k is the greatest number that divides 2996, 4752 and 7825 leaving the remainder in each case equal
I didn't understand as the answer done before
I was the one who answered you in this question.
h ttps://www.jiskha.com/questions/1894237/k-is-the-greatest-number-that-divides-2996-4752-and-7825-leaving-the-remainder-in-each
(delete the space at the front of the URL)
Both oobleck and i stated that we don't understand your actual question.
I don't expect you to follow the steps of how I found that highest
common divisor which leaves the same remainder, but I gave it to you
k = 439, common remainder = 362
Don't know what else you want us to do.
mathhelper's little BASIC program divided the numbers by 1,2,3... until a common remainder was found. You could just as easily iterate by subttracting 1,2,3... from the numbers, and then finding the GCF of the diminished numbers. When the GCF is greater than 1, then you have found your answer.
To find the greatest number that divides 2996, 4752, and 7825 leaving the same remainder in each case, we need to find the common factors of these three numbers.
First, we find the factors of each number:
Factors of 2996: 1, 2, 4, 7, 14, 28, 107, 214, 428, 749, 1498, 2996
Factors of 4752: 1, 2, 3, 4, 6, 8, 12, 16, 24, 32, 47, 48, 94, 141, 188, 282, 376, 564, 752, 1128, 1504, 2376, 4752
Factors of 7825: 1, 5, 25, 313, 1565, 7825
Next, we look for the common factors among these three lists. The only factor that appears in all three lists is 1.
Therefore, the greatest number that divides 2996, 4752, and 7825, leaving the same remainder in each case, is 1.