Hey guys! I'm back from a long hiatus from Brilliant. Here's a problem for you all:
How many four-digit integers are there for which the thousands digit equals the sum of the other three digits?
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
Just brute force computer work: Length [ Table [ With [ { l = IntegerDigits [ n ] } , If [ l [ [ 1 ] ] = Plus @@ Drop [ l , 1 ] , n , Nothing ] ] , { n , 1 0 0 0 , 9 9 9 9 } ] ] ⇒ 2 1 9
{1001,1010,1100,2002,2011,2020,2101,2110,2200,3003,3012,3021,3030,3102,3111,3120,3201,3210,3300,4004,4013,4022,4031,4040,4103, 4112,4121,4130,4202,4211,4220,4301,4310,4400,5005,5014,5023,5032,5041,5050,5104,5113,5122,5131,5140,5203,5212,5221,5230,5302, 5311,5320,5401,5410,5500,6006,6015,6024,6033,6042,6051,6060,6105,6114,6123,6132,6141,6150,6204,6213,6222,6231,6240,6303,6312, 6321,6330,6402,6411,6420,6501,6510,6600,7007,7016,7025,7034,7043,7052,7061,7070,7106,7115,7124,7133,7142,7151,7160,7205,7214, 7223,7232,7241,7250,7304,7313,7322,7331,7340,7403,7412,7421,7430,7502,7511,7520,7601,7610,7700,8008,8017,8026,8035,8044,8053, 8062,8071,8080,8107,8116,8125,8134,8143,8152,8161,8170,8206,8215,8224,8233,8242,8251,8260,8305,8314,8323,8332,8341,8350,8404, 8413,8422,8431,8440,8503,8512,8521,8530,8602,8611,8620,8701,8710,8800,9009,9018,9027,9036,9045,9054,9063,9072,9081,9090,9108, 9117,9126,9135,9144,9153,9162,9171,9180,9207,9216,9225,9234,9243,9252,9261,9270,9306,9315,9324,9333,9342,9351,9360,9405,9414, \9423,9432,9441,9450,9504,9513,9522,9531,9540,9603,9612,9621,9630,9702,9711,9720,9801,9810,9900}