What is the smallest number which when divided by 2 3 4 5 and 6 leaves a remainder of 1 each time but when divided by 11 leaves 0 as the remainder?
What is the smallest number which when divided by 2 3 4 5 and 6 leaves a remainder of 1 each time but when divided by 11 leaves 0 as the remainder?
2*3*5*6=180, so 181 will leave a 1 as a remainder when you divide it by any of the other numbers. But so will 31, 61, 91, 121, 151, for the same reason. The L.C.M. of 2,3,5&6 is 180. Hence add 1 and the required number is 181 which will leave remainder 1 when divided by each of them.
What is the smallest number divided by 2 and 3?
For example, for the numbers 2 and 3 it would be 6, which is the smallest number that’s evenly divisible by both 2 and 3.
What is the smallest multiple of 7 divisible by 9?
63
LCM of 7 and 9 by Listing Multiples Step 3: The smallest common multiple of 7 and 9 is 63.
What is the least multiple of 7 which when divided?
Let the number is (90k+4),which is multiple of 7. Put the value of k=1, 2, 3, 4, 5…. ∴ 364 is the least multiple of 7 which leaves 4 as remainder when divided by each of 6, 9, 15 and 18 .
What is the remainder when the same number is divided by 7?
If the same number is divided by 7 it leaves no remainder. The number is The least number which when divided by 2, 3, 4, 5 and 6 leaves the remainder 1 in each case. If the same number is divided by 7 it leaves no remainder.
What is the smallest number divisible by 2 3 4 5 6?
That is, if we subtract 1 from the number divisible by 2, 3, 4, 5, 6, we get the required number. The smallest number divisible by 2, 3, 4, 5, 6 will be the LCM of 2, 3, 4, 5, 6.
What is the least common multiple of 2 3 4 5 6?
If the same number is divided by 7 it leaves no remainder. The number is The least number which when divided by 2, 3, 4, 5 and 6 leaves the remainder 1 in each case. If the same number is divided by 7 it leaves no remainder. The number is The least common multiple of 2, 3, 4, 5, 6 is their LCM =60.
Which number is divisible by 7 with k = 1?
Put K = 1, number is 119 which is divisible by 7. HEnce answer is 119. NOTE – You might have observed it while solving part 1 only that 119 is divisible by 7. Such an elaborated Part 2 solution is required only if you are NOT able to observe the answer quickly.