Question

find the smallest number which when divided by 35, 56, 105 leaves a remainder 6 in each case

Answer 1

216 is ur answer

Hope this will help u.....

Answer 2

Answer:

The numbers which leave remainder 5 when divided by 35 are: 40, 75, 110,.. [(a multiple of 35)+5]

Similarly, the numbers which leave remainder 5 when divided by 56 are: 61, 117,... [(a multiple of 56)+5]

And finally, the numbers which leave remainder 5 when divided by 105 are: 110, 215,.. [(a multiple of 105)+5]

Here we want one number that does all three! So it has to be a number that is =[(multiple of 35 and 56 and 105)+5] = (A common multiple)+5.

But because we want the least such number, we take least common multiple of 35, 56 and 105 and then add 5.

Let’s now work on finding the lcm of these three numbers.

35=5*7

56=(2*2*2)*7

105=3*5*7.

Now the lcm is the number obtained by taking the highest available power of each prime factor.

So the lcm is = (2^3) * (3) * (5) * (7)=840.

[The Highest power available is 3 for 2(in 56), 1 for 3(in 105), 1 for 5(whether you see in 35 or 105), 1 for 7.]

Answer = 840+5 =845.