387.
A number 2 more than a multiple of 5, 7 and 11 is required.
The smallest such number is 2 more than the lowest common multiple of 5, 7 and 11.
The LCM of 5, 7 & 11 is 385
⇒ 385+2 = 387
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
How about 14 because 14/9 = 1 with a remainder of 5
It is 38.
The numbers divided by 5 gives 1 remainder means it should end either 1 or 6. But in case of number ends with 6 will be divided by 2. So only number ending with 1 have to consider. Also the number should be divisble by 7. So when number ending with 3 is multiplied by 7 will give result of number ending 1. So, possibilities are 7*3, 7*13, 7*23, etc In that least number which solves this problem 7*43 = 301. So the least possible number is 301.
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
301
419 Nachiket Joshi solution: LCM(7,6,5,4,3,2)=420 now subtract 1 frm answer ...............so the answer will be 419..... TRICK: Since the diff between the remainder and dividend is common(1 in this case ) u subtract 1 from the LCM