How do I know by looking at the number 1,862 that it is evenly divisible by 7?
142718
9 x 7 = 62+1 =63 (NOT 62)\n\nDividing by 7 (2 Tests)\n\n    * Take the last digit in a number.\n    * Double and subtract the last digit in your number from the rest of the digits.\n    * Repeat the process for larger numbers.\n    * Example: 1,862 (Double the 2 to get 4. Subtract 4 from 186 to get 182, again substract 4 from 18 to get 14, which is divisible by 7.
883360
if you mean: how I know that number N is dividable by number k with no residue?\n1)if the unit digit of N is even, then N is divisible by 2;\n2)if sum of digits of N is divisible by 3, then N is divisible by 3;\n3)if sum of digits of N is divisible by 9, then N is divisible by 9;\n4)if the unit digit of N is 0 or 5, then N is divisible by 5;\n5)if sum of odd digits is equal to sum of even digits, then N is divisible by 11;\n6)if the last two digits represent a number divisible by 4, then the whole N is divisible by 4;
860921
To show you how to do it yourself..\n\nfor 8/7: 7 "goes into" 8 once so separate it as 7/7 + 1/7, 7/7 becomes "1" and "1/7" remains as a fraction.\n\nTherefore "8/7" becomes "7/7 + 1/7" which becomes 1 1/7.\n\nSimilarily, 38/6 becomes 36/6 + 2/6 which becomes 6 2/6.\n\nYou get to do the rest.... happy learning!
728995
7 is a prime number.  It is divisible only by itself and 1.
578566
To check if a number is divisible by 36, use the divisibility rules for both 4 and 9.  If a number is divisible by both 4 and 9, it must be divisible by 36.\n\nIf you need to know the rules for 4 and 9 (or other numbers up to 13, as well as methods for larger numbers), visit the source site.
846607
I can give you a few "tests" for divisibility...\n\nIf the number in the ones position is divisible by 2, then the entire number is divisible by 2.\n\nIf all of the digits in the number add up to something that is divisible by 3, then the number is divisible by 3.\n\nIf the number ends in a zero or 5, it is divisible by 5.\n\nIf the number ends in a zero, it is divisble by 10.\n\nHere's a web page all about divisibility rules...\n\nhttp://www.mathgoodies.com/lessons/vol3/divisibility.html
407989
Just cross of like terms in the numerator and the denominator and see what is left.  like in the first problem the numerator and the denominator share two two's and a 3, so by crossing them off you are left with 4/7.  You can do that since 2/2 is one, and likewise with other similar fractions, and when you multiply by one you always get the same fraction.
169090
divide all the numbers(except 1) less than the prime number and see if they come out as a full number\netc. 19 19/2=9R1  **/3=6R1   **/4=4R3  **/5=3R4  **/6=3R1  **/7=2R5  **/8=2R3  **/9=2R1 **/10=1R9  **/11-18=1R8-1
146728
no, it is divisible by 1, 3, 23, 29, and itself (with the numbers being reduced as far as possible)
647670
Arbitrary is the only one with the correct answer.\n1+28-67x98+1001/7\nFirst multiply 67x98 = 6566 and divide 1001/7 = 143\n= 1+28-6566+143\n= 29-6566+143\n=-6537+143\n= -6394
279409
Find every number between 101 and 999 that is divisible by 7 and then add them together.  Hmmm......like I really am going to do that.
790673
count the number of decimal places\nin your example, its 7\nyour example becomes the numerator (remove the decimal point)\nand 1+ seven 0s be the denominator\nso its\n1285714/10000000
985702
86 divided by 24.\ntake it from there?
30019
28 - 7*2*2\n\n155 - 5*31
794955
It's simple arithmetic.  Let's break it up into different parts:\n\n7^2=49\nsq(16)=4, since 4^2=16\n\n49*4=196\n196/11 is your answer since you cannot simplify this any more.  Eleven is prime, and 196 is not divisible by 11.\n\n\nHope this helps.
678031
psbhowmick, 99 is divisible by 9, but 199 is not. You can't find the remainder simply by looking at the last 2 digits. For instance, the remainder of 22654785/19 is 2, not 9 as you suggest.\n\nDividing by 19, hmmm... the only shortcut I can think of is to use the mod function on your calcutor. Sorry.
806367
The best way to handle creating a program of that sort is to use the Mod function to find whether a number comes out evenly i.e. 7 Mod 3 returns 1 because it gives the remainder after 7 is divided by 3. Use a For...Next statement that take the number (intNum) that the user inputs and "Mod's" it by every number below it. This will give you some of the factors so you can find out the perfect, abundant, and deficient of numbers.