Questions :
Let x be the least number of 4 digits that when divided by 2, 3, 4, 5, 6 and 7 leaves a remainder 1 in each case. If x lies between 2000 and 2500, then what is the sum of the digits of x?
Options :
- 9
- 12
- 14
- 15
correct answer : b)
x + y = 25
xy = 144
(x – y)² = (x + y)² – 4xy
= (25)² – 4 × 144
= 625 – 576 = 49
x – y = 7