在一道有余数的除法中,已知被除数、除数、商与余数的和是391,余数是4,商是11,被除数是多少?
人气:132 ℃ 时间:2020-02-04 20:52:22
解答
设除数为x,则被除数为11x+4,由题意得:
11x+4+x+11+4=391,
12x+19=391,
12x=372,
x=31,
被除数为:11×31+4=345;
答:题目中的被除数是345.
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