Determina la suma de todos los números de tres cifras de la forma ba( 2a ) con b > a > 0, de manera que sean múltiplos de 4 y 11 POR FAVOR AYUDAAAAA
Respuestas a la pregunta
Respuesta:
Explicación paso a paso:
Multiplos de 4 de 3 Cifras: 100,104,108,112,116,120,124,128,132,136,140,144,148, 152,156,160,164,168,172,176,180,184,188,192,196,200,204,208, 212,216,220,224,228,232,236,240,244,248,252,256,260,264,268,272,276, 280,284,288,292,296,300,304,308,312,316,320,324,328,332,336,340,344, 348, 352,356,360,364,368,372,376,380,384,388,392,396,400,404,408,412,416, 420,424,428,432,436,440,444,448,452,456,460,464,468,472,476,480,484, 488,492,496,500,504,508,512,516,520,524,528,532,536,540,544,548,552, 556,560,564,568,572,576,580,584,588,592,596,600,604,608,612,616,620, 624,628,632,636,640,644,648,652,656,660,664,668,672,676,680,684,688, 692,696,700,704,708,712,716,720,724,728,732,736,740,744,748,752,756, 760,764,768,772,776,780,784,788,792,796,800,804,808,812,816,820,824, 828,832,836,840,844,848,852,856,860,864,868,872,876,880,884,888,892, 896,900,904,908,912,916,920,924,928,932,936,940,944,948,952,956,960, 964,968,972,976,980,984,988,992,996
Multiplos de 11 de 3 Cifras:
110; 121; 132; 143; 154; 165; 176; 187; 198; 209; 220; 231; 242; 253; 264; 275; 286; 297; 308; 319; 330; 341; 352; 363; 374; 385; 396; 407; 418; 429; 440; 451; 462; 473; 484; 495; 506; 517; 528; 539; 550; 561; 572; 583; 594; 605; 616; 627; 638; 649; 660; 671; 682; 693; 704; 715; 726; 737; 748; 759; 770; 781; 792; 803; 814; 825; 836; 847; 858; 869; 880; 891; 902; 913; 924; 935; 946; 957; 968; 979; 990
Respuesta:
2508
Explicación paso a paso:
ba(2a) = 11°
+ - +
b – a + 2a = 11°
b + a = 11°
↓ ↓
10 1 x
9 2 /
8 3 /
7 4 /
La suma es:
924 + 836 + 748 = 2508