Question

N = 7777 ....... 7777, where the digit 7 repeats itself 603 times. What is the remainder left when N is divided by 1144?___

Answer 1

1144 can be written as a product of 3 co-prime numbers - viz. $13\times 8\times 11$. Further, the given number when divided by 11, leaves a remainder of 7, when divided by 13 leaves a remainder of 10(because, if you divide 777777 by 13, there is no remainder. Hence, when you divide the given number by 13, the remainder would only depend on the remainder of the last three 7's i.e. 777 $÷$ 13). Also, the given number when divided by 8, leaves a remainder of 1. Hence, the given number is a number that is simultaneously 13n + 10, 8n + 1 and 11n + 7. If we create a series of 13n + 10, we can see that the series would be: 10, 23, 36, 49, 62 ..... At 62, the number is 13n + 10 as well as an 11n + 7 number. The next such number would be 62 + 143 (because 143 is the LCM of 13 and 11). Thus, writing down the series of numbers that belong to 13n + 10 and 11n + 7 and checking when it also simultaneously becomes 8n + 1, we can see that the series would be:
62, 205, 348, 491, 634, 777. The number 777 is also an 8n + 1 number. Hence, the correct remainder is 777.