explain how to add and subtract integers?
455504
I will explain a method that is the most easy on this planet, and a unique one.\n\nI will illustrate with the help of a simple example.\n\nEg. 4-2.\n\nDraw no. line as follows:\n\n<-6--5--4--3--2--1--0--1--2--3--4--5--6--7-->\n\nNumbers left to zero are negative integers\nand to the right are the positive ones.\n\nLocate 4 on this line. Now move 2 steps behind 4 as the operation to be carried out is subtraction.\n\nThis will give answer as 2. \n\nSimilarly for addition move the steps ahead.
615376
there are several methods,\n\nalmost all mathematical calculations like add, subtraction...... have many ways to calculate......\n\ncheck the website below to find alternate ways to calculate
917224
Simply use correct order of operations\n\nparentheses\nexponents\nmult/division (from left to right)\naddition/subtraction (from left to right)
596283
add them up like this \n4+2+(-3)+6+5+(-3)+(-2)+0+(-1)+(-4)+3\ndivide the result by 11 because that is how many numbers there are in your list.  That will be your answer!
878923
Make a program on your calculator.\nView the snippet in the link for code.
1020750
Assuming by "plus" and "minus", you are referring to "positive" and "negative"... they are fundamental rules for mathematics.  To understand them, you need to understand the definition of a "negative number", where subtraction of a positive number is the equivalent of addition of the number's negative form:\n\n5 - 3 = 5 + (-3) = 2\n\nYou can think of negative values as the "explicit absence of" a value or number of items.  With this definition, it should be clear why the multiplicative rules are as you stated.\n\n(positive) * (positive) = (positive) ...is just the obvious, since any positive number of items added multiple times will always result in a positive number of items.  Add 3 apples to a basket 4 times, and you will have 3*4=12 apples.  Mathematically:\n\n3 + 3 + 3 + 3 = 12\n\n(positive) * (negative) = (negative) ...you can use the "explicit absence of" idea I stated above to visualize this one.  If you have 2 holes into which a ball would fit and consider that number to be -2 (negative 2), and you had three rows of them, you will have 3 * (-2) = -6 holes, into which 6 balls would fit.  Mathematically:\n\n(-2) + (-2) + (-2) = -6\n\n(negative) * (negative) = (positive)  ...is derived directly from the 2nd rule above.  If a single negative number changes the sign of the product, then two of them should change the sign right back.  You can also use the "explicit absence of" idea.  If you were counting pairs of holes as negative values (-2), but you were to fill up 4 of these pairs with balls (the filling of which would be considered as "negating the hole", or creating a filled hole), then you would have (-2) * (-4) = 8 holes filled with balls.  Mathematically:\n\n-[ (-2) + (-2) + (-2) + (-2) ] = -(-8) = 8\n\nI hope this makes some sense, and that I didn't just create even more confusion.  ^.^
726208
I think the easiest way to know whether or not you add is by picturing a number line that is negative on the left and positive on the right with zero in the middle.\nif you subtract, then you move left\nif you add, then you move right.\n\n\n\nMemorize this!\n+ + makes +\n+ - makes -\n- + makes -\n- - makes +\n\nexample:\n1+(+2)=1+2\n1+(-2)=1-2\n1-(+2)=1-2\n1-(-2)=1+2\n\nanother example:\n(-1) + (+2) = (-1) + 2 = 2 - 1\n(-1) + (-2) = (-1) - 2 = -(1+2)\n(-1) - (+2) = (-1) - 2 = -(1+2)\n(-1) - (-2) = (-1) + 2 = 2 - 1\n\ntry these, and we'll tell you if you got it\n4 - (-5) =____\n6 - 7 = _____\n-9 - 8 = ______\n-3 - (-5) = _______
751233
I'm not sure why you are avoiding the loop, but you could use recursion instead. If I understand your question correctly this function in c++ should give you your desired results.\n\nint sum(int nbr){\n    if(nbr > 0){\n           return(nbr%10 + sum(nbr/10));       \n    }\n    else{\n         return(0);\n    }\n}\n\nIf you called it with a number such as 256, it would return an answer of 13, or the sum of each digit in the integer.
612486
This is pretty straighforward in C - I won't write the assignment for you, but the general proccess is as follows:  \n\n1.  Count the number of each coin - store that number in some variable (say, int num_dimes;  )  If you're reading from a file, it's a little more complicated, but if your user is entering something from a command line (as is the case with beginning programming assignments, usualy) it's just a matter of having them enter a number and using a function like get() to read the input.  \n\n2.  Multiply the number of each coins by the appropriate denomination.  For example - if you counted 6 dimes in the previous step, multiply 6 by .10.  Make sure that you store the result in a floating point variable, and it's generaly a good idea to ensure that you're multiplying two floats together in order to get a float answer.  Use something like this:\n  \nfloat dimes_total; \ndimes_total = (float)num_dimes * 0.10;   \n\nthe (float) in front of num_dimes casts that variable as a floating point number.  \n\n3.  After you've done this for each coin, simply add up the totals: \n\nfloat total; \ntotal = pennies_total + nickles_total  ... etc, etc.\n\n4.  print your answer to the screen - since dollars and cents use the same number base as arithmetic done by the compiler, nothing fancy is needed here.    In cases where you're dealing with different number bases, however, it gets to be a little more tricky.
947463
When you double a number, you just multiply it by 2.\ny = 2x\nx is the original number and y is the number after it's been doubled.
770189
Number will have to be the actual number it is working it out from..and it might not work as a long and definately not an integer.\n\nNumber as String\nOdd as Long (or Integer)\nEven as Long ( or Integer)\n\nIf Not (Number <1 or Number > 10) Then\nNumber = Number / 2\nIf InStr(Number, ".5") then\nOdd = Odd + 1\nElse\nEven = Even + 1\nEnd If\nEnd If\nMe.Print "Odd: " & Odd & " - " & "Even: " & Even
376319
Are you able to use any other numbers and operators?  If so this would be fairly easy.  Using only Three 8's and a subtraction symbol it is not possibe\n\nClosest example (8+8)/8-1=1
912127
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.
774139
For this, the simplest method is to add the input and output up:\n16 + 82 = 98\n34 + 64 = 98\n47 + 51 = 98\n51 + 47 = 98\n63 + 35 = 98\n\nSince they all add up to the same amount, the function is a straight addition/subtraction function. Thus, the function is: r = 98 - a.
1136386
subtracting a negative is the same as adding a positive\n\nmultiplying/dividing like signs (+,+ ; -,-) will give a positive product/quotient\n\nmultiplying/dividing opposite signs will give a negative answers
49715
use bitwise operators. something like:\n\nnum_bits = 0;\nfor(i = 0; i < 32; i++) {\nif number & (1<<i) num_bits++;\n}\n\nnow num_bits holds the number of bits that were set in number. The "<<" operator does bit shifting. So "1<<i" equals a number that's all 0's except for a 1 at position i. If you bitwise and that with something, that tells you if that bit is set in the number. So we do this for each of the bits and add up the results.\n\nThis is assuming 32-bit integers. Adjust accordingly for 64-bit integers.
697117
Subtract the cells and format it as numeric