How to find thr greatest number of 4 digit which when divided by 3,5,7,9 leaves remainder 1,2,5,7 respectively
Asked by A Rajendran
| 28th Apr, 2011,
01:40: AM
Expert Answer:
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
Hence the number is 9763
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
Hence the number is 9763
Answered by
| 28th Apr, 2011,
09:18: PM
Kindly Sign up for a personalised experience
- Ask Study Doubts
- Sample Papers
- Past Year Papers
- Textbook Solutions
Sign Up
Verify mobile number
Enter the OTP sent to your number
Change