The smallest number that can be subtracted from 1294, yielding a result that when divided by 9, 11, and 13 leaves a remainder of 6 is 1. To prove this, run the following snippit of a program
int i;
for (i=1294; i!= 0; i--)
if (i % 9 6) break;
printf("%u", 1294-i);
another way to prove it it to multiply 9 11 and 13, yielding 1287, adding six, yielding 1293, which is one away from 1294.
810: quotient 1, remainder 1
1005
How about 14 because 14/9 = 1 with a remainder of 5
It is 38.
27836/56 gives 497 and 4 as quotient and remainder Dividend- remainder =27836-4 = 27832 which is divisible by 56. So the least no that to be subracted is 4
127 is the least prime number greater than 25 that will have a remainder of 2 when divided by 25.
810: quotient 1, remainder 1
1005
The number to be subtracted is 11.The number to be subtracted is 11.The number to be subtracted is 11.The number to be subtracted is 11.
what least number must be subtracted from 13081 to get a number exactly divisible by 87
How about 14 because 14/9 = 1 with a remainder of 5
It is 38.
27836/56 gives 497 and 4 as quotient and remainder Dividend- remainder =27836-4 = 27832 which is divisible by 56. So the least no that to be subracted is 4
The smallest number which can be divided by both 4 and 5 without a remainder is 20. This is also known as the Least Common Multiple (LCM).
25
It is 1.1 = 0*12152128 r 1
121