Hi,

1)If 504k is a perfect square, find the least possible integer value of k

Solution:

504 = 3^2 x 7 x 2^3

504k = 3^2 x 7 x 2^3 x k...............[Note: perfect squares must have even powers]

so, smallest 504k = 3^2 x 7^2 x 2^4 = (3^2 x 7 x 2^3) x (7 x 2)

which means, k = 7 x 2 = **14**

2)Find a number which when divided by 3 gives a remainder of 1;when divided by 4 gives a remainder of 2;

when divided by 5 gives a remainder of 3;and when divided by 6 gives a remainder of 4?

Solution:

Let the no. be n, and a, b, c, d be integers such that

n = 3a + 1 = 4b + 2 = 5c + 3 = 6d + 4

Then, 3a + 1 = 4b + 2; i.e., a = (4b + 1)/3

and, 3a + 1 = 5c + 3; i.e., a = (5c + 2)/3

and, 3a + 1 = 6d + 4; i.e., a = (6d + 3)/3 = 2d + 1......[1]

From above, 4b + 1 = 5c + 2 = 6d + 3

when, 4b + 1 = 5c + 2; i.e., b = (5c + 1)/4..............[2]

and, 4b + 1 = 6d + 3; i.e., b = (6d + 2)/4 = (3d + 1)/2

again, from above, 5c + 1 = 6d + 2

leading to, c = (6d + 1)/5...............................[3]

i.e., c is a multiple of 5, which when 1 is subtracted from it will be a multiple of 6.

thus, 6d must be a multiple of 6 ending with 4 in ones place (can't end with 9 because multiples of 6 are even)

So, now with that fact,

if 6d = 24, d = 4; from [3], c = 5; a = 9, from [1]; b = 13/2! (no! because not integer)

next, if 6d = 54, d = 9; a = 19; c = 11; b = 14; giving n = 58 (for all cases of a, b, c, d)

A number satisfying the conditions is **58**.

[if there be more nos., try out 14 x 6, 19 x 6, etc., for 6d...have fun and enjoy!]

3)The whole number 239 has the following property: When divided by 2,3,4,5 and 6 it leaves a remainder of

1,2,3,4 and 5 respectively. How many 3-digit whole numbers ,up to 300 will have this property, excluding 239?

[*pardon me, I am out of time: try analysing this in similar manner as no. 2. thanks.*]