Find the divisors!
When 451 and 607 are divided by a positive integer , the remainders are same. Find the sum of all positive divisors of the largest possible value of .
The answer is 392.
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1 solution
When two numbers are being divided by a third number and leaves the same remainder.Then the difference of the two number is perfectly divisible by the third number.
So,607-451=156 is perfectly divisible by the number.
. Now,156=2 * 2 * 3 *13.
Thus,1-digit number=1,2,3,4,6
2-digit number=12,13,26,39,52,78
3-digit number=156 So sum=1+2+3+4+6+12+13+26+39+52 78+156=392