1! + 2! + 3! + ......... + 100! by 7?
214685
all factorials greater than 6! divide evenly since they contain 7\nso add 1! to 6! and divide by 7, find remainder\n=1!+2!+3!+4!+5!+6!\n=1+2+6+24+120+720\n=873/7\n=126 R 5\n\nanswer is 5
495315
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
520498
Just "pair up" the terms.\n\n1 + (5-3) + (9-7) + (13-11) + ... + (2001-1999)\n\n1 + 2 + 2 + 2 + .... + 2 = 1 + 2 * 500 = 1001
609593
(1) 1.3\n(4) 4\n(5) Right\n(6) 0\n(7) 0\n(8)0\n(9)12
242589
If you add the numbers together then they equal an even number.   1 1  1+1=2
639992
11,000 + 1,100 + 11.\n\n12,111.
284555
n + (n+1) + (n+2) = 102\n3n + 3 = 102\n3n = 99\nn = 33\nn+1 = 34\nn+2 = 35\nsecond integer = 34
36809
You must be very confused. \n\n1+1 = 2 plus one more is 3
209032
OH NO!\nLottery# 1,2,3,4,5,6
121048
I got 4 answers but there is more\n\n1 + 2 + 3 + 4 + 567 +89 = 666\n123 + 456 + 78 + 9 = 666\n9 + 87 + 6 + 543 + 21 = 666\n(975 - 864) x 1 x 2 x 3 = 666
482189
Look at it like this:\n\n100 + 1 = 101\n99 + 2 = 101\n98 + 3 = 101\n97 + 4 = 101\n...\n\nSo, if you add all of the 101's up, you have 100 of them, or 100 * 101.  But you added 100 twice, 99 twice, 98 twice, ... 2 twice and 1 twice.  So, you divide by 2 to get the final answer:\n\n100 * 101 / 2 = 5050
312142
Got it!!\n\n1, 1, 2, 4
68133
you are all crazy. 1+1= elevendy-six.
543147
Do you require mathematical signs, or are standard operations OK (like square root)?\n\n1: (1 + 1 + 1)!\n2: 2 + 2 + 2\n3: 3 * 3 - 3\n4: sqrt(4) + sqrt(4) + sqrt(4)\n5: 5 + 5/5\n6: 6 + 6 - 6\n7: 7 - 7/7\n8: 8 - sqrt(sqrt(8 + 8))\n9: 9/sqrt(9) + sqrt(9)\n\nAnd for a bonus: (0! + 0! + 0!)! = 6
11284
12,345,678,987,654,321\npretty awesome!
624001
27\n\n(+7 -4 +8 -3 +9 -2 +10)
483202
2+7\n3+6\n9+0\n\nThat's it. They all add up to 9... I think that's what the question is asking.