The sum of threee numbers is 98. The sum of the first and second is two more than the third number, and the second is four more than the first. Find the numbers.
213610
This is an algebra problem.\n\nX + Y + Z = 98\nX + Y = Z + 2\nY = X + 4\n\nSolve for X and you should be able to figure out the rest.
532647
Simply multiply\n9 * 8 * 7 * 6\nBecause the first number can be 1-9 but not zero and the second number can be 1-9 but one number has already been used and so on and so forth.\nHope it helps.
313979
These three sequntal clock wise numbers \n2+3+4 --> 9+4 --> 1 (adding the last no.(4)\nBut not works for all
431951
Consecutive integers are n, n + 1, n + 2, n + 3\n\nThe sum of the second and fourth equals the third increased by 17.\n\n(n + 1) + (n + 3) = (n + 2) + 17
413779
let the number be x and x+1\ntwice larger number 2(x+1)\nsum of small and twice large = x + 2(x+1) = 3x+ 2 = 38\nor 3x = 36\nor x =12\n\nnumbers are 12 and 13
426360
a sequence of numbers in which the two previous numbers add to give you the third.\n1,1,2,3,5,8,13...ect\nOne plus one is two.\nOne plus two is three\nTwo plus three is five\nFive plus three is Eight...\n\nIt 's being found increasingly in nature now as well.
303855
There is no simple trick for the addition of two numbers. You will have to calcuate their actual value and then add them. So for you example you would have to go to 4 + 8 = 12.
5694
35\n  16\n+ 47\n____\n  98\n+  2\n____\n 100
505059
"Q: 11) When three times a number, x, is added to a second number, y, the total is 33. The\nfirst number added to three times the second number is 19.\na)Form two equations connecting x and y.\nb)Find the two numbers."\n\neq 1: 3x + y = 33\neq 2: x + 3y = 19\n\nmultiply eq 2 by 3:\n3x + 9y = 57\nthen, subtr from eq 1:\n-8y = -24\ny = 3\nsolve for x using eq 2:\nx + 9 = 19\nx = 10\n\n"\nQ:11) Ahmed is x years old and his father is x2 years old. The sum of their ages is 56\nyears. Form an equation in x and solve it to find the age of each"\n\nx + 2x = 56\n3x = 56\nx = 18-2/3 years old for Ahmed\n2x = 37-1/3 years old for his father\n\nHope this helps!
284555
n + (n+1) + (n+2) = 102\n3n + 3 = 102\n3n = 99\nn = 33\nn+1 = 34\nn+2 = 35\nsecond integer = 34
373214
You have a billion numbers from 000,000,000 to 999,999,999.  These are all the possible combinations of digits 0 to 9 at each position (units, tens, hundreds, thousands, tens of thousands etc... to hundreds of millions).\n\nYou have 100,000,000 numbers starting with a 0.\n100,000,000 starting with a 1.\netc...\n100,000,000 starting with a 9.\n\nSince 0+1+2+3+4+5+6+7+8+9=45, this means the sum of all first digits is 45 x 100,000,000 = 4,500,000,000.\n\nSimilarly you will find that the sum of alll second digits (tens of millions) is also 4.5 billion.  Same for third digits, etc... and for ninth digits(units).\n\nSo the sum of all digits from 0 to 999,999,999 is 9 times 45 billion = 40,500,000,000.\n\nYou need to add 1 for the number 1,000,000,000, so the answer is 40.5 billion + 1 = 40,500,000,001.
288407
The sum is 2^9\n\nThis is the number of the subset of a set having n elements.
268505
1433\n1+4 = 5\n3+3 = 6\n5 is one less than 6\n4 is three more than 1
275332
it's 234.\nlet x be the unit, y the tens and z the hundreds digit.\n\nlet y = (x+z)/2\nz=x/2\nsubstitution: y=(x+x/2)/2\n\nx  +  (x+x/2)/2  +  x/2 = 9\n( (x+x/2) + x + 2x )/ 2 = 9\n(2x + x + 2x + 4x)  / 2 = 18\n     9 x = 36\n       x = 4\n\nhalf of 4 is 2\nhalf of the sum of 4 and 2 is 3.\n\n\nP.S. Ariesera's answer is much simpler and better!\nthat's very true... we must practice mind over hand hehe
384646
First, start listing the prime #'s starting with the number 2.\nAnd let's only go up to 50, because 2 x 50 = 100 and we need 3 #'s.\n\nPRIME #'S LESS THAN 50:\n\n2\n3\n5\n7\n11\n13\n17\n19\n23\n29\n31\n37\n41\n43\n47\n53 ........oops!  STOP, too far.\n\nOkay.  Now we want the product of the 3 #'s needs to be 100.\n\n100 is an even #, so one of the numbers must be even.  \nThat eliminates, quite a lot.\n\nActually, that eliminates all of the #'s but the number 2.\nSo, 2 is the 1st prime number less than 100.\n\nYa, know. . . .\n \nI gave it a try.\n\nAnd I can't seem to find the other two numbers.\n\nYou got me stumped.\n\nPerhaps I read the question wrong.\n\nDid you mean:\nWhat three different prime numbers, multiplied together, have a product of 100?\n\nIf so, you stumped the professor!  :o)
302953
1) Let n = the first integer, then\nn+1 = second integer\nThe product is n(n+1)\n2) Let n = the first even integer, then\nn+2 = second even integer and\nn+4 = third even integer.\nThe sum would be n+(n+2)+(n+4)
108772
22+2 = 24\n8+8+8 = 24\n3^3-3 = 24