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Math Help - [SOLVED] Year 8 maths. quick

  1. #1
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    Exclamation [SOLVED] Year 8 maths. quick

    Hi i know this may seem easy but it is confusing. please help!!

    Both the leftmost digit and the rightmost digit of a four-digit number N are equal to 1. When these digits are removed, the two-digit numbers thus obtained is N divided by 21. Find N

    even if u view this page please reply and either tell me to give up or u cant b bothered. or u can b bothered and will HELP ME!!
    Last edited by claudeluvstwilight; June 27th 2008 at 03:41 AM. Reason: it is nessassary
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by claudeluvstwilight View Post
    Hi i know this may seem easy but it is confusing. please help!!


    Both the leftmost digit and the rightmost digit of a four-digit number N are equal to 1. When these digits are removed, the two-digit numbers thus obtained is N divided by 21. Find N
    N times 21 is a four digit number such that the first and last digits are a 1. Looking only at the last digit, we see that the 3rd digit must also be a 1 else the four digit number would not end in 1.

    So we need to find a two digit number, ending in 1, such that 21 times it has the required properties.

    So we try 11, 21,... I get that 91 times 21 is 1911. As there are only nine two digit numbers ending in 1 we can easily work out which is the one.

    -Dan
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    Hello, claudeluvstwilight!

    Both the leftmost digit and the rightmost digit of a four-digit number N are equal to 1.
    When these digits are removed, the two-digit numbers thus obtained is N \div 21.
    Find N.

    The number is of the form: . 1ab1\;\text{ where }a\text{ and }b\text{ are digits.}
    . . Its value is: . N \:=\:1000 + 100a + 10b + 1 \:=\:100a + 10b + 1001

    When the 1's are removed, we have: . ab\:\text{ whose value is: }10a + b


    We are told that this is N \div 21

    We have: . 10a + b \:=\:\frac{100a + 10b + 1001}{21} \quad\Rightarrow\quad 210a + 21b \:=\:100a + 10b + 1001<br />

    . . which simplifies to: . 10a + b \:=\:91

    \text{Since }a\text{ and }b\text{ are digits, the only solution is: }\:a = 9,\:b = 1


    Therefore: . \boxed{N \;=\;1911}

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  4. #4
    Senior Member nikhil's Avatar
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    N=1ab1
    ab=N/21
    therfor N/21 must be a 2 digit number wher N is 4 digit number whici is also a multiple of 21.
    First 4 digit number divisible by 21 is 1008(21*48).
    Since N is in the form 1ab1 multiplication of 21 should be done by number which may yield 1 as rightmost digit. So options are 51,61,71,81,91.by multiplying them 4 digit number will be obtained.NOW take the middle two digits of each number and multiply it by 21. If the same number is obtained,that number will be the answer(21ab=N).only 1 number satisfy it and that is 1911(91*21=1911)
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