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13 Sep 2009 11:28:28 IST
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maths olmpiad problem : - find the greatest no. of 4 digits, which when divided by 3,5,7,and 9 lea
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maths olmpiad problem : - find the greatest no. of 4 digits, which when divided by 3,5,7,and 9 leaves remainders 1,3,5 and 7 respectively . PLZ give full solutions and step and not just answer , ratings promised to one who does it correct at first
Comments (15)
13 Sep 2009 18:46:55 IST
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Let's first look at the properties of ALL such
integers, not necessarily the greatest one with
four digits, not necessarily even one with four
digits, and not necessarily even one that is
positive.
Theorem:
If two integers p and q are such that p mod r = q mod r,
then they differ by a multiple of r.
Proof:
p mod r = q mod r
(p mod r) - (q mod r) = (0 mod r)
(p - q) mod r = 0 mod r
Thus they must differ by a multiple of r.
Therefore if p and q both satisfy the given
conditions, then they must differ by some
multiple of 3, which is also a multiple of 5,
which is also a multiple of 7, and which is
also a multiple of 9.
Since a multiple of 9 is also a multiple of 3,
we only need to multiply 5x7x9 = 315 to find the
smallest number two such numbers can differ by.
This suggests an arithmetic sequence of values
satisfying the given conditions.
Now let's find ANY integer meeting the
given requirements.
If p is such an integer then there exist
integers a,b,c,d such that
p = 3a + 1 = 5b + 3 = 7c + 5 = 9d + 7
Let's investigate to find out if there exists
a simple solution where a = b = c = d. If so then
p = 3a + 1 = 5a + 3 = 7a + 5 = 9a + 7
Now we are in luck here because setting any of
the 4 expressions on the right equal to each
other gives a = -1 and thus p = -2
So the arithmetic sequence with first term
a1 = p = -2 and common difference d = 315 will be
an arithmetic sequence of integers meeting the
given requirements.
The nth term of an arithmetic sequence is given by
an = a1 + (n-1)d
an = -2 + (n-1)(315)
an = -2 + 315n - 315
an = -317 + 315n
Now since the term of this sequence we are seeking
is the largest one with 4 digits, we require that it
be less than 10000.
So
an = -317 + 315n < 10000
315n < 10317
n < 32
So the largest value of n we can use
is n = 32.
So
a32 = -317 + 315(32)
a32 = 9763
That's it! Rate me people...as u promised...
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find the greatest no. of 4 digits, which when divided by 3,5,7,and 9 leaves remainders 1,3,5 and 7 respectively
observe that the remainder obtained in each case is 2 less than the divisor,
So, if the no. is X then X-2 is divisible by 3,5,7,9
So largest 4 digit no divisible by 3,5,7,9 is k*9*7*5 = 9765 (k is appropriately chosen)
So X-2 = 9765
=> X = 9767