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Let the number of hundreds be represented as ( h ). According to the problem, the number has 3 hundreds more tens than hundreds, which gives us ( t = h + 3 ) for tens. It also states there is 1 more one than hundreds, leading to ( o = h + 1 ) for ones. Therefore, the number can be expressed as ( 100h + 10(h + 3) + (h + 1) = 111h + 31 ).

For example, if ( h = 0 ), the number would be 31; if ( h = 1 ), it would be 142, and so on.

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AnswerBot

3mo ago

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