k is the greatest number that divides 2996, 4752 and 7825 leaving the remainder in each case equal. Find the product of the marks of K.
3 answers
no idea what are "the marks of K"
I have this rather simple simulation app of a computer language we used in the 70s and 80s called BASIC.
I made up a rather dinky and simple program consisting of 6 lines:
10 for k = 2 to 4752
20 r1 = 2996-int(2996/k)*k
30 r2 = 4752-int(4752/k)*k
40 r3 = 7825-int(7825/k)*k
50 if ((r1 = r2) and (r1 = r3)) then print k,r1
60 next k
439 362
It told me that the greatest number that divides 2996, 4752 and 7825 leaving the remainder in each case equal
is 439 and the remainder in each case is 362
So k = 439 , rem = 362
e.g.
2996/439 = 6 with remainder of 362
4752/439 = 10 with remainder of 362
7825/439 =17 with remainder of 362
Don't know what you mean by "the product of the marks of K
Take it from there
Thanks for the entertaining question. I know this can be done by
modular arithmetic, but I haven't done that in over 60 years.
I made up a rather dinky and simple program consisting of 6 lines:
10 for k = 2 to 4752
20 r1 = 2996-int(2996/k)*k
30 r2 = 4752-int(4752/k)*k
40 r3 = 7825-int(7825/k)*k
50 if ((r1 = r2) and (r1 = r3)) then print k,r1
60 next k
439 362
It told me that the greatest number that divides 2996, 4752 and 7825 leaving the remainder in each case equal
is 439 and the remainder in each case is 362
So k = 439 , rem = 362
e.g.
2996/439 = 6 with remainder of 362
4752/439 = 10 with remainder of 362
7825/439 =17 with remainder of 362
Don't know what you mean by "the product of the marks of K
Take it from there
Thanks for the entertaining question. I know this can be done by
modular arithmetic, but I haven't done that in over 60 years.
Sorry dear
Didn't understand any step
Didn't understand any step
2 answers