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 ?
Question 1
11 x 7 + 3 = 740
11 x 5 +3 = 608
11 x 8 + 3 = 861
Ans is 11
Question 2
315
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11 x 7 + 3 = 80 not 740.
So, answer cannot be 11
Originally posted by Seowlah:So, answer cannot be 11
I think he meant
11*67 + 3 = 740
11*55 + 3 = 608
11*78 + 3 = 861
Yes. You are right. I am sorry for the typos. Was rushing.
But this method of getting the answer of 11 is like using guess and check method.
This question is under the topic of HCF and LCM. Need to use these concepts to find the answer of 11.
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 861-740 = 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/What-is-the-greatest-number-that-will-divide-43-91-and-183-so-as-to-leave-the-same-remainder-in-each-case Or spend a few dollars and ask for help from experts to be 100% sure of the correctness of the answer.