在一道有余数的除法算式中,被除数,除数,商和余数的和是599.已知商是15,余数是12,求除数是多少?

2025-03-16 19:01:59
推荐回答(5个)
回答1:

除数是35

599-15-12=572得出被除数+除数=572
由于商的个位是5,余数个位是2,所以等出被除数的个位是5的倍数+2,即个位只能是2或7,即除数的个位就是0或5个,所以可以推断除数有可能是15、20、25、30、35、40....
而符合条件的除数就只有35了,即被除数是572-35=537

537/35=15.....12

用方程也可以,回答1的方程式二错了,应该是15Y+12=X,由于5年级没有学过2元1次方程,所以建议不要用方程。我们只要会推断出答案就行了

解:设除数是X,则根据 被除数=除数×商+余数,被除数是(15X+12),根据题意,有方程
15X+12+X+15+12=599
16X+39=599
16X=599-39
16X=560
X=560÷16
X=35
答:除数是35

回答2:

除数+被除数=599-15-12=572

除数=(572-12)÷(1+15)=35

回答3:

537/35=15……12
过程:(599-15-12-12)/(15+1)=35
35*15+12=537

回答4:

除数+被除数=599-15-12=572
除数=(572-12)÷(1+15)=35

回答5:

537 除数是35

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