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Let the number be represented by 'a'

a+(a+1)+(a+2)+(a+3)+(a+4)=465

a+a+1+a+2+a+3+a+4=465

a+a+a+a+a+1+2+3+4=465

5a+10=465

5a=465-10

5a=455

5a/5=455/5

a=91

So if a=91

a+1=92

a+2=93

a+3=94

a+4=95

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โˆ™ 2012-10-31 19:41:14
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A polynomial of degree zero is a constant term

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Q: How do you find five consecutive number whose sum is 465?
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