The sum of two numbers is 528 and their HCF is 33. Find the number of pairs of numbers satisfying condition?
955290
since both numbers have 33 as HCF theirfore the numbers will be \n(33a+33b)=528\na+b=16\n\nthus by putting a from 1 to 15 we can get the 15 pairs \n\na,b\n33,495\n66,462\n99,429\n132,396\n165,363\n198,330\n231,297\n264,264\nthe next 7 can be found out by reversing a,b
670891
54 - -2 = 54 + 2 = 56
598076
Sorry a12b13n5! - I think you are wrong on this one\n\nq +14 =8q + 8x7 (multplying out the brackets)\nq + 14 = 8q + 56 \nq - 8q = 56 - 14 (seperating the 'q's and numbers)\n\n-7q = 42\n\nthen devide by -7 to get\nq = -6\n\nto check\n-6 + 14 = 8 = 8(-6+1) = 8(1) = 8
688567
1 + 2 = 3\n3 + 4 = 7\n7 + 6 = 13\n13+ 8 = 21\n21+ 10= 31\n31+ 12= 43
16545
If we have only one number we have 50C1 = 50\nIf we have only one number we have 50C2 = 1225\nIf we have only one number we have 50C3 = 19600\n.\n.\n.\n.\nIf we have only one number we have 50C49 = 1225\nIf we have only one number we have 50C50 = 50\n\nSo the answer is 50C1 + 50C2 + 50C3 +........50C48 + 50C49 + 50C50
698707
Hynton C got #2)\nThe answer to #1 is 282.\nThe number of solutions to the equation\nx1+x2+x3+x4=10 in non-negative integers is C(13,3)\n[the number of solutions in non-negative integers in the equation\nx1 + x2 + x3 + ... + xn = m is C(m+n-1, n-1) where C(a,b) is the combinitorial function C(a,b) = a!/[(a-b)!b!]  If you need a proof of this, please see this similar question:\nhttp://answers.yahoo.com/question/index;_ylt=At9mkRc2TNqTuXuIxztgN1vsy6IX?qid=20061113231241AAF4cbE ]\n\nTherefore, there are C(13,3)=286 solutions to x1+x2+x3+x4=10 in non-negative integers.  4 of these are (10,0,0,0), (0,10,0,0), (0,0,10,0) and (0,0,0,10).  Therefore the total is 286-4 = 282.\n\nIf you doubt this, here they are generated by computer (first number is the count, second is the number who's digits add to 10):\n(1,19) (2,28) (3,37) (4,46) (5,55) (6,64) (7,73) (8,82) (9,91) (10,109) (11,118) (12,127) (13,136) (14,145) (15,154) (16,163) (17,172) (18,181) (19,190) (20,208) (21,217) (22,226) (23,235) (24,244) (25,253) (26,262) (27,271) (28,280) (29,307) (30,316) (31,325) (32,334) (33,343) (34,352) (35,361) (36,370) (37,406) (38,415) (39,424) (40,433) (41,442) (42,451) (43,460) (44,505) (45,514) (46,523) (47,532) (48,541) (49,550) (50,604) (51,613) (52,622) (53,631) (54,640) (55,703) (56,712) (57,721) (58,730) (59,802) (60,811) (61,820) (62,901) (63,910) (64,1009) (65,1018) (66,1027) (67,1036) (68,1045) (69,1054) (70,1063) (71,1072) (72,1081) (73,1090) (74,1108) (75,1117) (76,1126) (77,1135) (78,1144) (79,1153) (80,1162) (81,1171) (82,1180) (83,1207) (84,1216) (85,1225) (86,1234) (87,1243) (88,1252) (89,1261) (90,1270) (91,1306) (92,1315) (93,1324) (94,1333) (95,1342) (96,1351) (97,1360) (98,1405) (99,1414) (100,1423) (101,1432) (102,1441) (103,1450) (104,1504) (105,1513) (106,1522) (107,1531) (108,1540) (109,1603) (110,1612) (111,1621) (112,1630) (113,1702) (114,1711) (115,1720) (116,1801) (117,1810) (118,1900) (119,2008) (120,2017) (121,2026) (122,2035) (123,2044) (124,2053) (125,2062) (126,2071) (127,2080) (128,2107) (129,2116) (130,2125) (131,2134) (132,2143) (133,2152) (134,2161) (135,2170) (136,2206) (137,2215) (138,2224) (139,2233) (140,2242) (141,2251) (142,2260) (143,2305) (144,2314) (145,2323) (146,2332) (147,2341) (148,2350) (149,2404) (150,2413) (151,2422) (152,2431) (153,2440) (154,2503) (155,2512) (156,2521) (157,2530) (158,2602) (159,2611) (160,2620) (161,2701) (162,2710) (163,2800) (164,3007) (165,3016) (166,3025) (167,3034) (168,3043) (169,3052) (170,3061) (171,3070) (172,3106) (173,3115) (174,3124) (175,3133) (176,3142) (177,3151) (178,3160) (179,3205) (180,3214) (181,3223) (182,3232) (183,3241) (184,3250) (185,3304) (186,3313) (187,3322) (188,3331) (189,3340) (190,3403) (191,3412) (192,3421) (193,3430) (194,3502) (195,3511) (196,3520) (197,3601) (198,3610) (199,3700) (200,4006) (201,4015) (202,4024) (203,4033) (204,4042) (205,4051) (206,4060) (207,4105) (208,4114) (209,4123) (210,4132) (211,4141) (212,4150) (213,4204) (214,4213) (215,4222) (216,4231) (217,4240) (218,4303) (219,4312) (220,4321) (221,4330) (222,4402) (223,4411) (224,4420) (225,4501) (226,4510) (227,4600) (228,5005) (229,5014) (230,5023) (231,5032) (232,5041) (233,5050) (234,5104) (235,5113) (236,5122) (237,5131) (238,5140) (239,5203) (240,5212) (241,5221) (242,5230) (243,5302) (244,5311) (245,5320) (246,5401) (247,5410) (248,5500) (249,6004) (250,6013) (251,6022) (252,6031) (253,6040) (254,6103) (255,6112) (256,6121) (257,6130) (258,6202) (259,6211) (260,6220) (261,6301) (262,6310) (263,6400) (264,7003) (265,7012) (266,7021) (267,7030) (268,7102) (269,7111) (270,7120) (271,7201) (272,7210) (273,7300) (274,8002) (275,8011) (276,8020) (277,8101) (278,8110) (279,8200) (280,9001) (281,9010) (282,9100)
527322
1+1-21+30-5\n2-21+30-5\n-19+30-5\n11-5\n6
95089
1(1)+0(2)+1(4)+1(8)+1(16)+0(32)+0(64)+0(128)+1(256)=\n1+0+4+8+16+0+0+0+256=\n285
1118957
The first number has to be (x+6) and the second number is (x).  There are four possible combinations: 6,0 (60); 7,1 (71); 8,2 (82); 9,3 (93).  60 plus 8 (68) does not equal 06, or 6.  71 plus 8 does not equal 17.  82 plus 8 does not equal 28.  93 plus 8 does not equal 39.\nUnfortunately, this problem is impossible.\nSorry I couldn't be of more help.
523701
If you add the numbers together then they equal an even number.   1 1  1+1=2
5045
"HCF of A&B is 9& of B&C is 12"\n\nB is divisible by 36, because it's divisible by both 9 and 12.\nA is divisible by 9, but not by 2.  (If A were divisible by 2, then the HCF of A&B would be at least 18.)\nAnd C is divisible by 12, but not by 9.\n\n\n"LCM of A&C is 216"\n\n216=27*8\nA=27.\n\nOne solution is:  A=27, B=36, and C=24\nSo LCM of A & B is 108.\nBut B can be any value of 36*N, where N is not divisible by 2 or 3.\n\nSo another solution is: A=27, B=180 (36*5), and C=24.\nLCM of A & B is 540.
963823
LHS = (2n)!\nRHS = 2^n( n!)^2\n\nfor n=1,\n(2n)! = (2)! = 1.2 = 2\n2^n( n!)^2 = 2^1( 1!)^2 = 2\nSo LHS = RHS, condition is not satisfied.\n\nfor n = 2,\n(2n)! = (4)! = 1.2.3.4 = 24\n2^n( n!)^2 = 2^2( 2!)^2 = 16\nSo LHS > RHS\n\nYour problem has errors!!!\n\nI'll try to prove:\n (2n)! >= 2^n( n!)^2 .. .... (ii)\n\nfor n = 1, \n\n(2n)! = (2)! = 1.2 = 2\n2^n( n!)^2 = 2^1( 1!)^2 = 2\nSo LHS = RHS, condition is satisfied.\n\nlet, for any arbitrary integer m, (ii) is satisfied.\n\nSo, \n (2m)! >= 2^m(m!)^2 \n\nnow, if n = m+1, (ii) becomes,\n(2(m+1))! >=2^(m+1) ((m+1)!)^2  \n\nnow,\n(2(m+1))! = (2m+2))! = (2m+2)(2m+1) (2m)! \n= 2(m+1)(2m+1) (2m)! \n\n2^(m+1) ((m+1)!)^2  = 2.2^m  (m+1)^2 (m!)^2 \n\n\nwe have assumed, (2m)! >= 2^m(m!)^2 \nnow, multiplying by 2(m+1)(2m+1) we get,\n2(m+1)(2m+1) (2m)! >= 2^m 2(m+1)(2m+1) (m!)^2 \n\nnow, 2^m 2(m+1)(2m+1) (m!)^2  = 2.2^m (m+1)(2m+1) (m!)^2 \n\nagain 2m+1 > m+1\n\nso, 2(m+1)(2m+1) (2m)! >= 2^m 2(m+1)(m+1) (m!)^2 \n\nor,\n(2(m+1))! >=2^(m+1) ((m+1)!)^2  \n\n(ii) is true for n = m+1\n(Proved)
1050909
First choose which suit, C(4,1) = 4\nThen choose 4 of the 13, C(13,4) = 715\nMultiply everything together: 4*715 = 2860\nTotal number of possiblities = C(52,4) = 270725\nAnswer: 2860/270725 = 44/4165 = .01056422569
793036
19 + 7 = 26\n26 + 5 = 31\n31 + 3 = 34\n34 + 1 = 35\n35 + (-1) = 34
139634
Since the sum of each pair is 6, the function is f(x) = 6 - x
985783
Here's how you do these.\n\nTake one number and figure out all the numbers that evenly divide into your number. Write those down. Do the same for the other.\n\nFor 52... 1, 2, 4, 13, 26, 52\n\nFor 135... 1, 3, 5, 9, 15, 27, 45, 135\n\nLook at what numbers they have in common. In this case, they only have 1 in common. To find the GCF, find the largest number they had in common. In this case, it's one.\n\nHope that helped,\nUMRmathmajor
812747
a + b + c = 180\n52 + 34 + c = 180\nc = 94