what are the multiples of 5?
959413
numbers ending in -0 or -5
533127
2*2*2*5*5 or 2to the third power times 5 to the second power
523621
(10+15+23+37+45) / 5 = 26
16141
There are 6 combinations\n3,5,7\n3,7,5\n5,3,7\n5,7,3\n7,5,3\n7,3,5
390304
THE POSSIBILITIES:\n\n1.1:1→1*1=1\n2.1:2→1*2=2\n3.1:3→1*3=3\n4.1:4→1*4=4\n5.1:5→1*5=5 √\n6.1:6→1*6=6\n\n7.2:1→2*1=2\n8.2:2→2*2=2\n9.2:3→2*3=5\n10.2:4→2*4=8\n11.2:5→2*5=10 √\n12.2:6→2*6=12\n\n13.3:1→3*1=3\n14.3:2→3*2=6\n15.3:3→3*3=9\n16.3:4→3*4=12\n17.3:5→3*5=15 √\n18.3:6→3*6=18\n\n19.4:1→4*1=4\n20.4:2→4*2=8\n21.4:3→4*3=12\n22.4:4→4*4=16\n23.4:5→4*5=20 √\n24.4:6→4*6=24\n\n25.5:1→5*1=5    √\n26.5:2→5*2=10   √ \n27.5:3→5*3=15   √\n28.5:4→5*4=20   √\n29.5:5→5*5=25   √\n30.5:6→5*6=30   √\n\n31.6:1→6*1=6\n32.6:2→6*2=12\n33.6:3→6*3=18\n34.6:4→6*4=24\n35.6:5→6*5=30   √\n36.6:6→6*6=36\n\nProbability = No. of required possibilities/No. of total possible events =11/36\nSo the right answer is D
30589
2(5)-5(5)+2(5)=\n10-25+10\n20-25\n-5
619440
7 will be after five\nand eight will be after six.\n\nReason: The upper numbers in 'numerals' contains straight lines, the lower one contains curves.\nWell...
1106159
5 = 1\nYou told it already!
207303
We could change the second + to 4 by putting a vertical small line\n\n5 + 5 + 5 = 550\n5 + 5 4 5 = 550
911275
Quin 5\nSex 6\nSept 7\nOct 8\nNon 9\nDec 10
975944
5 would be (a)\n6 would be (b)\n7 would be (c)\n8 would be (b)
893743
8\n9\n7\n4\n4\n5\n7\n7\n9\n8\n5\n4\n8\n8
893652
with 1q, 2 ways (0,1 n)\nwith 3d, 1 ways (0 n)\nwith 2d, 3 ways (0,1,2 n)\nwith 1d, 5 ways (0,1,2,3,4 n)\nwith 0d, 7 ways (0,1,2,3,4,5,6 n)\n\nTotal of 18 ways
743455
one fourth (or one quarter), three sevenths, one tenth, five nineteenths, one one hundredth, four one thousandths (or one two-hundred and fiftieth if you simplify) ;)
789066
Firstly, it has to start with either a 1 or a 2, and end with a 2 or a 6. So we just take each case one at a time.\n1..2: that leaves 23556, so there are 5!/2! ways, ie 60 ways.\n1..6: that leaves 22355, so there are 5!/(2!2!) ways, ie 30 ways.\n2..2: that leaves 13556, so there are 5!/2! ways, ie 60 ways.\n2..6: that leaves 12355, so there are 5!/2! ways, ie 60 ways.\n\nNote that in each case there are 5! ways of arranging the other 5 numbers, but we need to divide by 2 if there is a pair of repeated numbers.\nThat gives a total of 210.\n\n(note that both of the other answers given so far are wrong - its not 360, because that person has forgotten they have to be even; and its not 120. They counted 2x2 possibilities for the first and last digits which is correct, but one of those cases already uses both 2s, so you *don't* want to divide by 2 in the end for that particular case (you're not counting them twice are you!)\nTheres really no other way of doing it than splitting them into cases like I did.
810912
105 or any number that equals 105 time any multiple of 3,5 or 7
907310
5\n\n1s,2s,2px,2py,2pz