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Topic: To find all remaining three-digit numbers

Example. We take, for example, 250 and it is attributed to it 2. We receive 2250. 2250 in 9 times more 250. It is necessary to find all remaining three-digit numbers, at attributing to which in the beginning any digit, the received result would be in 9 times more initial number. Direct search and generally programming do not roll. The common decision is necessary.

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Re: To find all remaining three-digit numbers

The answer: 125 250 375 500 625 750 875

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Re: To find all remaining three-digit numbers

Hello, VsevolodC, you wrote: the common decision Is necessary

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Re: To find all remaining three-digit numbers

Find all the 3-digit positive integer that, when a fourth digit is placed in there front, became nine times bigger.

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Re: To find all remaining three-digit numbers

Hello, Lepsik, you wrote: L> the common decision it idle time Is necessary, I think for 5-6 classes... We write down a condition for numbers with digits abcd and bcd: abcd=9*bcd or 1000a + 100b + 10c + d = 9 * (100b + 10c + d) whence a = (1/125) * (100b + 10c +d) means (100b + 10c +d) i.e. the initial number, shares on 125

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Re: To find all remaining three-digit numbers

Hello, VsevolodC, you wrote: VC> means (100b + 10c +d) i.e. the initial number, shares on 125 It is possible is even easier. If the decision to name pair from digit an and numbers bcd it is obvious enough that a difference of two various decisions - too the decision. And it, in turn means that all decisions are multiple to the smallest... Well and further it is obvious that (2, 250) it is necessary to divide the given decision on 2 what to receive minimum (1 125), and then to increase by remaining digits 3, 4, 5, 6. While we will be located in the three-digit

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Re: To find all remaining three-digit numbers

Hello, VsevolodC, you wrote: VC> Hello, Lepsik, you wrote: L>> common decision VC> it idle time Is necessary, I think for 5-6 classes... VC> we write down a condition for numbers with digits abcd and bcd: VC> abcd=9*bcd VC> or VC> 1000a + 100b + 10c + d = 9 * (100b + 10c + d) VC> whence VC> a = (1/125) * (100b + 10c +d) VC> means (100b + 10c +d) i.e. the initial number, shares on 125 It is possible algebraically, but hardly easier a - initial number x - digit added ahead Therefore 1000x + a = 9a 1000x = 8a 125x = an I.e. the initial number a shares on 125