.\n\nEven simpler:\nCount the number of odd digits in the number. (Ignore the even digits.) If the number of odd digits is odd, the number is odd, and if it's even the number is even.
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The steps for adding two fractions:\n\n1) reduce both fractions  (not strictly necessary, as long as you reduce in the end)\n2) give them both the same demoninator (the bottom number)\n3) add the top numbers (the numerators)\n4) reduce the answer\n\nSo, step one is already done.\n\nStep two is to give them a common (the same) denominator.  So what number should that be?\n\nIt's going to be the product of both denominators.  so 3 * 8 = 24.  So that's our common denominator.\n\nWhat you are going to do is to multiply the top and bottom half of each fraction by the same number (which is okay, it's equivilant to multiplying by 1) so that the bottom number is 24.  So for the left fraction, we will multiply the top and bottom by 3.  So we get (3 *3)/(8 * 3), which is 9/24.\n\nAnd now we multiply the second fraction, top and bottom, by 8.  so (2 * 8)/(3 * 8) = 16/24.  Now that both fractions have the same denominator, we add them by adding the top numbers, and putting that over the common denominator.  So 9 + 16 = 25, so we get 25/24.\n\nThis is technically acceptable, but teachers will probably want you write this as 1 and 1/24th.
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4) n is even.\nStarting with an even number (0) and ending with an even number, means you have an odd number off roots.