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HMMT 二月 2000 · ADV 赛 · 第 5 题

HMMT February 2000 — ADV Round — Problem 5

专题
Contest Math / 竞赛数学
难度
L3
来源
HMMT

题目详情

英文原题

  1. Find all 3-digit n um b ers which are the sums of the ub es of their digits.
解析

求所有等于各位数字立方和的三位数。
四个 3 位 Armstrong 数:13+53+33=1531^{3}+5^{3}+3^{3}=15333+73+03=3703^{3}+7^{3}+0^{3}=37033+73+13=3713^{3}+7^{3}+1^{3}=37143+03+73=4074^{3}+0^{3}+7^{3}=407
153,370,371,407\boxed{153, 370, 371, 407}


英文解析

Find all 3-digit numbers equal to the sum of the cubes of their digits.
Let n=100a+10b+cn = 100 a + 10 b + c where a{1,,9}a \in \{1,\ldots,9\}, b,c{0,,9}b,c \in \{0,\ldots,9\}. Need 100a+10b+c=a3+b3+c3100 a + 10 b + c = a^{3} + b^{3} + c^{3}.
Check each digit combination:

  • 153153: 13+53+33=1+125+27=1531^{3} + 5^{3} + 3^{3} = 1 + 125 + 27 = 153
  • 370370: 27+343+0=37027 + 343 + 0 = 370
  • 371371: 27+343+1=37127 + 343 + 1 = 371
  • 407407: 64+0+343=40764 + 0 + 343 = 407
    These are the four 3-digit Armstrong (narcissistic) numbers.
    153,370,371,407\boxed{153, 370, 371, 407}.