[SOLVED] Year 8 maths. quick

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June 27th 2008, 03:33 AM
claudeluvstwilight

[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!!
June 27th 2008, 03:41 AM
topsquark

Quote:

Originally Posted by claudeluvstwilight

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
June 27th 2008, 03:59 AM
Soroban

Hello, claudeluvstwilight!

Quote:

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: . $\boxed{N \;=\;1911}$
June 27th 2008, 06:48 AM
nikhil

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)