a+b+c+d=d+e+f+g=g+h+i=17 and a=4. All alphabets can only hold a value from 1 to 9. find d and g.?
93645
d=5, g=1\n\nFirst I found all the combinations of 3 numbers (no repeats and no fours) that added to 13 (17-4) for b+c+d. Then I found all the combinations of 3 numbers (no repeats and no fours that added to 17 for g+h+i. Then I figured out which of these pairs of numbers had digits that didn't overlap, and had a pair of digits (d & g) that added to 17 minus the digits that, between them, were still missing (e & f).\n\nAnd, I'm a gal. :)
682621
There are 11 letters in "THE ALPHABET"
604743
It is 9 as the solution is 5+17+1+2 = 8+2+11+4 = 2+5+4+14 = 25\nTherefore 10+1+?+5=25\nTherefore ?=9\nTherefore the answers is 9.
968306
well....\n\n1 F F 9\n\n...+ F 7\n\n9+7 is 16...so, 0 in the first place.\ncarry 1.....\n\nF + F + 1 = 15 +15 +1 = F (carry a 1)\n\nF+1 = 0 (carry 1)\n\n1 + 1 = 2\n\nso:    2 0 F 0
1154468
make a common denominator\n\na/b - c/d = a*d/b*d - c*b/d*b = (a*d-c*b)/b*d
172875
Obviously you are doing homework and don't want to figure it out yourself.  It is easy, just write them out.  Obviously you can disregard the B, because there is no choice (both dominant).  Now do all the combinations with a "A" and then double your answer to get the "a".\nACD\nACd\nAcD\nAcd\nThat is four combinations.  Double it for "a" and you get \n\nEIGHT\n\nTyping out the question is harder once you learn the rules.
330850
S A D + M A D \n A D = A D\n     = MS
527706
B is the only one.\n\nA has 9 which can be divided by 3\nC has 15 also divisible by 3\nD has 40, divisible by 5, 8, 10, 4, 20, and 2
690206
Just slide the decimal over for each multiple:\n\ngiga(G)=1,000,000,000\nmega(M)=1,000,000\nkilo(k)=1,000\nhecto(h)=100\ndeca(D)=10\n1=1 (duh!)\ndeci(d)=.1\ncenti(c)=.01\nmilli(m)=.001\nmicro(that "u" looking thing with a tail)=.000001\nnano(n)=.000000001
1147373
alphabets and numbers that's all
829703
I did check the above answerers answers, and they are wrong :P\na,b,c,d,e,f = 1,2,5,5,4,6.\nYou could go and solve this by the usual elimination methods, but for something this large its far better to use a calculator or online solver. Google for 'simultaneous equation solver' and you'll find  heaps.\n(Just to confirm above answers are wrong: for those numbers you get 4a + b + 5d + 2e + 3f = 69, not 57.)
812747
a + b + c = 180\n52 + 34 + c = 180\nc = 94
16389
x+9 < 16\n\nadd -9 for all\n===============> x+9 -9 < 16 -9    (I)\nx+9 -9 =x (II)\n16 -9 = 7 (III)\n\n(I) & (II) & (III)\n===================> x < 7\n\nif x is a natural number => \nx= 1 or 2 or 3 or 4 or 5 or 6
449383
98 33 79\n23 __ 55\n\n9+7  = 16\n8+7  = 17\n16+17= 33\n\n2+5= 7\n3+5= 8\n7+8= 15\n\nso Answer for the first puzzle is 15\n\n********************\n12 26 13\n11 __ 6\n\n12 26 13 12+13=25+1= 26\n11 _  6  11+6= 17+1= 18\n\nso Answer for the second puzzle is 18\n\n********************\n125 21 20\n70 __ 35\n\n125/5 = 25 20/5 = 4 25-4= 21\n70/5 = 14 35/5 = 7  14-7= 7\n\nso Answer for the third puzzle is 7\n\nI hope these are the required answers.\n\nPls post the correct answers.
743138
join AC\nnow youhavetwo triangles ABC and ADC\nin triangle ABC <CAB+<B+<ACB=180*.............(1)<s of a triangle\nin triangle ADC <CAD+<D+<ACD=180*............(2)--------do--------\nadding (1) and (2)\n<ABC+<ADC+<B+<D+ACB+<ACD=360*\n=><DAB+<B+<BCD+<D=360*\nhence proved
786782
The first letter can be any one of the 5 letters\nThe second one can be any of the 4 remaining letters\nand so on\n\nSo the answer is 5 x 4 x 3 x 2 x 1 = 120 different ways\n\nIf two of the letters are the same (eg green) then divide by (2 x 1) to give 60\n\nIf three of the letters are the same (eg eerie) then divide by (3 x 2 x 1) to give 20\n\nIf there are two pairs of letters (eg yuppy) then divide by (2 x 1) x (2 x 1) to give 30
622278
A B C D E * 4 = E D C B A\nA can be 1 or 2, larger numbers would turn the multiplication into a 6 digit number. But the result ends with A, so A has to be even (any digit multiplied by 4 gives an even number) => A = 2.\nE is at least 4 * A = 8, so E = 8 or E = 9. But we also know that 4 * E results in a number ending with 2 (A), so E = 3 or E = 8. Combining these two gives E = 8.\n2 B C D 8 * 4 = 8 D C B 2\nNow we can do the same for the rest.\nSince there was no carry-over to the first digit, 4 * B < 10 => B <= 2\nWe also know that B has to be odd, because 4 * D gives an even number and there is a carry-over from 3 (4 * 8 = 32), so B = 1.\nB is also the last digit of the result of 4 * D + 3, so 4 * D has to end with 8, so D = 2 or D = 7. Since A = 2, D has to be 7.\n2 1 C 7 8 * 4 = 8 7 C 1 2\nWe now know that 4 * C + 3 = 30 + C => 3 * C = 27 => C = 9\n\nA B C D E = 2 1 9 7 8