# 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 a_{1} = 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

a_{n} = a_{1} + (n-1)d

a_{n} = -2 + (n-1)(315)

a_{n} = -2 + 315n - 315

a_{n} = -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

a_{n} = -317 + 315n < 10000

315n < 10317

n < 32

So the largest value of n we can use is n = 32.

So

a_{32} = -317 + 315(32)

a_{32} = 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 a

_{1}= 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

a

_{n}= a

_{1}+ (n-1)d

a

_{n}= -2 + (n-1)(315)

a

_{n}= -2 + 315n - 315

a

_{n}= -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

a

_{n}= -317 + 315n < 10000

315n < 10317

n < 32

So the largest value of n we can use is n = 32.

So

a

_{32}= -317 + 315(32)

a

_{32}= 9763

Hence the number is 9763

### Answered by | 28th Apr, 2011, 09:18: PM

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