find the sum of integers from 1 to 100 which are not divisible by 3 and 5?
361589
Sum of all numbers 1 to 100:\nS = n[2a1 + (n - 1)d]/2\n= 100[2(1) + (100 - 1)]/2\n= 100(101)/2\n= 5050\n\nSum of all numbers 3 to 99, multiples of 3:\nS = n[2a1 + (n - 1)d]/2\n= 33[2(3) + (33 - 1)3]/2\n= 33(6 + 96)/2\n= 1683\n\nSum of all numbers 5 to 100, multiples of 5:\nS = n[2a1 + (n - 1)d]/2\n= 20[2(5) + (20 - 1)5]/2\n= 20(10 + 95)/2\n= 1050\n\nSum of all numbers 15 to 90, multiples of 15:\nS = n[2a1 + (n - 1)d]/2\n= 6[2(15) + (6 - 1)15]/2\n= 6(30 + 75)/2\n= 315\n\nSum of integers from 1 to 100 which are not divisible by 3 and 5:\nS = sum(1-100) - sum(3-99) - sum(5-100) + sum(15-90)\n= 5050 - 1683 - 1050 + 315\n= 2632\n\nWe have to add in the sum(15-90) because in removing sum(3-99) and sum(5-100), we're taking out multiples of 15 twice.  Adding sum(15-90) ensures that we only take out 15 once, 30 once, etc.
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.
648976
yes it's correct\nu can use the arithmetic series formula \nor do this\n\n1 + 100 = 101\n2 + 99 = 101\n......\n50 + 51  = 101\n\nSince there are 50 pairs of these numbers and their sum is always 101\nIt's 50 x 101 = 5050
750336
(100+a)(100+b) = 10,000 + 100(a+b) + ab\n 103*107\n=(100+3)*(100+7)\n=(100*100)+(100*7)+(3*100)+(3*7)\n=(10000)+(700)+(300)+(21)\n=11021
1014029
1,2,3,5,16,63, ...\n3=(1+2)*1\n5=(2+3)*1\n16=(3+5)*2\n63=(5+16)*3
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.
758828
formula=(100+96+92+.........)-(98+94+90+.........)\nsum of arithmetic progression=Sn=(1/2)n[2a+(n-1)d]\n(1/2)*25{[200+24*2]-[196+24*2]}\n(1/2)*25*(248-244)\n(1/2)*25*4\n=50
152234
1, 3, 5, 7, 11, 17....\n\n\nNumbers that can only be divided by that number or 1 with the result of a whole number.
913443
Test for divisible by 2 - last digit is 0, 2, 4, 6, 8\nTest for divisible by 4 - last *two* digits are divisible by 4\nTest for divisible by 5 - last digit is 0 or 5\nTest for divisible by 6 - divisible by 2 and sum of digits is a multiple of 3).\nTest for divisible by 9 - sum of digits is a multiple of 9\nTest for divisible by 10 - last digit is 0.\n\nIf you check 1800, you'll see that it passes ALL tests.
855714
Take the number, in this case 382, and divide it with the other one, 914.  Then multiply by 100 and that's it!
1139003
(n/(n - 3)) + (5/(2n - 6))\n(n/(n - 3)) + (5/(2(n - 3)))\nMultiply everything by 2(n - 3)\n(2n + 5)/(2n - 6)
906083
Sum = 1 + 2 + 3 + .............. + 98 + 99 + 100\nSum = 100 + 99 + 98 + .............. + 3 + 2 + 1\n--------------------------- -----------------------------\n2 x Sum = 101 + 101 + 101 + .... + 101 + 101 + 101\nSo Sum = ½ (100 * 101)\n= 5050
495849
Minimum number of 3s is the most elegant solution\n\n(33 x 3) + 3/3 = 100
403326
1+2+3+4+5+6+7+8+9/3+8*2-4\n=36+3+16-4\n=51
598513
n + (n+1) + (n+2) = 102\n3n + 3 = 102\n3n = 99\nn = 33\nn+1 = 34\nn+2 = 35\nsecond integer = 34
1015086
Simple!\n33 * 3 + (3 / 3) = 99 + 1 = 100.
810912
105 or any number that equals 105 time any multiple of 3,5 or 7