

Question 1
What is the greatest number that will divide 740, 608 and 861, leaving the same remainder in each case

Question 2
Form a three digit number with the following conditions
The digits in the number must be different
All the digits in the number must be prime numbers.
The number formed must be the product of four prime numbers.
What is the three digit number ?
Edited by Seowlah 16 Feb `17, 5:13PM



Actually the solution is not so difficult to arrive at. Since the remainer is the same for all the numbers, the differences between the numbers should be perfectly divisible by the HCF. In particular 861740 = 121, which is 11^2, readily identifying 11 as the HCF. Applying this to the 3 numbers confirms the solution..
alternatively you can subtract 3 from each number and write down all the factors for each, thereby identifying the HCF. For 608 it would have been very easy because 608  3 = 605, telling you 5 is one of the factors. Dividing 605 by 5 = 121, which is 11^2, thereby identifying 11 as the HCF.
Hope this helps.



You are absolutely right. I found several solutions this questions, for example here https://www.quora.com/Whatisthegreatestnumberthatwilldivide4391and183soastoleavethesameremainderineachcase Or spend a few dollars and ask for help from experts to be 100% sure of the correctness of the answer.
Edited by Lindamerrillme 11 Apr `17, 7:02PM
