help me to solve this : how we will have 5+5+5=550 and 71=1 ?
90985
You can have 4 cases:\n1) you wrote it wrong in your notebook;\n2) you wrote it wrong here;\n3) your teacher is nuts;\n4) you are kidding.\n\nWich one is?
823152
-3/5 + 8/35\nYou need a common denominator.\nSince 5 goes into 35 evenly, the common denominator will be 35.\n5*7 = 35, so wee need to multiply the top and bottom of the first fraction by 7\n-3/5 (7/7) = (-3*7)/(5*7) = -21/35\nNow we can add the numerators.\n-3/5 + 8/35\n= -21/35 + 8/35\n= (-21+8)/35\n= -13/35
186591
solve it by induction...\nlet n=1, then 5^2n-1 +1 =6 so divisible by 6\nlet n=k, then 5^2n-1 +1=5^2k-1 +1 and let it be divisible by 6\nlet n=k+1 then 5^2n-1 +1=5^(2(k+1)-1)) +1\nie 5^(2k+2-2) +1=(5^2k-1)(5^1)\nbut we know that 5^2k-1 is a multiple of 6,so (5^2n-1)*5 is also a multiple od 6. since it satisfies this for n=0, it should satisfy for all n
1159579
cos55=sin(90-55)=sin35\ncos^2 55° + cos^2 35° =\nsin^2 35° + cos^2 35° =  1,\nbecause for any number of degrees;\nsin^2 + cos^2 = 1
464956
use each number once and only once...\n\n2*3+4*5=26   \n\nThis is too easy.  Give us a hard one!\n...come on...?
272561
1 b = 5\n2 c = 7\n3 c = -8\n4 b = 9
253073
Each answer is 4 more than the sum of the previous two terms.  The 49 is a misprint and should be 59.\n\n(5+11) + 4 = 20\n(11+20) + 4 = 35\n(20+35) + 4 = 59\n\nso,\n\n(35+59) + 4 = 98\n(59+98) + 4 = 161
793036
19 + 7 = 26\n26 + 5 = 31\n31 + 3 = 34\n34 + 1 = 35\n35 + (-1) = 34
306637
There's a trick to figuring this out quickly.\n\nFor each number between 1 and 499 there is a corresponding number between 501 and 999 which, when added together, equals 1000.\n\nFor example:\n\n1 + 999 = 1000\n2 + 998 = 1000\n...\n499 + 501 = 1000\n\nNow, there are 499 pairs like that.  Add them together and you get 499,000\n\nAdd to that 1000 and 500, and you get 500,500, which is your total.
361299
2, 9, 16, 23, 30\n\nU(n+1) = Un + 7\n\nMoving through weeks on a calendar.\n\n\n\n50, 100, 200, 400\n\nU(n+1) = 2Un\n\nDoubling ingredients in a recipe.
537227
you do what is in the parenthesis first.   \n5(5+2)\ntake the 5+2 first which equals 7.  Then multiply that by 5 which is 35.\nNext take each side of the plus sign 5(5) and 5(2) and do them seperatly.  5(5)=25 and 5(2)=10  then put the plus in and 25+10= 35   Your done  35=35
157925
The problem asks for:\na natural number n wich: n = k*3+1, n=q*5+1, n=r*7+1, n=p*9+1, n=m*11. => n = cM(3,5,7,9)+1 and 11 divides n. (cM(3,5,7,9) = a common multiple of 3,5,7,9). \nYou take the less cM(3,5,7,9) of all, wich is 945, add 1, and you find out that 11 divides it.\nAnswer is:\n\n-- 946 --
859504
(x-7)(x-5)\nso x= +7, or +5
400630
Answer : 64\n\nExplanation : \n2+3=5\n5+5=10\n10+7=17\n17+9=26\n26+11=37\n37+13=50\n50+15=65\n\nSo the next number in the line should be 65, not 64.
166778
29786\n+  850\n+  850\n_____\n31486\n\nThe way to solve this is to first note that the units digit of Y + N + N is Y. Therefore N must be 5 or 0. If N is 5 then you'd need to carry the one so that now the units digit of 1 + T + E + E is T. But this is impossible because E + E is an even number. Therefore N is 0. Using similar logic, we find E must be 5 and we do carry a one to the next column. Let's look at what we have now:\n\n___1\nFORTY\n+__T50\n+__T50\n_____\nSIXTY\n\nNow, recognize that T50 and RTY are less than 1000 and FO thousand plus two thousand (and change) can only equal SI thousand if O is 9 and I is 1. This gives:\n\n___1\nF9RTY\n+_T50\n+_T50\n_____\nS1XTY\n\nF and S must be consecutive and R and X are not both odd or both even (since we carried the one). If R is 8,=> T is 7,=> X is 3, but there remain no consecutive numbers available for F and S, so R is not 8, yet 1 + R + T + T must be at least 22, so T is therefore 8. R = 6 also fails to allow consecutive values for F and S, so R = 7.\n\n___1\nF978Y\n+_850\n+_850\n_____\nS1X8Y\n\nFrom here it should be easy to see X is 4 and F and S are 2 and 3 respectively leaving Y to be 6 and solving the puzzle.
1089449
It's as simple as plugging in numbers.  The number 15 becomes the letter "a", the number 7 is "b", and the number 55 is "c".  Just substitute the number for the letter in the formula and solve.
251360
welcoming\n23 + 5 + 12 + 3 + 15 + 13 + 9 + 14 + 7 = 101